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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: it’s A and B
Explanation: everyone else on this post was giving you the wrong answer.
This question involves the concepts of Wein's displacement law and characteristic wavelength.
The blackbody temperature will be "3.22 x 10⁵ k".
<h3>WEIN'S DISPLACEMENT LAW</h3>
According to Wein's displacement law,

where,
= characteristic wavelength = 9 μm = 9 x 10⁻⁹ m- T = temperature = ?
- c = Wein's displacment constant = 2.897 x 10⁻³ m.k
Therefore,

T = 3.22 x 10⁵ k
Learn more about characteristic wavelength here:
brainly.com/question/14650107