What is the frequency of a photon that has the same momentum as a neutron moving with a speed of 1.90 × 103 m/s?

Answers

Answer 1

Answer: The mass of a neutron is:
$m=1.67 \cdot 10^(-27)kg$
Since we know its speed, we can calculate the neutron's momentum:
$p=mv=(1.67 \cdot 10^(-27)kg)(1.90 \cdot 10^3 m/s)=3.17 \cdot 10^(-24) kg m/s$

The problem says the photon has the same momentum of the neutron, p. The photon momentum is given by
$p= (h)/(\lambda)$
where h is the Planck constant, and $\lambda$ is the photon wavelength. If we re-arrange the equation and we use the momentum we found before, we can calculate the photon's wavelength:
$\lambda= (h)/(p)= (6.6 \cdot 10^(-34)Js)/(3.17 \cdot 10^(-24) kg m/s)=2.08 \cdot 10^(-10) m$

And since we know the photon travels at speed of light c, we can now calculate the photon frequency:
$f= (c)/(\lambda)= (3 \cdot 10^8 m/s)/(2.08 \cdot 10^(-10) m)= 1.44 \cdot 10^(18) Hz$

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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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On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snohomish River below. The swimmers stepped off the bridge, and I estimated that they hit the water 2.00 s later. A)How high was the bridge?B)How fast were the swimmers moving when they hit the water?

C)What would the swimmer's drop time be if the bridge were twice as high?