A 30 kg object has velocity v_x(t) = 1/2 t^2 - 3/4 t + 5. What is the net force

Question

A 30 kg object has velocity v_x(t) = 1/2 t^2 - 3/4 t + 5. What is the net force acting on the object after 5 s? Let N be the total number of letters in your first and last names, and then consider the distance N Times 200 miles. What must be the speed of a satellite that is in orbit at this altitude? Two planets with masses 7.5E + 23_kg and 2E + 24_kg are 410000000_m apart. Calculate the gravitational force between the planets. A force is applied to a 26_kg object and presses it against a wall. This force is just great enough to hold the object in place (to prevent it from sliding down the wall). If the coefficient of static friction between the object and the wall is mu_s = 0.29, then find the applied force.

Explanation / Answer

Vx(t) = 0.5 * t^2 - (3/4) * t + 5

initial veocity or velocity at 0 sec will be

Vx(0) = 0.5 * 0^2 - (3/4) * 0 + 5

Vx(0) = 5 m/s

velocity after 5 sec

Vx(5) = 0.5 * 5^2 - (3/4) * 5 + 5

Vx(5) = 13.75 m/s

acceleration = rate of change of velocity

acceleration = 13.75 - 5 / 5

acceleration = 1.75 m/s^2

force = mass * acceleration

force = 30 * 1.75

force = 52.5 N

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