Answer:
The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet
Explanation:
The given parameters are;
The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M
The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R
The weight of the cadet on Earth = 800 N

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.
Answer:
at t=46/22, x=24 699/1210 ≈ 24.56m
Explanation:
The general equation for location is:
x(t) = x₀ + v₀·t + 1/2 a·t²
Where:
x(t) is the location at time t. Let's say this is the height above the base of the cliff.
x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0
v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.
a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².
Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.
Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²
Stone: x(t) = 0 + 22·t - 1/2*9.8 t²
Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:
46 = 22·t
so t = 46/22 ≈ 2.09
Put this t back into either original (i.e., with the quadratic term) equation and get:
x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m
Answer:

Explanation:
We can use the following SUVAT equation to solve the problem:

where
v = 0 is the final velocity of the car
u = 24 m/s is the initial velocity
a is the acceleration
d = 196 m is the displacement of the car before coming to a stop
Solving the equation for a, we find the acceleration:

Answer:
speed = 7.9 m/s
Explanation:
speed = total distance / time taken
speed = 300 / 38
speed = 7.89473684 m/s
to the nearest tenth
speed = 7.9 m/s