Answer:
Kathmandu
Explanation:
As the altitude get higher, the gravitational pull of the earth on the object increases, therefore, the mass is higher up above.
Answer:
Kinetic energy is given by:
K.E. = 0.5 m v²
Susan has mass, m = 25 kg
Velocity with which Susan moves is, v = 10 m/s
Hannah has mass, m' = 30 kg
Velocity with which Hannah moves is, v' = 8.5 m/s
<u>Kinetic energy of Susan:</u>
0.5 m v² = 0.5 × 25 kg × (10 m/s)² = 1250 J
<u>Kinetic energy of Hannah:</u>
0.5 m v'² = 0.5 × 30 kg × (8.5 m/s)² = 1083.75 J
Susan's kinetic energy is <u>1250 J </u>and Hannah's kinetic energy is <u>1083.75 J</u>.
Since kinetic energy is dependent on mass and square of speed. Thus, speed has a greater effect than mass. As it is evident from the above example. Susan has greater kinetic energy due to higher speed than Hannah.
Answer:
The type of light and the material of lenz.
Explanation:
1) As the investigation is based on how the thickness of a lens effect the other variable. Thickness of the lenz is independent variable. So Lidia has to experiment with the different thicknesses in order to find the effect on dependent variable.
2) As the investigation is based to find the point where the beam of light is focused. It's a dependent variable and Lidia has no control over it. So the only thing she can do is to measure and observe how it respond to the changes in independent variable.
3) For conclusion, she has to make sure that the other variables are not effecting the output or results that is the beam point where the light is focused. So she must have to kept constant the type of light and material of lenz otherwise she won't be able to discriminate the effect of thickness of lenz from other causes.
Answer:
The y-component of the normal force is 45.74 N.
Explanation:
Given that,
Mass of the crate, m = 5 kg
Angle with hill, 
We need to find the y component of the normal force. We know that the y component of the normal force is given by :

So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.