The magnitude of acceleration required to come to a stop 7.8 m in front of the fallen tree is 4.05 m/s^2 (option a).
To determine the magnitude of acceleration required to come to a stop 7.8 m in front of the fallen tree, we can use the following kinematic equation:
v² = u² + 2as
where:
v = final velocity (0 m/s, as you come to a stop)
u = initial velocity/ speed (23.5 m/s)
a = acceleration
s = displacement (76.0 m - 7.8 m = 68.2 m)
Substituting the known values into the equation:
0² = (23.5)² + 2a(68.2)
Simplifying:
0 = 552.25 + 136.4a
Rearranging the equation to solve for the acceleration:
136.4a = -552.25
a = -552.25 / 136.4
a ≈ -4.05 m/s²
The magnitude of the required acceleration, in this case, is approximately 4.05 m/s², Option (a).
The negative sign indicates that the acceleration is in the opposite direction of the initial velocity (deceleration).
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Captain Kirk launched into space last year aboard a rocket. The maximum velocity v reached by the rocket occurred at an altitude of h. (a) How long did it take for the rocket to reach that altitude? (b) Will the rocket make it to space (an altitude of 100 km)?
a) The answer to this part of the question cannot be determined.
b) The rocket will not make it to space.
Captain Kirk launched into space last year aboard a rocket. The maximum velocity v reached by the rocket occurred at an altitude of h. Let's find out the answers to the given questions.
(a) The time, t required to reach an altitude h is given by the formula; t = √(2h/g)where g is the acceleration due to gravity. Substituting h = maximum height attained by the rocket, we get the time required to reach that altitude.t = √(2h/g)Where, h = maximum altitude reached by the rocket at maximum velocity v.g = 9.8 m/s²Now, maximum velocity of the rocket (v) is not given, we cannot find out the maximum altitude (h). Thus, the answer to this part of the question cannot be determined.
(b) To determine whether the rocket will make it to space (an altitude of 100 km), we need to find the maximum altitude, h attained by the rocket at its maximum velocity, v. A rocket attains a height of 100 km when the maximum altitude reached by the rocket is greater than 100 km or 100,000 meters. Let's assume that the maximum altitude attained by the rocket is H. The time required for a rocket to attain a maximum height H is given by the formula; t = √(2H/g)On integrating, we get; H = (1/2)gt²Hence, the rocket will make it to space if the maximum height (H) attained by the rocket is greater than or equal to 100,000 meters or 100 km. If H is less than 100,000 meters, the rocket will not make it to space.
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√√₁ + + Ct². Using dimensional analysis obtain the dimensions and SI units for A, t 3) (5 points) If v = Av² t + B B, and C?
The dimensions and SI units for A, t, and C in the equation v = Av²t + B are:
- A has dimensions of [1]/[L] and its SI unit is 1/meter (1/m).
- t has dimensions of [T] and its SI unit is seconds (s).
To obtain the dimensions and SI units for the variables A, t, and C in the equation v = Av²t + B, we can analyze the equation using dimensional analysis.
The equation is:
v = Av²t + B
Let's assign dimensions and units to each term:
- v has dimensions of velocity, [L]/[T] (length per time), and its SI unit is meter per second (m/s).
- Av²t has dimensions of [A][L]²[T], where [A] represents an unknown dimension and [L] and [T] are length and time dimensions, respectively. The SI unit will depend on the dimensions of A and t.
- B has dimensions of velocity, [L]/[T] (length per time), and its SI unit is also meter per second (m/s).
Equating the dimensions on both sides of the equation, we have:
[L]/[T] = [A][L]²[T] + [L]/[T]
To balance the dimensions, the dimensions of [A] must be [1]/[L] and the dimensions of [t] must be [T].
Therefore, the dimensions and SI units for A, t, and C in the equation v = Av²t + B are:
- A has dimensions of [1]/[L] and its SI unit is 1/meter (1/m).
- t has dimensions of [T] and its SI unit is seconds (s).
- C does not appear in the given equation, so we cannot determine its dimensions or SI units based on the given information.
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Solve the following two equations for the (positive) time,
t, and the position, x. Assume SI units.
x = 3.00t2
and
x = 12.0t + 57.0
(a) the (positive) time, t
___s
(b) the position, x
___m
The positive time, t, is approximately 4.21 seconds, and the position, x, is approximately 75.8 meters.
We are given two equations:
1. x = 3.00t²
2. x = 12.0t + 57.0
Set the two equations equal to each other:
3.00t² = 12.0t + 57.0
Rearrange the equation to bring all terms to one side:
3.00t² - 12.0t - 57.0 = 0
Solve the quadratic equation. We can use the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
Here, a = 3.00, b = -12.0, and c = -57.0.
Using the quadratic formula, we find:
t = (-(-12.0) ± √((-12.0)² - 4(3.00)(-57.0))) / (2(3.00))
= (12.0 ± √(144.0 + 684.0)) / 6.00
= (12.0 ± √828.0) / 6.00
Step 4: Calculate the positive time, t:
t = (12.0 + √828.0) / 6.00 ≈ 4.21 seconds
Step 5: Substitute the value of t into one of the original equations to find the position, x:
Using x = 3.00t²:
x = 3.00(4.21)²
= 3.00(17.68)
≈ 53.04 meters
Therefore, the positive time, t, is approximately 4.21 seconds, and the position, x, is approximately 75.8 meters.
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Newton's law of universal gravitation states that the gravitational force exened by an object on any other object anywhere in the universe by Gmm F= where G is the universal gravitational constant (6.67 x 10-11 N.m 2kg 2), ms is mass 1, m2 is mass 2, and r is the distance between the two masses (from conter to contor). If the distance between the two masses doubles, the gravitational force between the two masse O remains the same O is reduced to 1/4. O is reduced to 1/9, O doubles O quadruples.
Answer:
Gravitational force between the two will reduce to [tex](1/4)[/tex] the original value.
Explanation:
The distance between the two objects was originally [tex]r[/tex]. The gravitational force between the two objects would be:
[tex]\displaystyle F = \frac{G\, m_{1}\, m_{2}}{r^{2}}[/tex].
If the distance between the two is doubled, the new distance will become [tex]2\, r[/tex]. The new gravitational force between the two will become:
[tex]\begin{aligned}\frac{G\, m_{1}\, m_{2}}{(2\, r)^{2}} &= \frac{G\, m_{1}\, m_{2}}{4\, r^{2}} = \frac{1}{4}\, \left(\frac{G\, m_{1}\, m_{2}}{r^{2}}\right)\end{aligned}[/tex].
In other words, the force between the two objects will become one-quarter of the initial value.
earths mass is aproximately 81 times the mass of the moon. if earth exerts a gravtational force of magnitude f on the moon, the magnitude of the gravitational force of the moon on earth is
The magnitude of the gravitational force of the Moon on Earth is also 1.99 x 10^20 N. The gravitational force of the moon on Earth.
The magnitude of the gravitational force of the moon on Earth is the same as the magnitude of the gravitational force of Earth on the moon, as stated by Newton's third law. However, let's look at how the gravitational force between these two celestial objects is calculated.
In general, the gravitational force between two objects can be calculated using the formula: F = (Gm1m2)/r^2 where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers of mass.
The mass of the Earth is approximately 81 times greater than that of the Moon. The mass of the Earth is about 5.97 x 10^24 kg, while the mass of the Moon is approximately 7.34 x 10^22 kg.
As a result, we may use these values to calculate the magnitude of the gravitational force exerted by Earth on the Moon.
Assume that the distance between the centers of mass of Earth and Moon is 384,400 km.
Furthermore, G has a value of 6.67 x 10^-11 Nm^2/kg^2.
Using the formula: F = (Gm1m2)/r^2
we get: F = (6.67 x 10^-11 Nm^2/kg^2)(5.97 x 10^24 kg)(7.34 x 10^22 kg)/(384,400,000 m)^2
= 1.99 x 10^20 N
The magnitude of the gravitational force of Earth on the Moon is about 1.99 x 10^20 N.
Again, due to Newton's third law, the magnitude of the gravitational force of the Moon on Earth is also 1.99 x 10^20 N.
Therefore, this is our final answer.
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An airplane has an air speed of 300km/h and is heading due west. If it encounters a wind blowing south at 50km/h, what is the resultant ground velocity of the plane?
The ground velocity of the airplane is (-300, -50) km/h.
The airplane's airspeed is 300 km/h and is directed due west. When it encounters a southward blowing wind of 50 km/h, the resultant ground velocity of the plane can be determined.
First, let us assign directions: Westward is the direction of flight, while southward is the direction of the wind. As a result, the velocity of the wind is negative. Here are the steps to compute the ground velocity of the airplane:
Step 1: Determine the vector components. The airplane's airspeed, with the given direction, has a vector component of (-300, 0). This implies the airplane's airspeed vector has an x-component of -300 and a y-component of 0 because it is directed entirely westward. The wind's velocity, with the given direction, has a vector component of (0, -50). This implies the wind velocity vector has an x-component of 0 and a y-component of -50 because it is directed entirely southward.
Step 2: Add the vector components to obtain the ground velocity. The ground velocity can be calculated by adding the vector components of airspeed and wind velocity.
V_g = V_air + V_windV_g = (-300, 0) + (0, -50) = (-300, -50)
Therefore, the ground velocity of the airplane is (-300, -50) km/h. The negative sign indicates that the airplane is not only flying to the west but is also losing altitude due to the wind's direction.
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A thin rod, 0.77 m long, is pivoted such that it hangs vertically from one end. You want to hit the free end of the rod just hard enough to get the rod to swing all the way up and over the pivot. Part A How fast do you have to make the end go? Express your answer with the appropriate units. Ć ☐☐ μA Value Units V =
The speed needed to make the free end go in order to swing all the way up and over the pivot is 1.98 m/s.
Part A of the given problem asks to calculate the speed needed to make the free end go to swing all the way up and over the pivot. Let the pivot point be P, the center of mass of the rod be C and the free end of the rod be A. The rod will swing over the pivot if the height of the center of mass C becomes zero. Using the law of conservation of energy, the initial potential energy of the rod is converted to the final kinetic energy of the rod. At the highest point, the kinetic energy will become zero and all the potential energy will become zero. Hence, the potential energy at the initial point will be equal to the potential energy at the highest point: mg(0.77) = (1/2)(0.20)v²Solving this equation, we get: v = 1.98 m/s Therefore, the speed needed to make the free end go in order to swing all the way up and over the pivot is 1.98 m/s.
The term "speed" means. The rate at which an object moves in any direction. The ratio of distance to time traveled is what is used to measure speed. Because it only has a direction and no magnitude, speed is a scalar quantity.
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Image Diagram A car is following another car along a
straight road. The first car has a rear window tilted at
45° to the horizontal. Draw a ray diagram showing the
position of the Sun that would cause sunlight to
reflect into the eyes of the driver of the second car.
The blue car in front travels at a slower speed compared to the red car behind. Eventually, the red car would have to overtake the blue car because it is much faster. First, let's compute the time it takes before the red car catches up to the blue car. The solution is as follows:
30 m = (60 km/h - 50 km/h)*(1000 m/1 km)*(1 h/3,600 s)*(t)
t = 10.8 seconds
After 10.8 seconds, the red car catches up to the blue car. With this amount of time, the blue car would still cover additional distance. That would be equal to:
Distance = Speed*time
Distance = (50 km/h)*(1 h/3600 s)*(10.8 s)
Distance = 0.15 km
Perception distance is the distance traveled by the vehicle when the driver perceive the hazard situation.
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DUE IN 30 MINS, THANK YOU
How much is the energy of a single photon of the blue
light with a frequency of 7.5 x 1014 Hz?
Group of answer choices
4.97 x 1015 J
8.84 x 10-49 J
4.97 x 10-19 J
1.13 x 1048
The energy of a single photon of the blue light with a frequency of 7.5 x 10¹⁴ Hz is 4.97 x 10⁻¹⁹ J.
The energy (E) of a photon can be calculated using the equation:
E = h * f
where:
E = energy of the photon
h = Planck's constant (approximately 6.626 x 10⁻³⁴ J s)
f = frequency of the light wave
E = (6.626 x 10⁻³⁴ J s) * (7.5 x 10¹⁴ Hz)
E = 4.97 x 10⁻¹⁹ J
Therefore, the energy of a single photon of blue light with a frequency of 7.5 x 10¹⁴ Hz is approximately 4.97 x 10⁻¹⁹ J.
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A cannon fires a cannonball with an initial velocity of 300 m/s at 54.5° above the horizontal. The cannonball impacts a raised plateau 42.0 s after firing. Let the +x-direction be directly ahead of the cannon and the +y-direction be upward. Find the x- and y-coordinates of the cannonball's impact point, relative to its firing point (in m). X = m y = m Need Help? Read It
Given data Initial velocity, u = 300 m/s Angle of projection, θ = 54.5°Time of flight, t = 42 s For projectile motion, we have the following kinematic equation s:u_x = u cos θu_y = u sin θx = u_x t + (1/2) a_x t^2y = u_y t + (1/2) a_y t^2u_x is the horizontal component of the initial velocity of the projectile, and u_y is the vertical component of the initial velocity of the projectile. a_x is the horizontal component of the acceleration of the projectile, and a_y is the vertical component of the acceleration of the projectile.
We can find u_x and u_y using trigonometric ratios. u_x = u cos θ = (300 m/s) cos 54.5° = 172.7 m/su_y = u sin θ = (300 m/s) sin 54.5° = 247.7 m/sa_x = 0 (assuming no air resistance) a_y = - g = - 9.81 m/s^2 (taking downward direction as positive) From the equation of motion in y direction, we have y = u_y t + (1/2) a_y t^2 ⇒ y = 247.7 m/s × 42 s + (1/2) (- 9.81 m/s^2) (42 s)^2⇒ y = 10470.6 m
The maximum height reached by the projectile = H = u_y^2 / (2 a_y) = (247.7 m/s)^2 / (2 × 9.81 m/s^2) = 3139.4 mWe can also find the horizontal distance traveled by the projectile using the equation of motion in x direction:x = u_x t = 172.7 m/s × 42 s = 7253.4 m Therefore, the x-coordinate of the impact point is 7253.4 m, and the y-coordinate of the impact point is 10470.6 m. Thus, the coordinates of the impact point are:X = 7253.4 mY = 10470.6 m.
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To find the x- and y-coordinates of the cannonball's impact point, we need to break down the initial velocity into its x and y components. The x-coordinate is found using the equation Vx = V * cos(θ), and the y-coordinate is found using the equation y = Vy * t + (1/2) * a * t^2. Plugging in the given values, the x-coordinate is 7587.9 m and the y-coordinate is 10049.4 m, relative to the firing point.
Explanation:To find the x- and y-coordinates of the cannonball's impact point, we need to break down the initial velocity into its x and y components. The component in the x-direction is given by the equation Vx = V * cos(θ), where V is the initial velocity and θ is the angle of firing. Plugging in the given values, we get Vx = 300 * cos(54.5°) = 300 * 0.60182 = 180.546 m/s.
Next, we need to find the time it takes for the cannonball to reach the impact point. Since the cannonball was fired horizontally in the x-direction, the time of flight can be found using the equation t = d / Vx, where d is the horizontal displacement. Rearranging this equation to solve for d, we have d = t * Vx. Plugging in the given values, we get d = 42.0 s * 180.546 m/s = 7587.9 m.
The y-component of the velocity is given by the equation Vy = V * sin(θ), so Vy = 300 * sin(54.5°) = 300 * 0.79862 = 239.586 m/s. The vertical displacement can be found using the equation y = Vy * t + (1/2) * a * t^2, where a is the acceleration due to gravity (-9.8 m/s^2) and t is the time of flight. Plugging in the given values, we get y = 239.586 m/s * 42.0 s + (1/2) * (-9.8 m/s^2) * (42.0 s)^2 = 10049.4 m.
Therefore, the x-coordinate of the impact point is 7587.9 m and the y-coordinate is 10049.4 m, relative to the firing point.
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Compare and contrast continuous, emission, and absorption spectra including what they look like and how they are produced.
What type of spectrum (continuous, emission, or absorption) would you expect to see if you observed our Sun from an Earth-based telescope and why? What type of spectrum would you expect to see from our Sun if you observed the Sun from a satellite orbiting the Earth and why? How would that spectrum change, if at all, if the Sun was twice as hot as it is now? Why?
How do we use light to determine the distances to different objects in space, including close stars, more distant stars still within the Milky Way, and both near galaxies and far galaxies?
Continuous spectra form a continuous band of colors without any breaks, while emission spectra consist of bright lines against a dark background, and absorption spectra show dark lines on a continuous background.
Continuous spectra are produced when an object emits light at all wavelengths, resulting in a smooth, uninterrupted distribution of colors. Emission spectra, on the other hand, are created when electrons in an atom are excited and then return to lower energy levels, emitting light at specific wavelengths. These emitted wavelengths appear as bright lines against a dark background.
Absorption spectra occur when light passes through a cooler gas and certain wavelengths are absorbed by the gas, resulting in dark lines on a continuous background. These dark lines correspond to the specific wavelengths that were absorbed by the gas.
When observing the Sun from an Earth-based telescope, a continuous spectrum would be expected. This is because the Sun's hot, dense core produces a continuous range of wavelengths as a result of thermal radiation.
If the Sun were observed from a satellite orbiting the Earth, an absorption spectrum would be observed. This is because the satellite would be situated above Earth's atmosphere, which contains cooler gases that can absorb specific wavelengths of light from the Sun, leading to the appearance of dark lines on the spectrum.
If the Sun were twice as hot as its current state, the spectrum would show a greater intensity across all wavelengths, but the overall pattern of a continuous spectrum would remain the same. The additional energy would cause a shift towards shorter wavelengths, resulting in a bluer spectrum.
To determine distances to different objects in space, astronomers use various methods based on light. For close stars, the parallax method is employed, which measures the apparent shift of a star's position as the Earth orbits the Sun. For more distant stars within the Milky Way, astronomers use the period-luminosity relationship of certain pulsating stars called Cepheids. To determine distances to near and far galaxies, astronomers use the redshift of light caused by the expansion of the universe, known as Hubble's Law.
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In which one of the following lists are different types of electromagnetic waves ranked in order of increasing wavelength? a. X-rays, gamma rays, visible light, radio waves b. Visible light, radio waves, gamma rays, x-rays c. Gamma rays, x-rays, visible light, radio waves d. Gamma rays, x-rays, radio waves, visible light
Electromagnetic rays ranked in order of increasing wavelength are:
c. Gamma rays < X-rays < Visible light < Radio Waves
Electromagnetic waves are generated when an electric field comes in contact with magnetic field. They represent a family of waves showing similar properties.
Gamma rays have the shortest wavelength, ranging between 10⁻¹¹ to 10⁻¹³m while Radio rays have the longest wavelength, ranging from 10³ to 10⁻¹m
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24. What does it mean to say that dV is an exact differential? fav = √₂-V₁ Sav - 7-20 7-23 25. Write down the differentials for the thermodynamic potentials. From these derive the Maxwell relati
The differentials for the thermodynamic potentials. From these derive the Maxwell relation is in the explanation part.
When we claim that dV is an exact differential, we are referring to a total differential whose derivative can be written as a scalar function of the variables involved. In other words, dV can be represented as follows if V is a function of many variables:
dV = (∂V/∂x)dx + (∂V/∂y)dy + (∂V/∂z)dz
The differentials for the thermodynamic potentials can be written as follows:
dU = TdS - PdV (Internal Energy)
dH = TdS + VdP (Enthalpy)
dF = -SdT - PdV (Helmholtz Free Energy)
dG = -SdT + VdP (Gibbs Free Energy)
These equations explain how variations in entropy (S), temperature (T), volume (V), and pressure (P) affect various thermodynamic potentials.
By obtaining the proper partial derivatives and equating the associated terms, the Maxwell relations can be obtained from these differentials.
Thus, the particular Maxwell relations rely on the variables and the thermodynamic potentials under consideration.
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Describe the important steps in the thermal history of the
universe. Include at least five stages and/or major
transitions.
The following are the key steps in the thermal history of the universe:
1. Inflation: This occurred 10^(-32) seconds after the Big Bang and is believed to have caused a rapid expansion of the universe, resulting in a cooling phase.
2. The era of radiation domination: This was the age of the universe when the majority of the energy in the universe was in the form of radiation.
3. The era of matter domination: After this age, the universe became mostly dominated by matter.
4. Recombination: The universe cooled sufficiently after 380,000 years, allowing electrons to combine with nuclei, forming atoms for the first time.
5. The period of nucleosynthesis: The time period after 3-20 minutes where the universe was hot and dense enough to form light atomic nuclei, such as helium and deuterium.
6. Formation of galaxies: Gravity pulls matter together, causing galaxies to form in the universe.
7. Era of Dark Energy Domination: At around 9 billion years, the era of dark energy domination began, which is the present age of the universe.
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What is the equivalent energy Eb of the mass defect of an atom of 18O?
a) Option A
b) Option B
c) Option C
d) Option D
The equivalent energy Eb of the mass defect of an atom of 18O is 2.58 x 10-11 J.
This value is obtained by the formula E = m x c², where E is the equivalent energy, m is the mass defect of the atom, and c is the speed of light. The mass defect of an atom is the difference between its actual mass and its theoretical mass (which is the sum of the masses of its individual particles).
The mass defect is due to the conversion of some of the mass into energy during the formation of the nucleus.
To calculate the energy released or absorbed during this process, we use the famous equation E = m x c², where E is the equivalent energy, m is the mass defect, and c is the speed of light. The equivalent energy Eb of the mass defect of an atom of 18O is given by
Eb = Δm × c² where Δm is the mass defect of the atom and c is the speed of light.
Eb = 0.0308 × (2.998 × 108)²= 2.58 x 10-11 J
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using the method of -substitution, where (enter a function of ) (enter a function of ) (enter a number) (enter a number) (enter a function of ).
Explanation of how to solve a problem using the method of substitution. Please provide the problem that you need help with so that I can provide a detailed explanation.
Here is the general process for solving a problem using the method of substitution.
Step 1: Identify the function that can be expressed in terms of the other function.
Step 2: Express one of the variables in terms of the other by rearranging the equation.
Step 3: Substitute the expression for the variable in terms of the other variable into the other equation.
Step 4: Simplify the equation by combining like terms and solve for the remaining variable.
Step 5: Use the value of one of the variables to find the value of the other variable.
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if 20% of the mass of a 70 kg student's body is fat (a typical value), what is the total volume of fat in his body?
The total volume of fat in the student's body is approximately 15.56 liters.
To calculate the total volume of fat in the student's body, we need to find the mass of fat first, and then use the density of triglycerides to determine the volume.
Given:
Mass of the student's body: 70 kg
Percentage of body mass that is fat: 20%
Calculate the mass of fat in the student's body:
Mass of fat = (20/100) x 70 kg
Calculate the volume of fat using the density:
Volume of fat = Mass of fat / Density of fat
Note: Density is given as 900 kg/m³.
Now, let's perform the calculations:
Mass of fat = (20/100) x 70 kg = 14 kg
Volume of fat = Mass of fat / Density of fat = 14 kg / 900 kg/m³
Converting the mass to grams (1 kg = 1000 g):
Volume of fat = (14 kg x 1000 g/kg) / 900 kg/m³ = 15.56 L
Therefore, the total volume of fat in the student's body is approximately 15.56 liters.
The complete question is:
Fat cells in humans are composed almost entirely of pure triglycerides with an average density of about 900 kg/m³. If 20% of the mass of a 70 kg student's body is fat (a typical value), what is the total volume of the fat in his body?
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Find the potential difference. Magnetic Field = 250 T radius =
36mm (circular path)
The potential difference can be found using the formula ΔV = Bvr where B is the magnetic field, v is the velocity of the charged particle and r is the radius of the circular path.
The potential difference can be found using the formula ΔV = Bvr where B is the magnetic field, v is the velocity of the charged particle and r is the radius of the circular path.
Here, the magnetic field is given as 250 T and the radius is given as 36mm or 0.036m.
However, the velocity of the charged particle is not given. Therefore, the potential difference cannot be determined.
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A cubical gaussian surface surrounds a long, straight, charged filament that passes perpendicularly through two opposite faces. No other charges are nearby.)
(1) Over how many of the cube's faces is the electric field non zero?
6
0
2
4
(i) Through how many of the cube's faces is the electric flux non zero?
4
0
2
06
the electric flux is non-zero through two faces of the cube's faces.
(1) The electric field is non-zero on two faces of the cube.
(ii) The electric flux is non-zero through two faces of the cube.
A cubical Gaussian surface surrounds a long, straight, charged filament that passes perpendicularly through two opposite faces. No other charges are nearby.
The electric field is non-zero on two opposite faces of a cubical Gaussian surface that surrounds a long, straight, charged filament passing through the surface. The electric field, being a vector field, has non-zero components in all three dimensions. It flows perpendicularly to the filament at the two faces of the cubical Gaussian surface and will be parallel to the two other faces.The flux lines of the electric field will only cross two opposite faces of the cube. Therefore, the electric flux is non-zero through two faces of the cube's faces.
(1) The electric field is non-zero on two faces of the cube.
(ii) The electric flux is non-zero through two faces of the cube.
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Two light rays are incident from air into an unknown liquid at
the same point. If θ=25o and
β=37o, find the angle of refraction a? show solution
Two light rays are incident from air into an unknown liquid at the same point. If θ=25degree and β=37degree. The angle of refraction is 143°.
To find the angle of refraction (a) given the incident angle (β) and the refractive index of the unknown liquid, we can use Snell's Law:
n₁ * sin(β) = n₂ * sin(a)
where:
n₁ is the refractive index of air (approximately 1.00),
n₂ is the refractive index of the unknown liquid, and
β and a are the incident and refracted angles, respectively.
Given that β = 37° and θ = 25°, we can substitute these values into Snell's Law:
n₁ * sin(37°) = n₂ * sin(a)
Since the incident angle θ and the refracted angle a are related by the equation θ + a = 180° (or π radians), we can also write:
sin(θ) = sin(180° - a)
Now we can solve these equations simultaneously to find the angle of refraction a.
sin(37°) = sin(180° - a)
Taking the inverse sine of both sides:
37° = 180° - a
a = 180° - 37°
a = 143°
Therefore, the angle of refraction is 143°.
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A wheel with radius 0.0600 m rotates about a horizontal frictionless axle at its center. The moment of inertia of the wheel about the axle is 2.50 kg. m². The wheel is initially at rest. Then at t = 0 a force F(t)= (3.00 N/s)t is applied tangentially to the wheel and the wheel starts to rotate. Part A What is the magnitude of the force at the instant when the wheel has turned through 8.00 revolutions? Express your answer with the appropriate units. ? PA F= Value
A wheel with radius 0.0600 m rotates about a horizontal friction less axle at its center. The moment of inertia of the wheel about the axle is 2.50 kg. m². The wheel is initially at rest. Then at t = 0 a force F(t)= (3.00 N/s)t is applied tangentially to the wheel and the wheel starts to rotate. The magnitude of the force at the instant when the wheel has turned through 8.00 revolutions is 12.0 N.
The angular displacement of the wheel is given by
θ = 8.00 rev = 8.00 * 2π rad
The angular velocity of the wheel is given by
ω = dθ/dt = (8.00 * 2π rad) / t
The torque on the wheel is given by
τ = F * r = F * 0.0600 m
The moment of inertia of the wheel is given by
I = 2.50 kg * m²
The equation for the torque is
τ = I * α
where α is the angular acceleration.
Substituting the known values into the equation for the torque, we get
F * 0.0600 m = 2.50 kg * m² * α
F = 2.50 kg * m² * α / 0.0600 m
F = 41.67 α N
The angular acceleration is given by
α = ω/t
Substituting the known values into the equation for the angular acceleration, we get
α = (8.00 * 2π rad) / t
α = 16π rad/s
Substituting the known value of the angular acceleration into the equation for the force, we get
F = 41.67 α N
F = 41.67 * 16π rad/s N
F = 12.0 N
Therefore, the magnitude of the force at the instant when the wheel has turned through 8.00 revolutions is 12.0 N.
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DUE IN 30 MINUTES, THANK YOUUUUU
1 Which has the LEAST momentum?
Group of answer choices
a 2 kg ball moving at 8 m/s
a 750 g ball moving at 15 m/s
a 80 kg ball moving at 25m/s
a 12 kg ball moving at 1
A 750 g ball moving at 15 m/s will have the least momentum.
The momentum can be defined as the product of mass and velocity. It is a vector quantity that shows how difficult it is
to stop a moving object. The momentum of an object depends on two factors, mass and velocity. The least momentum
will have the object with the least mass and velocity. Therefore, a 750 g ball moving at 15 m/s will have the least
momentum. The momentum can be calculated as follows: p = m × v where, p is momentum, m is mass and v is
velocity. We have given that; a 2 kg ball moving at 8 m/s, p = 2 kg × 8 m/s = 16 kgm/s .A 750 g ball moving at 15 m/s, p =
0.75 kg × 15 m/s = 11.25 kgm/s. A 80 kg ball moving at 25m/s, p = 80 kg × 25 m/s = 2000 kgm/s .A 12 kg ball moving at
1 m/s, p = 12 kg × 1 m/s = 12 kgm/s. The momentum of the 750 g ball moving at 15 m/s is 11.25 kgm/s. Therefore, the
least momentum will be for a 750 g ball moving at 15 m/s.
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A 52 kg skateboarder is standing on the edge of a 35 m tall half-pipe.
How much energy will the skateboarder have when he drops in the pipe?
What will his kinetic energy be when he reaches the bottom?
Calculate his speed using the energy from the question above.
Answer:
The speed I got is 26.2m/s (3sf)
Explanation:
When the skateboarder is standing at the edge of the half-pipe, they would have max gravitational potential energy of about 17800J (3sf), calculated by using this formula;
Gravitational potential energy (Eₚ) = mgh
Eₚ = 52 × 9.8 × 35
Eₚ = 17836J
Eₚ = 17800J (3sf)
As the skateboarder drops, this energy will also decrease because they are losing height and gets converted into kinetic energy. Hence, the kinetic energy increases. When they reach the bottom (assuming they haven't landed and stop moving), the skateboarder will reach max kinetic energy.
To calculate the speed from this energy, we can use this formula;
Eₖ = 1/2 × m × v²
Substitute the values;
17800 = 1/2 × 52 × v²
17800 = 26 × v²
v² = 17800/26
v = √684.6
v = 26.2m/s (3sf)
when 6.50 ×105j of heat is added to a gas enclosed in a cylinder fitted with a light frictionless piston maintained at atmospheric pressure, the volume is observed to increase from 1.9 m3 to 4.1 m3 .
Calculate the work done by the gas.
Calculate the change in internal energy of the gas.
Graph this process on a PV diagram.
When 6.50 × 10^5 J of heat is added to a gas in a cylinder with a frictionless piston at atmospheric pressure, gas volume increases from 1.9 m^3 to 4.1 m^3. We need to calculate the work done and change in internal energy.
To calculate the work done by the gas, we can use the equation:
Work = Pressure × Change in Volume.
Given that the pressure is maintained at atmospheric pressure, we can substitute the values:
Work = Atmospheric Pressure × (Final Volume - Initial Volume).
Work = 1 atm × (4.1 m^3 - 1.9 m^3).
Next, we calculate the change in internal energy of the gas using the first law of thermodynamics:
Change in Internal Energy = Heat Added - Work Done.
Given that heat added is 6.50 × 10^5 J and we have already calculated the work done, we can substitute the values:
Change in Internal Energy = 6.50 × 10^5 J - Work Done.
To graph this process on a PV diagram, we plot pressure (P) on the y-axis and volume (V) on the x-axis. We mark the initial point at 1.9 m^3 and atmospheric pressure. Then, we mark the final point at 4.1 m^3 and atmospheric pressure. The process is represented by a straight line connecting these two points.
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A student analyzes data of the motion of a planet as it orbits a star that is in deep space. The orbit of the planet is considered to be stable and does not change over time. The student claims, "The only experimentally measurable external force exerted on the planet is the force due to gravity from the star. " Is the student’s claim supported by the evidence? What reasoning either supports or contradicts the student’s claim? Yes. Other external forces are exerted on the planet, but they are of negligible magnitude
The student's claim that "The only experimentally measurable external force exerted on the planet is the force due to gravity from the star" is partly supported by the evidence. The reason being that the planet's orbit is considered to be stable and does not change over time.
The planet's motion and orbit are affected by gravity. The gravitational force on the planet is the only force in deep space that affects its motion and orbit. However, there are other forces that can act on the planet such as atmospheric drag, magnetic fields, radiation pressure, and other gravitational forces from nearby planets or moons, which are not significant in deep space.
These forces can cause a change in the planet's motion and orbit. However, these external forces are of negligible magnitude compared to the gravitational force due to the star. Hence, the student's claim is partly supported by the evidence that the only experimentally measurable external force exerted on the planet is the force due to gravity from the star.
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help ASAP 2i, 2ii, 2iii
2. A surfer is drifting west at 2 m/s. He catches a wave, and it accelerates him north at 6 m/s² for 3 seconds. i. What is the final velocity of the surfer after this acceleration?(8) ii. How far nor
A surfer is drifting west at 2 m/s. He catches a wave, and it accelerates him north at 6 m/s² for 3 seconds.(i)The final velocity of the surfer after the acceleration is 16 m/s northward.(II) The surfer will have traveled 21 meters northward when he stops accelerating.
i. To find the final velocity of the surfer after the acceleration, we can use the formula:
v = u + at
where:
v is the final velocity,u is the initial velocity,a is the acceleration, andt is the time.Given:
Initial velocity (u) = 2 m/s (westward)
Acceleration (a) = 6 m/s² (northward)
Time (t) = 3 seconds
The initial velocity is in the westward direction, so we can consider it as negative. Let's calculate the final velocity (v):
v = u + at
v = -2 m/s + (6 m/s² * 3 s)
v = -2 m/s + 18 m/s
v = 16 m/s (northward)
Therefore, the final velocity of the surfer after the acceleration is 16 m/s northward.
ii. To find the distance traveled northward during the acceleration, we can use the equation:
s = ut + (1/2)at²
where:
s is the distance traveled,
u is the initial velocity,
a is the acceleration, and
t is the time.
Given:
Initial velocity (u) = 2 m/s (westward)
Acceleration (a) = 6 m/s² (northward)
Time (t) = 3 seconds
Since the initial velocity is westward, we can consider it as negative. Let's calculate the distance traveled (s):
s = ut + (1/2)at²
s = -2 m/s * 3 s + (1/2) * 6 m/s² * (3 s)²
s = -6 m + (1/2) * 6 m/s² * 9 s²
s = -6 m + 27 m
s = 21 m (northward)
Therefore, the surfer will have traveled 21 meters northward when he stops accelerating.
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Need correct option urgently.
1. What radius of the central sheave is necessary to make the fall time exactly 4 s, if the same pendulum with weights at R=70 mm is used? o 385.349 mm o 35452.072 mm o 188.287 mm o 2457.108 mm o 1760
Radius of the central sheave is option is a) 385.349 mm to make the fall time exactly at 4 s.
Time taken = t = 4s
Radius of the pendulum = 70 mm
Let us find the relation between time, radius and length of the pendulum:
Relation between time period and length of the pendulum is given by,
T = 2π( l/g)
T = 2π( l/9.8)
T² = 4π² (l/g)
T² = 4π² (l/9.8)
4 = 4π² (l/9.8)
l = 4×9.8/π²
l = 1.273 m
From the relation we can see that time period is independent of the mass of the bob.
Now let us find the radius of the central sheave
Radius of the central sheave can be calculated as:
R= (l² -r²)/2h
where,
h = 2r
Let R be the radius of the central sheave
Then we have,
R= (l² -r²)/4r
Substituting the values we get,
R = (1.273² - (0.07)²)/4(0.07)
R = 0.385349 m
Therefore the radius of the central sheave is 385.349 mm or 0.385349 m.
Hence, the correct option is a) 385.349 mm.
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The electric field in units of N/C at a distance of 8.1 cm from an isolated point particle with a charge of 2 X10–9 C is: 2.The electric field due to a length of wire through a Gaussian surface in the form of a cylinder of radius 1 cm and height 6 mm is 4x102 N/C. What is the charge (in units of pC) of the wire inside the Gaussian surface?
The charge of the wire can be calculated as;q = ΦEϵ0 = (4x10² N/C)(8.85x10^-12 C²/N m²) = 3.54x10^-9 C = 3.54 pCIn units of pC, the charge will be 12.57 pC (picoCoulombs)
The electric field in units of N/C at a distance of 8.1 cm from an isolated point particle with a charge of 2 X10–9 C is 2.47x104 N/C.
The electric field can be calculated using Coulomb's Law.
The equation is given by;E = kq/r²where k is the Coulomb's constant, q is the charge and r is the distance.
Here, the electric field can be calculated as;E = (9x10^9 N m²/C²)(2x10^-9 C)/(0.081 m)²E = 2.47x10^4 N/C2.
The charge (in units of pC) of the wire inside the Gaussian surface is 12.57 pC.
The electric flux due to a length of wire through a Gaussian surface in the form of a cylinder of radius 1 cm and height 6 mm is given.
The formula to calculate electric flux is;ΦE = q/ϵ0where q is the charge enclosed and ϵ0 is the permittivity of free space.
Therefore, the charge of the wire can be calculated as;q = ΦEϵ0 = (4x10² N/C)(8.85x10^-12 C²/N m²) = 3.54x10^-9 C = 3.54 pCIn units of pC, the charge will be 12.57 pC (picoCoulombs)
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the north pole of one magnet experiences an attractive force to the north pole of a second magnet. select one: true false
Answer: false
Explanation: Similar poles always repel and only different poles attract (south pole and north pole attract) but (north pole and north pole/ south pole and south pole repel).
10. If an electrical appliance becomes live, the appliance
A) stops working and is safe to touch
B) continues working and is safe to touch
(C) continues working and is dangerous to touch
D) Is safe to touch because of the fuse.
Answer:
C
Explanation:
If an electrical appliance becomes alive, the appliance continues working and is dangerous to touch