How does Mercury's close proximity to the sun and thin atmosphere affect its ability to maintain liquid water
1 answer:
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Answer:
V = 11.83 m/s
Explanation:
Given the following data;
Mass = 2000 kg
Force = 10000N
Distance = 14 m
To find the final velocity of the car;
First of all, we would determine the acceleration of the car;
Acceleration = force/mass
Acceleration = 10000/2000
Acceleration = 5 m/s²
Next, we would use the third equation of motion to find the final velocity;
Where;
V represents the final velocity measured in meter per seconds.
U represents the initial velocity measured in meter per seconds.
a represents acceleration measured in meters per seconds square.
S represents the displacement measured in meters.
Substituting into the equation, we have;
V² = 0² + 2*5*14
V² = 0 + 140
V = √140
V = 11.83 m/s
We have the equation of motion , where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.
Here displacement = 135 m, Initial velocity = 0 m/s, acceleration = 9.81
Substituting
A box falls out of a stationary helicopter hovering 135 m above the ground will take 5.25 seconds to reach the ground.