Answer:
-30m/s
Explanation:
Given:
Initial velocity of object = 200 feet/second
Final velocity of object = 50 feet/second
Time of travel = 5 seconds
To calculate acceleration of the object we will find the rate of change of velocity with respect to time.
So, acceleration "a" is given by:
where vf represents final velocity, vi represents initial velocity and is time of travel.
Plugging in values to evaluate acceleration.
The acceleration of the object is -30m/s
Answer:
ΔU = 5.21 × 10^(10) J
Explanation:
We are given;
Mass of object; m = 1040 kg
To solve this, we will use the formula for potential energy which is;
U = -GMm/r
But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.
Thus;
ΔU = -GMm((1/r_f) - (1/r_i))
Where;
M is mass of earth = 5.98 × 10^(24) kg
r_f is final radius
r_i is initial radius
G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²
Since, it's moving to altitude four times the Earth's radius, it means that;
r_i = R_e
r_f = R_e + 4R_e = 5R_e
Where R_e is radius of earth = 6371 × 10³ m
Thus;
ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)
× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))
ΔU = 5.21 × 10^(10) J
The smaller surface area
Hope this helps!
157.5J
KE=1/2 mv^2
KE= 1/2(35kg)(3m/s)^2
KE=(17.5kg)(9m^2/s^2)
KE= 157.5J
