A race car has a mass of 708 kg It starts from rest and travels 35 m in 3.2 s The car is uniformly accelerated during the entire time What net force is applied to it?
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Total distance moved = Vi * t + ½ * at^2
It starts from rest
Initial velocity = 0
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And travels 35 m
Distance = 35 m
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in 3.2 s
time = 3.2 s
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Total distance moved = Vi * t + ½ * at^2
35 = 0 + ½ * a *3.2^2
35 = ½ * a * 10.24
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a = (35 * 2) ÷ 10.24
a = 6.836 m/s^2
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Force = mass * acceleration
Mass = 708 kg
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Force = 708 * 6.836=
Force = 4839.9 N
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193.3653333..newton
Acceleration means speeding up or slowing down, a change in velocity. Since the velocity was constant, the acceleration was. 0
A catalyst is a substance that accelerates or increases the rate of a chemical reaction. In principle a catalyst should remain unaffected during the process. However normally it loses some activity during it resuses.
Yes. A gelatinous cover, over tiny hairs in the saccule, which contain otholiths (ear stones) detects acceleration when they are bent during this movement.
The vibration at speed is more than likely a tire out of balance or a tire with a broken belt. Have this checked ASAP. The pulsation during acceleration can be many things, from a vacuum leak to a clogged fuel filter. Take it to a tech for service.
A solar eclipse is when the moon blocks out the sun.
So it's acceleration is 4m/s2. So at any point because it says uniformly, it will be accelerating at 4m/s2 each second
a few things missing: - How fast is the acceleration? - How long does it take to go from 4ms to 20ms?
160 m
it's approximate 35ms....
Not enough information. You also need to know how much the acceleration is. Once you know that, calculate the final speed, then calculate the average speed as (initial speed + final speed) / 2, and multiply that by the time to get the distance.
Using the definition of acceleration as change of speed / time, you basically need to know: * A time interval during which the object accelerates. * The velocity at the beginning of this time interval. * The velocity at the end of this time interval.
The formula for distance covered during uniform acceleration isd = 1/2 * (vf + vi) * t (1)Time, t, is given; initial velocity, vi, is 0; but final velocity, vf, is unknown and must be computed from given information. Knowing the rate of acceleration, initial velocity and time, The final velocity may be computed using the formula for average acceleration (actual acceleration under uniform motion) which isa = (vf - vi) / t (2)Rewriting to solve for vf with vi = 0 we havevf = a * tvf = 6m/s2 * 12svf = 72m/sPlugging this value into equation (1) with the other given values we haved = 1/2 * (72m/s + 0 m/s) * 12sd = 432mSo the airplane will travel 432m from rest in 12 seconds under 6m/s uniform acceleration.
Speed is constant. Acceleration is zero.
Acceleration is zero velocity is constant at 8m every second acceleration is the gradient of velocity and the gadient of a constant = 0
It's called stop cheating on your math papers.
That's true throughout any period of time during which the acceleration is constant.
None! The ball is DE-ccelerating as it rises, stops for an infinitesimal time at peak, then accelerates back down.