Answer:
a) T = 1342.6 cos θ, b) v = 13.93 √(sin θ tan θ)
Explanation:
We can solve this problem using Newton's second law
Let's fix a reference system with a horizontal axis and the other vertical, therefore the only force to decompose is the tension, in these problems the most common is to measure the angle with respect to the vertical. Let's use trigonometry to find the components of the dispute
cos θ = [tex]T_{y}[/tex] / T
T_{y} = T cos tea
sin θ = Tₓ / T
Tₓ = T sin θ
let's write Newton's second law
axis and vertical
T cos θ - W = 0
T = mg / cos θ
let's calculate
T = 137 9.8 cos θ
T = 1342.6 cos θ
unfortunately there is no drawing or indication of the angle
Axis x Horizontal
T sin θ = m a
acceleration is centripetal
a = v² / R
T sin θ = m v² / R
v² = (g / cos θ) R sin θ
v = √ (gR tan θ)
let's use trigonometry to find radius of gyration
sin θ = R / L
R = L sin θ
v = √ (g L sin θ tan θ)
let's calculate
v = √(9.8 19.8 sin θ tant θ)
v = 13.93 √(sin θ tan θ)
they do not give the angle for which the calculation cannot be finished
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]
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]
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.
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
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
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 :)
PLEASE MARK IT THE BRAINLIEST!
Which of the following statements best describes the visible spectrum of light as seen by the human eye? The lowest frequency appears , and the highest frequency appears violet. B. The lowest frequency appears red, and the highest frequency appears yellow. C. The lowest frequency appears green, and the highest frequency appears violet. D. The lowest frequency appears green, and the highest frequency appears yellow.
Answer:
The Answer is red is the lowest and violet is the highest frequency
Explanation:
I think that means A, because the red isn't in the question. But I'm sure red is the lowest frequency and violet is the highest in the visible light spectrum
The visible spectrum as it appears to the human eye is that A. the lowest frequency appears red, and the highest frequency appears violet.
Humans can only view a portion of the electromagnetic spectrum and this portion is known as visible light.
The colors in this visible light have different frequencies which include:
Violet with a frequency range of 700 - 790 THzBlue with a frequency range of 600 - 700 THzGreen with a frequency range of 530 - 580 THz Yellow with a frequency range of 510–530 THzOrange with a frequency range of 480–510 THz and, Red with a frequency range of 400–480 THzNotice how red is the lowest frequency and violet is the highest so we can conclusively say that the lowest frequency appears red, and the highest frequency appears violet.
Find out more at https://brainly.com/question/15091042.
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.
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 .
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 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
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.
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
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.
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!!!
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]
On average, both arms and hands together account for 13% of a person's mass, while the head is 7.0% and the trunk and legs account for 80%. We can model a spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally. Suppose a 74.0 kg skater is 1.80 m tall, has arms that are each 70.0 cm long (including the hands), and a trunk that can be modeled as being 35.0 cm in diameter. If the skater is initially spinning at 68.0 rpm with her arms outstretched, what will her angular velocity 2 be (in rpm ) after she pulls in her arms and they are at her sides parallel to her trunk? Assume that friction between the skater and the ice is negligble.
Answer:
176.38 rpm
Explanation:
mass percentage of arms and legs = 13%
mass percentage of legs and trunk = 80%
mass percentage of head = 7%
Total mass of the skater = 74.0 kg
length of arms = 70 cm = 0.7 m
height of skater = 1.8 m
diameter of trunk = 35 cm = 0.35 m
Initial angular momentum = 68 rpm
We assume:
The spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally.friction between the skater and the ice is negligible.We split her body into two systems, the spinning hands as spinning rods
1. Each rod has moment of inertia = [tex]\frac{1}{3} mL^{2}[/tex]
mass m of the arms is 13% of 74 kg = 0.13 x 74 = 9.62 kg
mass of each side will be assumed to be 9.62/2 = 4.81 kg
L = length of each arm
therefore,
I = [tex]\frac{1}{3}[/tex] x 4.81 x [tex]0.7^{2}[/tex] = 0.79 kg-m for each arm
2. Her body as a cylinder has moment of inertia = [tex]\frac{1}{2} mr^{2}[/tex]
r = radius of her body = diameter/2 = 0.35/2 = 0.175 m
mass of body trunk = (80% + 7%) of 74 kg = 0.87 x 74 = 64.38 kg
I = [tex]\frac{1}{2}[/tex] x 64.38 x [tex]0.175^{2}[/tex] = 0.99 kg-m
We consider each case
case 1: Body spinning with arm outstretched
Total moment of inertia = sum of moments of inertia of both arms and moment of inertia of body trunk
I = (0.79 x 2) + 0.99 = 2.57 kg-m
angular momentum = Iω
where ω = angular speed = 68.0 rpm = [tex]\frac{2\pi }{60}[/tex] x 68 = 7.12 rad/s
angular momentum = 2.57 x 7.12 = 18.29 kg-rad/m-s
case 2: Arms pulled down parallel to trunk
The momentum of inertia will be due to her body trunk alone which is 0.91 kg-m
angular momentum = Iω
= 0.99 x ω = 0.91ω
according to conservation of angular momentum, both angular momentum must be equal, therefore,
18.29 = 0.99ω
ω = 18.29/0.99 = 18.47 rad/s
18.47 ÷ [tex]\frac{2\pi }{60}[/tex] = 176.38 rpm
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.
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
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
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
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]
Consider a blackbody that radiates with an intensity I1I1I_1 at a room temperature of 300K300K. At what intensity I2I2I_2 will this blackbody radiate when it is at a temperature of 400K400K
Answer:
Explanation:
We shall apply Stefan's formula
E = AσT⁴
When T = 300
I₁ = Aσ x 300⁴
When T = 400K
I₂ = Aσ x 400⁴
I₂ / I₁ = 400⁴ / 300⁴
= 256 / 81
= 3.16
I₂ = 3.16 I₁ .
______ ______ are created when something is caused to vibrate.
Answer: the answer is MIRANDA COOPER
Explanation:
jfhudgfuisgfuidsgid
The energy delivered to the resistive coil is dissipated as heat at a rate equal to the power input of the circuit. However, not all of the energy in the circuit is dissipated by the coil. Because the emf source has internal resistance, energy is also dissipated by the battery as heat. Calculate the rate of dissipation of energy PbatPbatP_bat in the battery.
Answer:
P = I²r
Explanation:
ε= IR + Ir
where r is the internal resistance
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.
You are standing on a large sheet of frictionless ice and holding a large rock. In order to get off the ice, you throw the rock so it has velocity 12.0 m/s relative to the earth at an angle of 35.0∘ above the horizontal. Your mass is 72.0 kg and the rock’s mass is 3.50 kg . What is your speed after you throw the rock?
Answer:
0.4778 m/s
Explanation:
To solve this question, we will make use of law of conservation of momentum.
We are given that the rock's velocity is 12 m/s at 35°. Thus, the horizontal component of this velocity is;
V_x = (12 m/s)(cos(35°)) = 9.83 m/s.
Thus, the horizontal component of the rock's momentum is;
(3.5 kg)(9.83 m/s) = 34.405 kg·m/s.
Since the person is not pushed up off the ice or down into it, his momentum will have no vertical component and so his momentum will have the same magnitude as the horizontal component of the rock's momentum.
Thus, to get the person's speed, we know that; momentum = mass x velocity
Mass of person = 72 kg and we have momentum as 34.405 kg·m/s
Thus;
34.405 = 72 x velocity
Velocity = 34.405/72
Velocity = 0.4778 m/s
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]
Tired of being chased by a jaguar, you set a trap. Hoping to drop it on the jaguar, you try to push a
44.0 kg stone boulder off of the edge of a cliff that slopes down at an angle of 15.0°. Being weak with
hunger, the best you can do is push the boulder with a force of 222 N. The coefficient of kinetic friction
between the boulder and the ground is is 0.700. (Ignore static friction.)
What is the acceleration of the boulder while you push it down the incline?
Answer: acceleration = 3.27m/s^2
Explanation:
Given that the
Mass M = 44kg
Angle Ø = 15 degree
Coefficient of friction ų = 0.7
Force F = 222N
F - Fr = ma ...... (1)
Where Fr = frictional force
Fr = ųN
N = normal reaction = mg
Fr = ųmgsinØ
Fr = 0.7 × 44 × 9.81 × sin 15
Fr = 78.2N
Substitutes Fr, F and M into equation one.
222 - 78.2 = 44a
143.79 = 44a
Make a the subject of formula
a = 143.79/44
Acceleration a = 3.27 m/s^2
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.