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Inessa [10]
3 years ago
14

A car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s. At this time, the driver suddenly realizes that she m

ust stop the car. If it takes 0.200 s for the driver to apply the brakes, what must be the magnitude of the constant acceleration of the car after the brakes are applied so that the car will come to rest at the stop sign?
Physics
1 answer:
victus00 [196]3 years ago
6 0

Answer:

The acceleration of the car will be a=9600m/sec^

Explanation:

We have given that distance from stop sign s = 200 m

Time t = 0.2 sec

We have to find the constant acceleration

Now from second equation of motion s=ut+\frac{1}{2}at^2

200=40\times 0.2+\frac{1}{2}\times a\times 0.2^2

a=9600m/sec^

So the acceleration of the car will be a=9600m/sec^

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Average speed of the runner is the rate at which the runner covers the total distance. Average speed of the runner in the race is given by,

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Total time = time taken by the runner to cover entire distance

Instantaneous speed is the speed of the runner at the particular moment in the given time. Instantaneous speed is given by,

Instantaneous speed = \frac{dx}{dt}

x = position of the runner at time t

t = time taken to cover distance x

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3 years ago
What forces contribute to density
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3 years ago
Reflecting telescopes are popular because they're blank than a regarding telescope
olganol [36]
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3 years ago
If an object travels at a constant speed in a circular path, the acceleration of the object is:
Vilka [71]

Answer:

1- The acceleration of the object is larger in magnitude the smaller the radius of the circle.

Explanation:

The acceleration of an object in a circular path is

a = \frac{v^2}{r}

As can be seen from the equation, if the radius of the circle is decreases, the magnitude of the acceleration increases.

As for the direction of the acceleration, it is always towards the center, and it is always perpendicular to the direction of the velocity.

6 0
3 years ago
A 2kg hockey puck is sliding across the ice skating rink at 2 m/s. A player hits the puck so it's velocity increases to 10 m/s.
konstantin123 [22]

The work done on the puck is 96 J

Explanation:

According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.

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W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2

where

K_f = \frac{1}{2}mv^2 is the final kinetic energy of the puck, with

m = 2 kg being the mass of the puck

v = 10 m/s is the final speed

K_i = \frac{1}{2}mu^2 is the initial kinetic energy of the puck, with

u = 2 m/s being the initial speed of the puck

Substituting numbers into the equation, we find the work done by the player on the puck:

W=\frac{1}{2}(2)(10)^2 - \frac{1}{2}(2)(2)^2=96 J

Learn more about work and kinetic energy:

brainly.com/question/6763771  

brainly.com/question/6443626  

brainly.com/question/6536722

#LearnwithBrainly

6 0
3 years ago
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