______ ______ are created when something is caused to vibrate.
Answer: the answer is MIRANDA COOPER
Explanation:
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What is the effect on the period of a pendulum if you decrease its length by 6.35%? (Answer this question in terms of the initial period T.) T' = 0.87703 Incorrect: Your answer is incorrect. T
Answer:
T' = 0.9677T
Explanation:
The period of a pendulum is given by the following formula:
[tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]
l: length of the pendulum
g: gravitational acceleration
If the length of the pendulum is decreased in 6.35% the length of the pendulum becomes:
[tex]l'=l-0.0635l=0.9365l[/tex]
The new period for a length of l' is:
[tex]T'=2\pi \sqrt{\frac{l'}{g}}=2\pi \sqrt{\frac{0.9365l}{g}}=\sqrt{0.9365}(2\pi \sqrt{\frac{l}{g}})=0.9677(2\pi \sqrt{\frac{l}{g}})\\\\T'=0.9677T[/tex]
hence, the new period is 0.9677T
A coat rack weighs 65.0 lbs when it is filled with winter coats and 40.0 lbs when it is empty. The base of the coat rack has an area of 452.4 in2. How much more pressure, in psi (pounds per square inch), is exerted by the coat rack on the floor when it is filled with winter coats than when it is empty
Answer:
0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.
Explanation:
First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:
P₁ = F/A
where,
P₁ = Pressure exerted by empty rack = ?
F = Force exerted by empty rack = Weight of Empty Rack = 40 lb
A = Base Area = 452.4 in²
Therefore,
P₁ = 40 lb/452.4 in²
P₁ = 0.088 psi
Now, we calculate the pressure exerted by the rack along with the coat.
P₂ = F/A
where,
P₂ = Pressure exerted by rack filled with coats= ?
F = Force exerted by filled rack = Weight of Filled Rack = 65 lb
A = Base Area = 452.4 in²
Therefore,
P₂ = 65 lb/452.4 in²
P₂ = 0.144 psi
Now, the difference between both pressures is:
ΔP = P₂ - P₁
ΔP = 0.144 psi - 0.088 psi
ΔP = 0.056 psi
A vertical spring-mass system undergoes damped oscillations due to air resistance. The spring constant is 2.65 ✕ 104 N/m and the mass at the end of the spring is 11.7 kg. (a) If the damping coefficient is b = 4.50 N · s/m, what is the frequency of the oscillator? Hz
Answer:
f = 7.57 Hz
Explanation:
To find the frequency of the damping oscillator, you first use the following formula for the angular frequency:
[tex]\omega=\sqrt{\omega_o-(\frac{b}{2m})^2}=\sqrt{\frac{k}{m}-(\frac{b}{2m})^2}\\\\[/tex] (1)
k: spring constant = 2.65*10^4 N/m
m: mass = 11.7 kg
b: damping coefficient = 4.50 Ns/m
You replace the values of k, m and b in the equation (1):
[tex]\omega=\sqrt{\frac{2.65*10^4N/m}{11.7kg}-(\frac{4.50Ns/m}{2(11.7kg)})^2}\\\\\omega=47.59\frac{rad}{s}[/tex]
Finally, you calculate the frequency:
[tex]f=\frac{\omega}{2\pi}=\frac{47.59}{2\pi}Hz=7.57\ Hz[/tex]
hence, the frequency of the oscillator is 7.57 Hz
An ideal gas in a balloon is kept in thermal equilibrium with its constant-temperature surroundings. How much work is done if the outside pressure is slowly reduced, allowing the balloon to expand to 50 times its original size
Answer:
w = 252.32 N
Explanation:
given data
balloon expand = 50 times its original size
we consider here initially pressure and volume
pressure = 645 pa
volume = 0.10 m³
solution
as in isothermala process ideal gas
PV = mRT
P = [tex]\frac{mRT}{v}[/tex]
P = [tex]\frac{c}{v}[/tex]
here c is constant
so work done is express as
[tex]w = c \int\limits^{V2}_{V1} {\frac{dv}{v}}[/tex]
w = [tex]c \times ln( \frac{v2}{v1})[/tex]
and we know c = p1 × v1
so
w = p1 × v1 × [tex]ln (\frac{50v1}{v1} )[/tex]
w = 645 × 0.1 × ln(50)
w = 252.32 N
A small block is released from rest at the top of a frictionless incline. The block travels a distance 0.633 m in the first second after it is released. How far does it travel in the next second
Answer:1.89 m
Explanation:
Given
Block travels [tex]0.63\ m[/tex] in first second
It is released from rest i.e. initial speed is zero (u=0)
using
[tex]s=ut+\frac{1}{2}at^2[/tex]
where a=acceleration
here acceleration is the component of gravity on incline plane (say [tex]\theta [/tex])
so
[tex]s_1=\frac{1}{2}\times g\sin \theta (1)^2[/tex]
[tex]0.633\times 2=9.8\sin \theta \times 1^2[/tex]
[tex]\sin\theta =0.1291[/tex]
[tex]\theta =7.41^{\circ}[/tex]
So distance traveled in [tex]2\ sec[/tex]
[tex]s=\frac{1}{2}\times g\sin \theta (2)^2[/tex]
[tex]s=0.5\times 9.8\times \sin (7.41)\times 4[/tex]
[tex]s=2.52\ m[/tex]
So distance traveled in [tex]2^{nd}\ sec[/tex] is
[tex]s-s_1=2.52-0.633=1.89\ m[/tex]
A peak with a retention time of 407 s has a width at half-height (w1/2) of 7.6 s. A neighboring peak is eluted 17 s later with a w1/2 of 9.4 s. A compound that is known not to be retained was eluted in 2.5 s. The peaks are not baseline resolved. How many theoretical plates would be needed to achieve a resolution of 1.5?
Answer:
2.46 x 104
Explanation:
Solution
Recall that:
The retention time of a peak = 407 s
with a width at half-height of = 7.6 s
A compound is retained in 2.5 s.
resolution to be achieved = 1.5
Thus,
The number of plates (theoretical)= 16(tr2 / w2)
The R Resolution R= 0.589 Δtr / w1/2av = 0.589(17s) / 1/2(7.6s + 9.4s) = 1.18
Supposed that applied column contains 10,000 theoretical plates and the resolution of two peaks is 1.18
So if the column is replaced to obtain 1.5 resolution, the number of theoretical plates is needed is stated below;
width at the base = 9.4 - 7.6 = 1.8; tr = 0.786
N = 5.55tr2 / w21/2 = 5.55 (0.7862/ 1.182) x 104
= 2.46 x 104
Therefore, required theoretical plates to achieve a resolution of 1.5 is 2.46 x 104
Do wave properties affect wave speed
Answer:
Nope!
Explanation:
The amplitude of a wave does not affect the speed at which the wave travels. Both Wave A and Wave B travel at the same speed. The speed of a wave is only altered by alterations in the properties of the medium through which it travels.
HOPE IT HELPS :)
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In a particular lab, a cube of ice (Tice = -5.5˚C) is taken and dropped into a calorimeter cup (98g) partially filled with 326 g of water (Water = 20˚C). The cup was at the same initial temperature as the water and is perfectly insulating. The final temperature of the system is 15˚C. What was the mass of ice added?
Answer:
The mass of the ice added = 16.71 g
Explanation:
The heat gained by the ice is equal to the heat lost by the calorimeter cup and the water in the cup.
But for this question, the cup is said to be perfectly insulated, hence, there is no loss of heat from the calorimeter cup.
Heat gained by the ice = Heat lost by the 326 g of water.
Let the mass of ice be m
The heat gained by the ice = (Heat gained by ice in temperature from -5.5°C to 0°C) + (Heat used by the ice to melt at 0°C) + (Heat required for the melted ice to rise in temperature from 0°C to 15°C)
Heat gained by ice in temperature from -5.5°C to 0°C = mCΔT
m = unknown mass of ice
C = Specific Heat capacity of ice = 2.108 J/g°C
ΔT = change in temperature = 0 - (-5.5) = 5.5°C
Heat gained by ice in temperature from -5.5°C to 0°C = m×2.108×5.5 = (11.594m) J
Heat used by the ice to melt at 0°C = mL
m = unknown mass of ice
L = Latent Heat of fusion of ice to water = 334 J/g
Heat used by the ice to melt at 0°C = m×334 = (334m) J
Heat required for the melted ice or water now, to rise in temperature from 0°C to 15°C = mCΔT
m = unknown mass of water (which was ice)
C = Specific Heat capacity of water = 4.186 J/g°C
ΔT = change in temperature = 15 - 0 = 15°C
Heat required for the melted ice or water now, to rise in temperature from 0°C to 15°C = m×4.186×15 = (62.79m) J
Total heat gained by the ice = 11.594m + 334m + 62.79m = (408.384m) J
Heat lost by the water in the calorimeter cup = MCΔT
M = mass of water in the calorimeter cup = 326 g
C = specific heat capacity of water = 4.186 J/g°C
ΔT = change in temperature = 20 - 15 = 5°C
Heat lost by the water in the calorimeter cup = 326×4.186×5 = 6,823.18 J
Heat gained by the ice = Heat lost by the 326 g of water.
408.384m = 6,823.18
m = (6,823.18/408.384)
m = 16.71 g
Hope this Helps!!!
How much work is done (by a battery, generator, or some other source of potential difference) in moving Avogadro's number of electrons from an initial point where the electric potential is 6.70 V to a point where the electric potential is -8.90 V? (The potential in each case is measured relative to a common reference point.)
Answer:
W = 1.5 x 10⁶ J = 1.5 MJ
Explanation:
First, we calculate the potential difference between the given 2 points. So, we have:
V₁ = Electric Potential at Initial Position = 6.7 V
V₂ = Electric Potential at Final Position = - 8.9 V
Therefore,
ΔV = Potential Difference = V₂ - V₁ = -8.9 V - 6.7 V = - 15.6 V
Since, we use magnitude in calculation only. Therefore,
ΔV = 15.6 V
Now, we calculate total charge:
Total Charge = q = (No. of Electrons)(Charge on 1 Electron)
where,
No. of Electrons = Avagadro's No. = 6.022 x 10²³
Charge on 1 electron = 1.6 x 10⁻¹⁹ C
Therefore,
q = (6.022 x 10²³)(1.6 x 10⁻¹⁹ C)
q = 96352 C
Now, from the definition of potential difference, we know that it is equal to the worked done on a unit charge moving it between the two points of different potentials:
ΔV = W/q
W = (ΔV )(q)
where,
W = work done = ?
W = (15.6 V)(96352 C)
W = 1.5 x 10⁶ J = 1.5 MJ
A carousel has a radius of 1.70 m and a moment of inertia of 110 kg · m2. A girl of mass 44.0 kg is standing at the edge of the carousel, which is rotating with an angular speed of 3.40 rad/s. Now the girl walks toward the center of the carousel and stops at a certain distance from the center d. The angular speed of the carousel is now 5.4 rad/s. How far from the center, in meters, did the girl stop?
Answer:
Explanation:
Initial moment of inertia of the carousel + girl
I₁ = 110 + 44 x 1.7²
= 110 + 127.16
= 237.16 kgm².
final moment of inertia of carousel + girl
I₂ = 110 + 44 x d²
applying law of conservation of angular momentum
I₁ ω₁ = I₂ω₂
ω₁ and ω₂ are angular velocities of the carousel before and after .
237.16 x 3.4 = (110 + 44 x d²)x 5.4
806.34 = 594 + 237.6 d²
237.6 d² = 212.34
d²= .8936
d = .9453 m
The manufacturer of a 9 [V] flashlight battery says that the battery will deliver 20 [mA] for 80 continuous hours. However, during that time the voltage will drop from 9 [V] to 6 [V]. Assume the voltage drop is linear in time, but that the current is constant. How much energy does the batteiy deliver in 80 [h]? Battery "capacity" can be stated in terms of the total charge the battery will deliver, with units of milliamp-hours [mA-h]. If the battery can be considered "dead" when the voltage reaches 6 V, what is the capacity of the flashlight battery in mA-h?
Answer:
a. Energy ≅ 43.2 kJ
b. Capacity of battery = 1600-mAh
Explanation:
average voltage value = (9 + 6)/2 = 15/2 = 7.5
current = 20 mA = 20 x [tex]10^{-3}[/tex]
time duration = 80 hrs = 80 x 60 x 60 = 288000 sec
If the current is assumed to be constant, then the energy delivered is the product of voltage across, current delivered and the time duration during which it is delivered.
Energy delivered by battery = 7.5 x 20 x [tex]10^{-3}[/tex] x 288000 = 43200 J
Energy ≅ 43.2 kJ
If the battery is considered dead when it reaches 6-v, then that means that at 6-v, there is no potential difference between them.
capacity of flashlight battery is the product of current delivered and the time duration of delivery.
capacity = 20-mA x 80-hr = 1600-mAh
Measure Your Reaction Time Here's something you can try at home-an experiment to measure your reaction time. Have a friend hold a ruler by one end, letting the other end hang down vertically. At the lower end, hold your thumb and index finger on either side of the ruler, ready to grip it. Have your friend release the ruler without warning. Catch it as quickly as you can.If you catch the ruler 5.7 cm from the lower end, what is your reaction time?
Express your answer using two significant figures.
Answer:
Explanation:
I catch the ruler 5.7 cm from lower end that means my reaction time is equal to time of fall of ruler as free fall under gravity .
h = 1/2 gt²
t = [tex]\sqrt{\frac{2h}{g} }[/tex]
= [tex]\sqrt{\frac{2\times 5.7}{9.8} }[/tex]
= 1.078 s
= 1.1 s .
A plane is flying to a city 756 km directly north of its initial location. The plane maintains a speed of 203 km/h relative to the air during its flight. (a) If the plane flies through a constant headwind blowing south at 53.5 km/h, how much time (in h) will it take to reach the city
Answer:
The answer is 5.05 hours.
Explanation:
If the plane has an airspeed of 203 km/h which only applies for air and not the ground speed, we can subtract the speed of the wind since it is a headwind in the directly opposite direction.
So the speed of the plane becomes 203 - 53.5 = 149.5 km/h which will give us the true airspeed of the plane and the ground speed as well.
From here we can calculate the time it will take to reach the city as
756 km / 149.5 km/h = 5.05 hours.
I hope this answer helps.
4–72 A person puts a few apples into the freezer at 215°C to cool them quickly for guests who are about to arrive. Initially, the apples are at a uniform temperature of 20°C, and the heat transfer coefficient on the surfaces is 8 W/m2·K. Treating the apples as 9-cm-diameter spheres and taking their properties to be r 5 840 kg/m3, cp 5 3.81 kJ/kg·K, k 5 0.418 W/m·K, and a 5 1.3 3 1027 m2/s, determine the center and surface temperatures of the apples in 1 h. Also, determine the amount of heat transfer from each apple. Solve this problem using analytical one-term approximation method (not the Heisler charts).
Complete and Clear Question:
A person puts a few apples into the freezer at -15°C to cool them quickly for guests who are about to arrive. Initially, the apples are at a uniform temperature of 20°C, and the heat transfer coefficient on the surfaces is 8 W/m2·K. Treating the apples as 9-cm-diameter spheres and taking their properties to be [tex]\rho =[/tex] 840 kg/m3, [tex]c_{p} =[/tex] 3.81 kJ/kg·K, k = 0.418 W/m·K, and [tex]\alpha = 1.3 * 10^{-7} m^{2} /s[/tex], determine the center and surface temperatures of the apples in 1 h. Also, determine the amount of heat transfer from each apple. Solve this problem using analytical one-term approximation method (not the Heisler charts).
Answer:
Temperature at the center of the apple, T(t) = 11.215°C
Temperature at the surface of the apple, T(r,t) = 2.68°C
Amount of heat transfer from each apple, Q = 21.47 kJ
Explanation:
For clarity and easiness of expression, the calculations are handwritten and attached as a file. Check the attached files for the complete calculation.
A NFL linebacker runs the 100m sprint in 12s. What is his final velocity?
Answer:
Final velocity of NFL line backer is 16.67 m/s.
Explanation:
From the question, we have following data about the NFL line backer:
Initial Speed of line backer = Vi = 0 m/s (Since, he starts from rest)
Distance covered by NFL line backer = s = 100 m
Time taken by the NFL line backer to complete 100 m sprint = t = 12 s
Acceleration of NFL line backer during sprint = a
Final Velocity of NFL line backer = Vf = ?
First we need to find the acceleration of the NFL line backer. For that purpose we will use 2nd equation of motion:
s = (Vi)(t) + (0.5)at²
using values:
100 m = (0 m/s)(12 s) + (0.5)(a)(12 s)²
100 m/72 s² = a
a = 1.39 m/s²
Now, we use 1st equation of motion to find Vf:
Vf = Vi + at
Vf = 0 m/s + (1.39 m/s²)(12 s)
Vf = 16.67 m/s
A radar antenna is rotating and makes one revolution every 24 s, as measured on earth. However, instruments on a spaceship moving with respect to the earth at a speed v measure that the antenna makes one revolution every 44 s. What is the ratio v/c of the speed v to the speed c of light in a vacuum
Answer:
0.838
Explanation:
The ratio v/c of the speed v to the speed c of light in a vacuum is shown below:
Given that
[tex]\triangle t_0 = 24\ seconds[/tex] = time interval for one revolution
[tex]\triangle t = 44\ seconds[/tex] = time interval measured with speed v
based on the given information, the ratio v/c of the speed v to the speed c of light in a vacuum is
[tex]\triangle t = \frac{\triangle t_0}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]
[tex]{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{\triangle t_0}{\triangle t}[/tex]
Now squaring both the sides
[tex]\frac{v^2}{c^2} = 1 - \frac{(\triangle t_0)^2}{(\triangle t)^2}[/tex]
Now remove the squaring root from both the sides and putting the values
[tex]\frac{v}{c} = {\sqrt{1 - \frac{(\triangle t_0)^2}{(\triangle t)^2}[/tex]
[tex]= {\sqrt{1 - \frac{(24)^2}{(44)^2}[/tex]
= 0.838
a 350g mass as attached to a spring of constant 5.2N/m and set into oscillation with amplitude of 10 cm. what is the frequency, period, maximum velocity and the maximum force in the spring?
Explanation:
It is given that,
Mass of the object, m = 350 g = 0.35 kg
Spring constant of the spring, k = 5.2 N/m
Amplitude of the oscillation, A = 10 cm = 0.1 m
Frequency of a spring mass system is given by :
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{5.2}{0.35}} \\\\f=0.613\ Hz[/tex]
Time period:
[tex]T=\dfrac{1}{f}\\\\T=\dfrac{1}{0.613}\\\\T=1.63\ s[/tex]
Maximum velocity in the spring is given by :
[tex]v=A\omega[/tex]
[tex]v=A\sqrt{\dfrac{k}{m}} \\\\v=0.1\times \sqrt{\dfrac{5.2}{0.35}} \\\\v=0.38\ m/s[/tex]
The maximum force acting in the spring is :
[tex]F=-kx\\\\F=kA\\\\F=5.2\times 0.1\\\\F=0.52\ N[/tex]
Hence, this is the required solution.
What effect does the velocity of a rotating object have on the centripetal acceleration?
Answer:
as centripatal force acts upon an object moving at a circle in constant speed de force acts always inwards as de velocity of object is directed tangent to de circle de force can accelerate de object by changing its direction bt not actually de speed
The two quantities are closely related, but the cause/effect is the other way around.
-- The centripetal force is caused by something outside this discussion, not by the object.
-- The centripetal force acting on the object determines the object's centripetal acceleration.
-- The centripetal acceleration is the cause of whatever the object's velocity (speed and direction) turns out to be.
-- It's the centripetal acceleration that has the effect on the object's velocity.
As an example, you wouldn't say that the orbiting of a TV satellite is what causes the Earth's centripetal force that acts on it.
Question 1 [7]
Hydrogen gas is used in a Carnot cycle having an efficiency of 60% with a low temperature of 300K. During the heat rejection the pressure changes from 90 kPa to 120 kPa. Find the high and low temperature heat transfer and the net cycle work per unit mass of hydrogen.
Question 2 [8]
A rigid insulated container has two rooms separated by a membrane. Room A contains 1 kg of air at 200°C and Room B contains 1.5 kg of air at 20°C, both rooms are at 100 kPa. Consider two different cases
A. The Heat transfer between A and B creates a final uniform T
B. The membrane breaks and air comes to a uniform state.
For both cases find the final temperature. Are the two-process reversible and different? Explain.
Imagine an isolated positive point charge Q (many times larger than the charge on a single proton). There is a charged particle A (whose charge is much smaller than charge Q) at a distance from the point charge Q. On which of the following quantities does the magnitude of the electric force on this charged particle A depend:_____.
1. the size and shape of the point charge Q
2. the specific location of the point charge Q (while the distance between Q and A is fixed)
3. the mass of the charged particle A
4. the size and shape of the charged particle A
5. the distance between the point charge Q and the charged particle A
6. the type of charge on the charged particle A
7. the mass of the point charge Q
8. the amount of the charge on the point charge Q
9. the magnitude of charge on the charged particle A
10. the specific location of the charged particle A (while the distance between Q and A is fixed)
11. the relative orientation between Q and A (while the distance between Q and A is fixed)
Answer:
Explanation:
The magnitude of the electric force on this charged particle A depends upon the following
5. the distance between the point charge Q and the charged particle A
8. the amount of the charge on the point charge Q
9. the magnitude of charge on the charged particle A
Suppose you catch and hold a baseball, and then someone invites you to catch and hold a bowling ball with either the same momentum or the same kinetic energy as the baseball. Which would you choose
Answer:
The same momentum will be the best option.
Explanation:
Let's recall that the force will be express in terms of the momentum. We can write the force as the variation of the momentum over time.
[tex]F=\frac{dp}{dt}[/tex]
This is the force needed to stop the base ball or the bowling ball.
if we will choose the same kinetic energy it would imply an increase of momentum, because of the difference of the masses, and therefore an increase of the force. We do not want this.
Now, if we choose the same momentum the kinetic energy will increase, but the force will the same. We want the less force as possible to stop it, and we have the same at least.
Therefore the same momentum would be the best option.
I hope it helps you!
The best choice to catch and hold the bowling ball will be; with the same momentum
We know that formula for impulse is;
Impulse = Force x Time
And we know that change in momentum is equal to impulse. Thus;
Change in momentum = F × t
ΔP = F × t
F = ΔP/t
This formula represents the force required to stop the baseball or the bowling ball.
Now, momentum is proportional to the square root of kinetic energy.
Now, since momentum is directly proportional to velocity, while kinetic energy is proportional to the square of the velocity, it means that if kinetic energy is quadrupled, then the momentum will become double.
Now, the collision in the question is completely inelastic and as such, all the bowling balls kinetic energy will be in inelastic collision, the kinetic energy is lost.
Formula for the kinetic energy in terms of the momentum here is;
K = p²/2m
Looking at it overall, we can say that the best choice to catch and hold the bowling ball will be with the same momentum since it results in lesser force.
Read more at;https://brainly.com/question/13994440
What is most often given a value of zero to describe an object's position on a straight line?
O displacement
O reference point
O distance
O ending location
Answer:
O reference point
Explanation:
A reference point is often given the value of zero to describe an object position on a straight line, or when it didn't move. If the object doesn't move, that means that there is no displacement, and it is a reference point. The answer to the question is reference point.
Um corpo de massa m= 2,0Kg é lançado horizontalmente, de uma altura h= 125m, com velocidade de módulo Vo =10m/s, como mostra a figura. Desprezando a resistência do ar e adotando g= 10m/s2 , determine: a) A energia mecânica total do corpo; b) A energia cinética do corpo a meia altura em relação ao solo; c) O tempo gasto até que o corpo atinja o solo; d) O alcance do movimento.
Answer:
A) E = 2550 J
B) K = 1325 J
C) t = 5,05 s
Explanation:
A) The total mechanical energy is given by the sum of the gravitational potential energy and the kinetic energy of the body:
[tex]E=U+K=mgh+\frac{1}{2}mv^2[/tex] (1)
m: mass of the body = 2,0 kg
g: gravitational acceleration = 9,8 m/s^2
h: height = 125 m
v: initial velocity of the body = 10 m/s
You replace the values of all variables h, m, g and v in the equation (1):
[tex]E=(2,0kg)(9,8m/s^2)(125m)+\frac{1}{2}(2,0kg)(10m/s)^2=2550\ J[/tex]
the total mechanical energy is 2550 J
B) The kinetic energy of the corp, when it is at a height of h/2 is given by:
[tex]K=\frac{1}{2}mv^2[/tex]
where
[tex]v=\sqrt{(v_x)^2+(v_y)^2}[/tex]
The x component of the velocity is constant in the complete trajectory, which is the initial velocity, that is, vo = vx
The y component is given by:
[tex]v_y^2=v_{oy}^2+2gy[/tex]
voy: vertical initial velocity = 0m/s
y: height = h/2 = 125/2 = 62.5 m
[tex]v_y=\sqrt{2g\frac{h}{2}}=\sqrt{2(9.8m/s^2)(62.5m)}=35m/s[/tex]
Then, you can calculate the velocity of the body and next, you can calculate the kinetic energy:
[tex]v=\sqrt{(10m/s)^2+(35m/s)^2}=36,40\frac{m}{s}\\\\K=\frac{1}{2}(2,0kg)(36,40m/s)^2=1325\ J[/tex]
C) The time that body takes in all its trajectory is:
[tex]t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(125m)}{9,8m/s^2}}=5,05s[/tex]
Light bulb 1 operates with a filament temperature of 2800 K, whereas light bulb 2 has a filament temperature of 1700 K. Both filaments have the same emissivity, and both bulbs radiate the same power. Find the ratio A1/A2 of the filament areas of the bulbs. A1/A2
Answer:
A₁/A₂ = 0.136
Explanation:
The power radiated by a filament bulb is given by the following formula:
E = σεAT⁴
where,
E = Emissive Power
σ = Stephen Boltzman Constant
ε = emissivity
A = Area
T = Absolute Temperature
Therefore, for bulb 1:
E₁ = σε₁A₁T₁⁴
And for bulb 2:
E₂ = σε₂A₂T₂⁴
Dividing both the equations:
E₁/E₂ = σε₁A₁T₁⁴/σε₂A₂T₂⁴
According to given condition, the emissive power and the emissivity is same for both the bulbs. Therefore,
E/E = σεA₁T₁⁴/σεA₂T₂⁴
1 = A₁T₁⁴/A₂T₂⁴
A₁/A₂ = (T₂/T₁)⁴
where,
T₁ = 2800 K
T₂ = 1700 K
Therefore,
A₁/A₂ = (1700 K/2800 K)⁴
A₁/A₂ = 0.136
Which of the following are true?
a) the total momentum of an isolated system is constant.
b) the total momentum of any number of particles is equal to the algebraic sum of the momenta of individual particles.
c) the total momentum of any number of particles is equal to the vector sum of the momenta of individual particles.
d) the vector sum of forces acting on a particle equals the rate of change of momentum of the particle with respect to time.
e) the total momentum of any system is constant.
f) the vector sum of forces acting on a particle equals the rate of change of velocity of the particle with respect to time.
The electric potential in a region that is within 2.00 mm of the origin of a rectangular coordinate system is given by V=Axl+Bym+Czn+DV=Axl+Bym+Czn+Dwhere AA, BB, CC, DD, ll, mm, and nn are constants. The units of AA, BB, CC, and DD are such that if xx, yy, and zz are in meters, then VV is in volts. You measure VV and each component of the electric field at four points and obtain these results:Point (x,y,z)(m) V(V) Ex(V/m) Ey(V/m) Ez(V/m) 1 (0, 0, 0) 10.0 0 0 0 2 (1.00, 0, 0) 4.0 16.0 0 0 3 (0, 1.00, 0) 6.0 0 16.0 0 4 (0, 0, 1.00) 8.0 0 0 16.01. Use the data to calculate A.2. Use the data to calculate B3. Use the data to calculate C4. Use the data to calculate D5. Use the data to calculate E6. Use the data to calculate l7. Use the data to calculate m8. Use the data to calculate n
Answer:
Given the potential, [tex] V = Ax^l+By^m+Cz^n+D [/tex]
The components of the electric field are:
[tex]E_x = \frac{-dV}{dx} = -Alx^l^-^1[/tex]
[tex]E_y = \frac{-dV}{dy} = - Bmy^m^-^1[/tex]
[tex]E_z = \frac{-dV}{dz} = - nCzn^n^-^1[/tex]
Let's calculate the potential difference for all given points.
[tex] V(0, 0, 0) = 10V => Ax^l+By^m+Cz^n+D = 10 [/tex]
[tex]=> D = 10[/tex]
[tex] V(1, 0, 0) = 4V => A + 10 = 4 [/tex]
Solving for A, we have:
[tex] A = 4 - 10 [/tex]
[tex] A = -6 [/tex]
[tex] V(0, 1, 0) = 6V => B + 10 = 6 [/tex]
Solving for B, we have:
[tex] B = 6 - 10[/tex]
[tex] B = -4 [/tex]
[tex] V(0, 0, 1) = 8V => C + 10 = 4 [/tex]
Solving for C, we have:
[tex] C = 8 - 10 [/tex]
[tex] C = -2 [/tex]
For all given points, let's calculate the magnitude of electric field as follow:
[tex]E_x (1, 0, 0) = 16 => - Alx^l^-^1 = 16[/tex]
[tex]Al = -16[/tex]
Solving for l, we have:
[tex]l = \frac{-16}{A}[/tex]
From above, A = -6
[tex]l = \frac{-16}{-6}[/tex]
[tex]l = \frac{8}{3}[/tex]
[tex] E_y (0, 1, 0) = 16=> Bmy^m^-^1 = 16 [/tex]
[tex]Bm = -16[/tex]
Solving for m, we have:
[tex]m = \frac{-16}{A}[/tex]
From above, B = -4
[tex]m = \frac{-16}{-4}[/tex]
[tex]m = 4[/tex]
[tex] E_y (0, 0, 1) = 16=> nCz^n^-^1 = 16 [/tex]
[tex]nC = - 16[/tex]
Solving for n, we have:
[tex]n = \frac{-16}{C}[/tex]
From above, C = -2
[tex]n = \frac{-16}{-2}[/tex]
[tex]n = 8[/tex]
Two vectors A and B are such that A =1,B=2,A.B=1 find angle
Answer:[tex]60^{\circ}[/tex]
Explanation:
Given
[tex]\mid\Vec{A}\mid=1[/tex]
[tex]\mid\Vec{B}\mid=2[/tex]
And [tex]A\cdot B=1[/tex]
We know [tex]\vec{A}\cdot \vec{B}=\mid\Vec{A}\mid\mid\Vec{B}\mid\cos \theta[/tex]
Where [tex]\theta[/tex] is the angle between them
Substituting the values
[tex]1=1\times 2\cos \theta[/tex]
[tex]\cos \theta =\dfrac{1}{2}[/tex]
[tex]\theta =60^{\circ}[/tex]
Thus the angle between [tex]A[/tex] and [tex]B[/tex] is [tex]60^{\circ}[/tex]
An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast will he be moving backward just after releasing the ball?PLEASE SORT QUESTION BELOWSort the following quantities as known or unknown. Take the horizontal direction to be along the x axis.mQ: the mass of the quarterbackmB: the mass of the football(vQx)i: the horizontal velocity of quarterback before throwing the ball(vBx)i: the horizontal velocity of football before being thrown(vQx)f: the horizontal velocity of quarterback after throwing the ball(vBx)f: the horizontal velocity of football after being thrown
Answer:
a
The speed of the quarterback backward is [tex]v_q = 0.08 \ m/s[/tex]
b
Known are
[tex]m_Q , m_B , (v_{Bx})_i (v_{Qx})_f, (v_{Bx})_f[/tex]
Unknown
[tex](v_{Qx})_f[/tex]
Explanation:
From the question we are told that
The mass of the quarterback is [tex]m_Q = 80 \ kg[/tex]
The mass of the ball is [tex]m_B = 0.43 \ kg[/tex]
The speed of the ball is [tex]v_{B x}= 15 \ m/s[/tex]
The law of momentum conservation can be mathematically represented as
[tex]m_Q u_{Qx} + m_Bu_{Bx} = - m_{Q} v_{Qx} + m_B v_{Bx}[/tex]
Now at initial both ball and quarterback are at rest and the negative sign signify that the quarterback moved backwards after throwing the ball
So
[tex]m_Q v_{Qx} = m_B v_ {Bx}[/tex]
=> [tex]v_{Qx} = \frac{m_Bv_{Bx}}{m_Q}[/tex]
substituting values
[tex]v_q = \frac{0.43 * 15}{80}[/tex]
[tex]v_q = 0.08 \ m/s[/tex]
A 15 g toy car moving to the right at 24 cm/s has a head-on nearly elastic collision with a 21 g toy car moving in the opposite direction at 31 cm/s. After colliding, the 15 g car moves with a velocity of 41 cm/s to the left. Find the speed of the second car after the collision.
Answer:
The speed of the second toy car after collision is [tex]v_2 = 0.155 \ m/s[/tex]
Explanation:
Let movement to the right be positive and the opposite negative
From the question we are told that
The mass of the car is [tex]m_1 = 15 \ g = \frac{15}{1000} = 0.015 \ kg[/tex]
The initial velocity of the car is [tex]u_1 = 24 \ cm /s = 0.24 m/s[/tex]
The mass of the second toy car [tex]m_2 = 21 g = 0.021 \ kg[/tex]
The initial velocity of the car is [tex]u_2 = 31 \ cm/s =- 0.31 m/s[/tex]
The final velocity of the first car is [tex]v = 41cm/s = - 0.41 m/s[/tex]
From law of momentum conservation we have that
[tex]m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2[/tex]
substituting values
[tex](0.015* 0.24) +( 0.021 * -0.31) = (0.015 * -0.41 ) + 0.021 v_2[/tex]
[tex]-0.00291 = -0.0615 + 0.021 v_2[/tex]
[tex]v_2 = 0.155 \ m/s[/tex]