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When the brakes of a car are applied a deceleration of 6 meters per second squared is produced How much time will it take to bring the car to a halt from an initial speed of 110 kilometers per hour?
Divide 110 km hr-1 by 3600 to s hour-1 get 30.6 m s-1. This is needed to get the car's speed in the same units as the deceleration. Now, divide 30.6 m s-1 by 6 m s-2 to get 5.1 s. That is how long it will take to stop the car.
What is the effect of deceleration on an object in force?
Deceleration is the rate at which an object slows down. In a force, deceleration can cause the object to come to a stop, change direction, or reduce its speed. The force applied during deceleration will act in the opposite direction of the object's motion to bring it to a halt.
A car goes forward along a level road at constant velocity The additional force needed to bring the car into equilibrium is?
The additional force needed to bring the car into equilibrium would be equal in magnitude and opposite in direction to the force that was providing the constant velocity. This force is the force of friction that acts to oppose the motion of the car.
What should be the spring constant k of a spring designed to bring a 1200 kg car to rest from a speed of 95 kph so that the occupants undergo a maximum acceleration of 4.0 g?
To find the spring constant k, we first calculate the deceleration needed to bring the car to rest from 95 kph using a = (v^2 - u^2) / (2 * s), where u is initial velocity, v is final velocity, s is displacement. Converting 95 kph to m/s, we get u = 26.39 m/s. We then use the maximum acceleration of 4.0 g to find the deceleration, a_max = 4.0 * 9.81 m/s^2. Finally, we can calculate the spring constant k by equating F_spring = m * a = k * x, where x is the displacement and solve for k.
What would be the momentum of the passenger bus its velocity were doubled?
a) doubled b) tripled c) doubled (assuming the engine is used to bring it back to the same speed) d) quadrupled e) halved
When throwing a ball straight up when in the air with an initial velocity of 10 meters per second what high will it go and how long will it take to return to the ground?
The initial velocity is 10 meters/sec and is thrown up against the gravitational pull of the earth. This means that the ball is experiencing a deceleration at the rate of 9.8 meters/sec/sec to bring its final velocity to zero. v^2 - u^2 = 2gs where u is the initial velocity, v the final velocity, g is the acceleration or deceleration, and s is the distance traveled. 0^2 - 10^2 = 2 x (-9.8) x s -100 = -19.6s 100 = 19.6s s = 100/19.6 = 5.102 meters Now v = u + gt where v is the final velocity, u is the initial velocty, g is the acceleration or deceleration, and t is the time. When the ball is thrown up with 10 meters/sec velocity it is acted upon by the deceleration of gravity until its velocity becomes zero. So 0 = 10 - 9.8t or 9.8t = 10 t = 1.020 seconds The time for the ball to go up is 1.020 seconds and the same time is taken for the ball to come back for a total of 2.040 seconds.
When the brakes of a car are applied a deceleration of 6 meters per second squared is produced How much time will it take to bring the car to a halt from an initial speed of 110 kilometers per hour?
Divide 110 km hr-1 by 3600 to s hour-1 get 30.6 m s-1. This is needed to get the car's speed in the same units as the deceleration. Now, divide 30.6 m s-1 by 6 m s-2 to get 5.1 s. That is how long it will take to stop the car.
What is the effect of deceleration on an object in force?
Deceleration is the rate at which an object slows down. In a force, deceleration can cause the object to come to a stop, change direction, or reduce its speed. The force applied during deceleration will act in the opposite direction of the object's motion to bring it to a halt.
A car goes forward along a level road at constant velocity The additional force needed to bring the car into equilibrium is?
The additional force needed to bring the car into equilibrium would be equal in magnitude and opposite in direction to the force that was providing the constant velocity. This force is the force of friction that acts to oppose the motion of the car.
What should be the spring constant k of a spring designed to bring a 1200 kg car to rest from a speed of 95 kph so that the occupants undergo a maximum acceleration of 4.0 g?
To find the spring constant k, we first calculate the deceleration needed to bring the car to rest from 95 kph using a = (v^2 - u^2) / (2 * s), where u is initial velocity, v is final velocity, s is displacement. Converting 95 kph to m/s, we get u = 26.39 m/s. We then use the maximum acceleration of 4.0 g to find the deceleration, a_max = 4.0 * 9.81 m/s^2. Finally, we can calculate the spring constant k by equating F_spring = m * a = k * x, where x is the displacement and solve for k.
What is the magnitude of the net force needed to bring a 2175 kg car to rest from 24.2 meters per second in 7.4 seconds?
approximately 7000 newtons
Which side needed blockade running?
The Confederates needed blockade-runners to bring in much-needed war supplies.
Do vets bring work home?
If needed they will, but it depends.
Why did King Charles 1 bring back parliament?
He needed them to fund his wars.
What is the meaning of unnecessary?
In means "not needed" or "not required."Example:"It is unnecessary for you to bring your monitor to our shop when you bring your computer since we have our own."
How much force is needed to bring a 3200 lb car from rest to a velocity of 44ftsec in 8 sec?
How much force is needed to bring a 3200 lb car from rest to a velocity of 44ftsec in 8 sec? a = (vf - Vi)/time a = (0 - 44)/ 8 a = -5.5 m/s^2 F = m * a Since the weight is given in pounds, gravity = 32.2 m/s^2 W = m*g 3200 = m * 32.2 m = 99.4 Slugs At least that is what we used to call mass in the English system F =99.4 * -5.5 F = -546.7 lbs.
What type of change can the balanced forces bring about in an object?
Balanced forces do not bring about any change in motion. If forces are balanced, an object is either at rest, or moving in a straight line at constant velocity.
