PHYSICS COLLEGE

A boy throws a 15 kg ball at 4.7 m/s to a 65 kg girl who is stationary and standing on a skateboard. After catching the ball, the girl is travelling at: a) 0.88 m/s b) 0 m/s c) 1.1 m/s d) 3.2 m/s

Answers

Answer 1

Answer:

Answer:

$a)v_(f)=0.88m/s$

Explanation:

To solve this problem we use the Momentum's conservation Law, before and after the girl catch the ball:

$\n p_(1)=p_(2)\nm_(ball)*v_(o.ball)+m_(girl)*v_(o.girl) = m_(ball)*v_(f.ball) + m_(girl)*v_(f.girl)$ (1)

At the beginning the girl is stationary:

$v_(o.girl)=0m/s$ (2)

If the girl catch the ball, both have the same speed:

$v_(f.girl)=v_(f.ball)=v_(f)$ (3)

We replace (2) and (3) in (1):

$m_(ball)*v_(o.ball) = (m_(ball)+m_(girl))*v_(f) \n$

We can now solve the equation for v_{f}:

$v_(f)=(m_(ball)*v_(o.ball))/((m_(ball)+m_(girl)))=(15*4.7)/(15+65)=0.88m/s$

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Uranium-235 undergoes fission, forming krypton-92, barium-141, and 3neutrons. The mass of the uranium-235 is greater than the total mass of the
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O B. Mass was destroyed and disappeared during the process.
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D. Mass was transformed into energy during the process.

Answers

Answer:

D. Mass was transformed into energy during the process.

Answer:

Explanation:

Some of the mass

If the frequency of a system undergoing simple harmonic motion doubles, by what factor does the maximum value of acceleration change?a. 4
b. 2/pi
c. 2
d. (2)^1/2

Answers

Answer:

the answers the correct one is a 4

Explanation:

The centripetal acceleration is by

a = v² / R

angular and linear velocities are related

v = w R

let's substitute

a = w² R

for initial condition

a₀ = w₀² R

suppose the initial angular velocity is wo, suppose the angular velocity doubles

a = (2w₀)² R

a = 4 (w₀² R)

a = 4 a₀

when reviewing the answers the correct one is a

A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.

Answers

The work ( $W_f$ ) done by the friction force between the ramp and the skateboarder is given by $-\mu_k \cdot m_s \cdot g \cdot h_y$ .

The workdone by the friction force ( $W_f$ ) can be calculated using the formula for work, which is the product of the force applied ( $F_f$ ) and the displacement (d) over which the force is applied:

$W_f = F_f \cdot d$

In this scenario, the frictionforce works against the skateboarder's momentum down the ramp, therefore it does no good.

Given:

Mass of skateboarder ( $m_s$ ) = 54 kg

Height of the ramp ( $h_y$ ) = 3.3 m

Final velocity ( $v_f$ ) = 6.2 m/s

Coefficient of kineticfriction ( $\mu_k$ ) between skateboarder and ramp

Acceleration due to gravity (g) = $9.81 \, \text{m/s}^2$

The normal force ( $F_{\text{normal}}$ ) is equal to the weight of the skateboarder:

$F_{\text{normal}} = m_s \cdot g$

The displacement (d) is the vertical distance ( $h_y$ ) that the skateboarder descends down the ramp.

Now we can write the expression for the work done by the friction force ( $W_f$ ):

$W_f = -\mu_k \cdot F_{\text{normal}} \cdot d$

Substitute the expression for the normal force:

$W_f = -\mu_k \cdot (m_s \cdot g) \cdot h_y$

Thus, this expression represents the work done by the friction force between the ramp and the skateboarder in terms of the given variables.

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Your question seems incomplete, the probable complete question is:

Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.

Final answer:

The momentum of the box with respect to the floor can be found by multiplying its mass by its velocity. When the box is put down on the frictionless skating surface, its velocity becomes zero and its momentum with respect to the floor is also zero.

Explanation:

To find the momentum of the box, we can use the formula:

Momentum = mass x velocity

a. The momentum of the box with respect to the floor is: 5 kg x 5 m/s = 25 kg·m/s

b. When the box is put down on the frictionless skating surface, its velocity becomes zero. So, the momentum of the box with respect to the floor is also zero.

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An electron has a velocity of 3.2 x 10^6 m/s. What is its’ momentum? (b) What is its’ wavelength? (c) What other objects/materials have this space/size? (d) Assuming that we can measure the velocity to an accuracy of 10%. Use the Heisenberg uncertainty principle to calculate the uncertainty in the position.

Answers

Answer:

P = 2.91*10^{-24} kg m/s

$\lambda = 2.73 *10^(-10) m$

size of atom hat lie in range of 1 to 5 Angstrom

$\Delta x = 0.2272$ Angstrom

Explanation:

A) MOMENTUM

p = mv

where m is mass of electron

so momentum p can be calculated as

p = 9.11*10^{-31} *3.2*10^{6}

P = 2.91*10^{-24} kg m/s

b) wavelength

$\lambda = (h)/(mv)$

where h is plank constant

so $\lambda = (6.626*10^(-34))/(2.91*10^(-24))$

$\lambda = 2.73 *10^(-10) m$

c) size of atom hat lie in range of 1 to 5 Angstrom

d) from the information given in the question we have

$(\Delta v)/(v) = 0.1$

$\Delta v = 0.1 v$

we know that

$\Delta p *\Delta x = (h)/(4\pi)$

$m \Delta v \Delta x =(h)/(4\pi)$

$\Delta x = (h)/(m \Delta v)$

$\Delta x = (2.272)/(0.1)$ [ $\Delta v = 0.1 v$ ]

$\Delta x = 0.2272$ Angstrom

Suppose that 600 W of radiation in a microwave oven is absorbed by 250 g of water in a very lightweight cup. Approximately how long will it take to heat the water from 20 C to 80 C?(A) 50 s
(B) 100 s
(C) 150 s
(D) 200 s

Answers

Answer:

option (B)

Explanation:

Power, P = 600 W

mass of water, m = 250 g = 0.250 kg

T1 = 20° C

T2 = 80° C

ΔT = 80 - 20 = 60

specific heat of water, c = 4200 J/kg °C

Let the time taken is t.

Power x time = mass of water x specific heat of water x rise in temperature

600 x t = 0.250 x 4200 x 60

t = 105 second

option (B)

Final answer:

To heat 250g of water from 20°C to 80°C using a 600W microwave, it would approximately take 100 seconds.

Explanation:

In order to solve this problem, we first need to know the specific heat capacity of water, which is approximately 4.18 J/g°C. This value represents the amount of energy required to raise 1 gram of water by 1 degree Celsius. Given this value, we'll need to use the formula q = mcΔT, where q is the energy transferred (in joules), m is the mass of the water (in grams), c is the specific heat capacity (in J/g°C), and ΔT is the change in temperature (in °C).

We're given that the initial temperature of water is 20°C and we want to heat it to 80°C, so ΔT = 80°C - 20°C = 60°C. Substituting the known values into the formula, we get: q = 250g * 4.18 J/g°C * 60°C = 62700 J. Now, we know that power (P) = q/t. Given that the microwave oven operates at 600 W (or 600 J/s), we can solve for t: 62700 J ÷ 600 J/s ≈ 104.5 seconds. So, the closest answer would be (B) 100 seconds, considering the approximate value.

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Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit. The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm. What is the width of the slit?