Answer:
The energy carried by an electromagnetic wave is proportional to the frequency of the wave. The wavelength and frequency of the wave are connected via the speed of light: Electromagnetic waves are split into different categories based on their frequency (or, equivalently, on their wavelength).
Explanation:
Answer:
v = 40 [m/s].
Explanation:
Linear momentum is defined as the product of mass by Velocity. In this way, by means of the following equation, we can calculate the momentum.

where:
m = mass [kg]
v = velocity [m/s]
![P =20*10\\P =200 [kg*m/s]](https://tex.z-dn.net/?f=P%20%3D20%2A10%5C%5CP%20%3D200%20%5Bkg%2Am%2Fs%5D)
Since all momentum is transferred, we can say that this momentum is equal for the mass of 5 [kg]. In this way, we can determine the speed after the impact.
![v = P/m\\v = 200/5\\v = 40 [m/s]](https://tex.z-dn.net/?f=v%20%3D%20P%2Fm%5C%5Cv%20%3D%20200%2F5%5C%5Cv%20%3D%2040%20%5Bm%2Fs%5D)
Let's assume that ground level is the height 0 meters. The change in potential energy is going to be gravitational potential energy, which is given by PE=mgh.
ΔPE=mgh-mgy
=mg(h-y)
=50(28-0)
=1400 J
The question is incomplete! The complete question along with answer and explanation is provided below.
Question:
A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.
What is the change in the potential energy (in Joules) of the mass as it goes up the incline?
If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?
Given Information:
Mass = m = 0.5 kg
Horizontal distance = d = 40 cm = 0.4 m
Vertical distance = h = 7 cm = 0.07 m
Normal force = Fn = 1 N
Required Information:
Potential energy = PE = ?
Work done = W = ?
Answer:
Potential energy = 0.343 Joules
Work done = 0.39 N.m
Explanation:
The potential energy is given by
PE = mgh
where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.
PE = 0.5*9.8*0.07
PE = 0.343 Joules
As you can see in the attached image
sinθ = opposite/hypotenuse
sinθ = 0.07/0.4
θ = sin⁻¹(0.07/0.4)
θ = 10.078°
The horizontal component of the normal force is given by
Fx = Fncos(θ)
Fx = 1*cos(10.078)
Fx = 0.984 N
Work done is given by
W = Fxd
where d is the horizontal distance
W = 0.984*0.4
W = 0.39 N.m
Answer:
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