Answer:
Half: 6 cm^2 Whole: 12 cm^2
Explanation:
First, we know that the edges of the cube are 2 cm long. So there are 6 faces on a cube. We do 2x6=12 cm^2 as our total surface area. Then it asks for each half. So you would divide it by 2 and get 6 cm^2 as your half.
Answer:
9.6m/s
Explanation:
Using the equation S=d/t where s=speed, d=distance, and t=time
plug in the known variables
S=120m/12.5s
S=9.6m/s
Answer:
a) 
b)
c) 
d) Treat the humans as though they were points or uniform-density spheres.
Explanation:
Given:
- mass of Mars,

- radius of the Mars,

- mass of human,

a)
Gravitation force exerted by the Mars on the human body:

where:
= gravitational constant


b)
The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.


c)
When a similar person of the same mass is standing at a distance of 4 meters:


d)
The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.
- Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.
- Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.
Answer:
the horizontal distance covered by the cannonball before it hits the ground is 327.5 m
Explanation:
Given;
height of the cliff, h = 210 m
initial horizontal velocity of the cannonball, Ux = 50 m/s
initial vertical velocity of the cannonball, Uy = 0
The time for the cannonball to reach the ground is calculated as;
The horizontal distance covered by the cannonball before it hits the ground is calculated as;

Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m
The strength of the electric and magnetic fields there is no physical "distance" of oscillation here. nothing is actually moving up and down if you draw light as a sinusoidal wave, the up and down motion is the strength of the EM fields cheers