During a climb UP the mountain, gravity does NO work on the climber.
Actually, it's more correct to say that gravity does NEGATIVE work
on him. The climber has to DO the positive work to haul himself up.
Work = (mass) x (gravity) x (height) .
For the guy in this problem:
Work = (67 kg) x (9.8 m/s²) x (3,500 meters)
= 2,298,100 joules.
If he eats no candy bars on the way, and completely depends on
his stored body fat for the energy, then he'll burn off
(2,298,100 joules) / (3.8 x 10⁷ joules/kg)
= 0.06 kg of fat.
That's only about 2.1 ounces. We KNOW he'll lose more weight than that,
climbing 11,000 feet. That's because climbing is pretty inefficient.
In addition to the potential energy you have to give your body weight,
you also have to expend energy breathing, digesting, metabolizing,
and sweating.
First Law of Thermodynamics: Energy can be changed from one form to another, but it cannot be created or destroyed. The total amount of energy and matter in the Universe remains constant, merely changing from one form to another.
Answer:
Explanation:
<u>Properties of a virtual image:</u>
1. Image formed cannot be projected or focused on a screen.
2. The distance of the object to the mirror is the same as the distance from the image to the mirror.
3. The size of the image formed is the same as the size of the object.
4. The image formed is laterally inverted. That is the right becomes left and vice versa.
5. The image is upright.
<u>Properties of a real image:</u>
1. Image formed can be projected on a screen.
2, The distance from the image to the mirror is not the same as the distance from the object to the mirror.
3. The size of the image is not the same as the size of the object.
4. Image formed is upside down.
The answer is 75 * Cos (45) = ..... Km
OK, the wedge is accelerating (a) at Theta = 180 degrees (to the right) and the wedge is inclined theta = 75 degrees. For the m = 2 kg block to remain at rest all we need is a net force f = W cos(theta) - F sin(theta) = 0; where F = ma and W = mg the weight of the block. That is, the weight component along the incline is offset by the acceleration component along the surface; so the block does not slide.
Solving we have W cos(theta) = mg cos(theta) = ma sin(theta) = F sin(theta); such that a = g cos(theta)/sin(theta) = g cot(theta). Assuming g ~ 9.81 m/sec^2, you can now plug and chug to find the answer.
<span>The physics is this...when the net force on a body is f = 0, that body will not accelerate and start to move if it is already still. So when the block's weight component along the surface of the wedge is offset by the equal but opposite force along the surface of the accelerating wedge, the still block will not move.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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