What force would accelerate a 10kg bowling ball down an alleyway at a rate of 3ms2?

What force would accelerate a 10kg bowling ball down an alleyway at a rate of 3ms2?

This is given by the familiar equation: →F=m→a . Given a mass of 10kg and an acceleration of 3ms2 , we can calculate the net force on the bowling ball from the above equation. Therefore, 30N of force is required to accelerate the bowling ball down the alleyway at a rate of 3ms2 .

What is the acceleration of a 5 kg bowling ball when a 10 N force is applied?

Here the force is 10N and the mass is 5 kg. Dividing both sides by 5kg, we get a = 2 m/s^2.

How much impulse is needed to stop a 10 kg bowling ball moving at 6m s?

Answer and Explanation: Therefore, the impulse needed to stop the bowling ball is − 60 k g m / s .

How do you solve elastic collision problems?

What is the acceleration of a 50 kg object pushed with a force of 500 newtons?

What is the acceleration of a 50 kg object pushed with a force of 500 newtons? The acceleration can be found by Newton’s second law: a = F / m = 500 / 50 = 10 m/s 2 .

What is the acceleration of a 6.4 bowling ball if a force of 12 N is applied to it?

1 Answer. The acceleration of the bowling ball is 1.875 ms2 .

How much acceleration is produced by a 10 and force on a 5 kg mass?

= 10 5 = 2 m / s 2. Was this answer helpful? A force of 10 N acts on a body of mass 5 kg.

How to calculate acceleration?

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2).

What is acceleration of 5kg mass?

Answer: Thus, the acceleration of 5kg mass is 5. 44m/s2.

What is the momentum of a 6.0 kg bowling ball with a velocity?

p=6×2.2=13.2kg m/s. Q.

What is the formula for the impulse of a ball?

Momentum and Impulse The impulse of a force is I=Ft I = F t – when a constant force F acts for a time t . The units are Ns . The Impulse-Momentum Principle says I=mv−mu I = m v − m u which is final momentum – initial momentum so Impulse is the change in momentum.

How much force is required to stop a bus in 10 seconds of mass?

Expert-Verified Answer Force used to stop the bus = mass of the bus in kg× deceleration = 5000 kg× – 2 m/s² = – 10,000 N. The minus sign implies force is supplied by the decelerating agency.

How to calculate velocity?

Determine the object’s original velocity by dividing the time it took for the object to travel a given distance by the total distance. In the equation V = d/t, V is the velocity, d is the distance, and t is the time.

How to find momentum?

Step 1: List the mass and velocity of the object. Step 2: Convert any values into SI units (kg, m, s). Step 3: Multiply the mass and velocity of the object together to get the momentum of the object.

What is the formula of collision?

What is the formula of collision? From the conservation of momentum, the formula during a collision is given by: m1v1 + m2v2 = m1v’1 + m2v’2. If the collision is perfectly inelastic, the final velocity of the system is determined using v’ = (m1v1 + m2v2)/m1 + m2.

How much force in newtons is required to accelerate a five kg bowling ball at 2 m s 2?

Expert-Verified Answer 10 Newtons of the force is required to accelerate a 5 kg bowling ball at 2 m / s², as we know that the force is calculated by the multiplication of the mass and the acceleration of the object.

How much momentum will a ball of mass 10 kg transfer to the floor if it falls from a height of 80 cm?

Momentum transferred to the floor is 40kgm/s.

What net force is required to accelerate a car at a rate of 2m s if the car has a mass of 3000 kg?

What net force is required to accelerate a car at a rate of 2 m/s if the car has a mass of 3,000 kg? F= 6,000 N m 3.000 kg as 2m/s/s 2.

What is the acceleration at its highest point if a situation in which a ball is thrown straight up with no air resistance

Acceleration from gravity is always constant and downward, but the direction and magnitude of velocity change. At the highest point in its trajectory, the ball has zero velocity, and the magnitude of velocity increases again as the ball falls back toward the earth (see figure 1).