A car with a mass of 1100-kg moves at 24 m/s. what breaking force is needed to bring the car to a halt in 20 seconds?
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Answer:
A car of mass 1100 kg moves at 24 m/s. What braking force is needed to bring the car to a halt in 20 s? Breaking force = mv/t = (1100 kg)(24 m/s)/20 s = 1.32x10^3 N What average force is exerted on a 25-g egg by a bed sheet if the egg hits the sheet at 4 m/s and takes 0.2 s to stop? p=mv, = (0.025 kg)(4 m/s), = 0.1 kg (m/s) pfinal = 0 deltap = pfinal - pinitial = 0 - 0.1 kg (m/s) f = -.01 kg (m/s) / .02 = -(1/2)kN a 100-kg quarterback is traveling 5 m/s and is stopped by a tackler in 1 s. Calculate (a) the initial momentum of the quarterback, (b) the impulse imparted by the tackler, and (c) the average force exerted by the tackler. (a) momentum = mv = (100kg)(5 m/s) = 500 kg (m/s) (b) p = 500 kg (m/s), deltap = 0 - 500 kg (m/s) = -500 kg (m/s) (c)average force - deltap/t = -500 kg (m/s)/1s = -500 kN A 40-kg football player going through the air at 4 m/s tackles a 60-kg player who is heading toward her at 3 m/s, in the air. What is the speed and direction of the entangled players? I'm not really sure how to solve this problem... I know they are vector forces, but are they drawn like this: ----> <------ or are they at a different angle? And to find the speed, you would use c^2 = a^2 + b^2, but this isn't a right triangle, or I don't think, so what do I use. And I know to find the angle you usually use inverse tangent, but, can you in this case? A jet engine gets its thrust by taking in air, heating and compressing it, adn then ejecting it at a high speed. If a particular engine takes in 20 kg of air per second at 100 m/s, and ejects it at 500 m/s, calculate the thrust of the engine. What is the equation I use to find the thrust of the engine? A 40-kg projectile leaves a 2000-kg launcher with a speed of 400 m/s. What is the recoil speed of the launcher? I'm not sure, but here is my work: momentum = mv = (40kg)(400 m/s) = 16,000 kg (m/s) mlauncher*vlauncher = -mprojectile*vprojectile = -(16,000 kg (m/s)) vlauncher = -16,000 kg (m/s) / 2000 kg = -8 m/s A car of mass 1400 kg travels at 20 m/s and collides with a stationary truck of mass 2800 kg, with its parking brake off. The two vehicles interlock as a result of the collision and slide along the icy road. What is the velocity of the car-truck system? (1400 kg)(20 m/s) = (1400 kg + 2800 kg)*v 28,000 kg (m/s) = 4200kg*v v = 28,000 kg (m/s) / 4200 kg = 6.7 m/s
Source https://www.physicsforums.com/threads/momentum-and-breaking-force-of-car.65676/
Answer:
1320 N of braking force is needed to bring the car to a halt.
Explanation:
Step-by-step explanation:
Given: The mass of the car (m) is 1100 kg.
Initial velocity (u) of the car is 24 m/s.
The time in which the car comes to a halt, t= 20 s.
Final velocity (when the car comes to a stop), v= 0 m/s.
We know that: Force, F= ma given by Newton and
acceleration, a= (v-u)/t
Here,
F=1100*((0-24)/20)= -1320 N
In this case, the negative sign implies that force is applied in a direction opposite to the direction of motion (of the car).
Therefore, 1320 N of braking force is needed to stop the vehicle.
Definition necessary for solving the question,
Force: Newton's second law states that a force is defined as the change in momentum (defined as the mass times the velocity) per change in time.
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