A car’s brakes decelerate it at a rate of -1.70 m/s2. What is the time, in seconds, required to slow the car from 15 m/s to 9.0
1 answer:
Answer:
t = 3.52 s
Explanation:
Given that,
The deceleration of a car is, a = -1.7 m/s²
The initial velocity of the car, u = 15 m/s
Final velocity of the car, v = 9 m/s
We need to find the time that is required to slow the car from 15 m/s to 9 m/s. The definition of acceleration is :
So, it will take 3.52 seconds to slow the car from 15 m/s to 9 m/s.
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Answer:
0.031 W
Explanation:
The power used is equal to the rate of work done:
where
P is the power
W is the work done
t is the time taken to do the work W
In this problem, we have:
W = 900 J is the work done by the motor
t = 8 h is the time taken
We have to convert the time into SI units; keeping in mind that
1 hour = 3600 s
We have
And therefore, the power used is
Answer:
<em>The velocity after the collision is 2.82 m/s</em>
Explanation:
<u>Law Of Conservation Of Linear Momentum </u>
It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is
P=mv.
If we have a system of two bodies, then the total momentum is the sum of the individual momentums:
If a collision occurs and the velocities change to v', the final momentum is:
Since the total momentum is conserved, then:
P = P'
Or, equivalently:
If both masses stick together after the collision at a common speed v', then:
The common velocity after this situation is:
There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.
After the collision, both cars stick together. Let's compute the common speed after that:
The velocity after the collision is 2.82 m/s