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erastova [34]
3 years ago
6

Two cars are traveling along a straight line in the same direction, the lead car at 24.7m/s and the other car at 29.9m/s. at the

moment the cars are 38.1m apart, the lead driver applies the brakes, causing her car to have an acceleration of -1.96m/s2. how long does it take for the lead car to stop? (in s)
Physics
1 answer:
sattari [20]3 years ago
7 0

Answer: 12.6 s

We would use the equation of motion:

v-u=at

where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

It is given that the lead car has an initial velocity, u=24.7\hspace{1mm} m/s. It decelerates with a=-1.96 \hspace{1mm} m/s^{2} to stop (v=0). we need to calculate the time taken by the lead car to stop.

\Rightarrow t=\frac{v-u}{a}\\ \Rightarrow t=\frac{0-24.7}{-1.96}=12.6\hspace{1mm} s

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A 4.76 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the en
Romashka-Z-Leto [24]

Answer:

The magnitude of the force is 64.634 newtons.

Explanation:

According to the statement, the crate is a constant mass system, whose upward force is described by the following expression:

F(t) = m\cdot \ddot{y} (t) (1)

Where:

F(t) - Force, in newtons.

m - Mass, in kilograms.

\ddot {y}(t) - Acceleration, in meters per square second.

The function acceleration is obtained by deriving the function position twice in time:

\dot y (t) = 2.80 + 1.83\cdot t^{2} (2)

\ddot y(t) = 3.66\cdot t (3)

And we expand (1) by applying (3):

F(t) = 3.66\cdot m \cdot t

Where t is the time, in seconds.

If we know that m = 4.76\,kg and t = 3.71\,s, then the magnitude of the force is:

F = 3.66\cdot (4.76)\cdot (3.71)

F = 64.634\,N

The magnitude of the force is 64.634 newtons.

6 0
3 years ago
A 50.0 g toy car is released from rest on a frictionless track with a vertical loop of radius R (loop-the-loop). The initial hei
Mariana [72]

Answer:

the speed of the car at the top of the vertical loop  v_{top} = 2.0 \sqrt{gR \ \ }

the magnitude of the normal force acting on the car at the top of the vertical loop   F_{N} = 1.47 \ \ N

Explanation:

Using the law of conservation of energy ;

mgh = mg (2R) + \frac{1}{2}mv^2_{top}\\\\mg ( 4.00 \ R) = mg (2R) + \frac{1}{2}mv^2_{top}\\\\g(4.00 \ R) = g (2R) + \frac{1}{2}v^2 _{top}\\\\v_{top} = \sqrt{2g(4.00R - 2R)}\\\\v_{top} = \sqrt{2g(4.00-2)R

v_{top} = 2.0 \sqrt{gR \ \ }

The  magnitude of the normal force acting on the car at the top of the vertical loop can be calculated as:

F_{N} = \frac{mv^2_{top}}{R} \ - mg\\\\F_{N} = \frac{m(2.0 \sqrt{gR})^2}{R} \ - mg\\\\F_{N} = [(2.0^2-1]mg\\\\F_{N} = [(2.0)^2 -1) (50*10^{-3} \ kg)(9.8 \ m/s^2]\\\\

F_{N} = 1.47 \ \ N

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