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2 years ago

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A car is moving with a speed of 32.0 m/s. the driver sees an accident ahead and slams on the brakes, causing the car to slow dow

n with a uniform acceleration of magnitude 3.50 m/s2. how far does the car travel after the driver put on the brakes until it comes to a stop? a car is moving with a speed of 32.0 m/s. the driver sees an accident ahead and slams on the brakes, causing the car to slow down with a uniform acceleration of magnitude 3.50 m/s2. how far does the car travel after the driver put on the brakes until it comes to a stop? 112 m 292 m 4.57 m 146 m 9.14 m

1 answer:

UkoKoshka [18]2 years ago

5 0

The car will travel up to a distance of 146.28 m, hence the correct answer is 146 m.

Since the car is decelerating with the constant acceleration, so we can apply the third equation of motion.

v^2=u^2-2aS

here, v is the final speed of the car, which is 0 as the car stops,

u is the initial speed=32 m/s

a is the constant acceleration=3.5 m/s^2

and S is the distance cover by the car before it stops.

Now plugging the values in the third equation of motion

0=32^2-(2*3.5*S)

S=146.28 m

Therefore the car will cover a distance of 146 m before it stops.

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Answer:

The distance the ball moves up the incline before reversing its direction is 3.2653 m.

The total time required for the ball to return to the child’s hand is 3.2654 s.

Explanation:

When the girl is moving up:

The final velocity (v) = 0 m/s

Initial velocity (u) = 4 m/s

a = -0.25g = -0.25*9.8 = -2.45 m/s². (Negative because it is in opposite of the velocity and also it deaccelerates while going up).

Let time be t to reach the top.

Using

v = u + a×t

0 = 4 - 2.45*t

t = 1.6327 s

Since, this is the same time the ball will come back. So,

<u>Total time to go and come back = 2* 1.6327 = 3.2654 s </u>

To find the distance, using:

v² = u² + 2×a×s

0² = 4² + 2×(-2.45)×s

s = 3.2653 m

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2 years ago

Two weights are connected by a very light cord that passes over an 80.0Nfrictionless pulley of radius 0.300m. The pulley is a so

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Answer:

The force does the ceiling exert on the hook is 269.59 N

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From the attached diagram, the net force in object 1 is:

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In object 2:

$m_{2} a=W_{2} -T_{2}$

Adding the two equations:

$m_{2} a+m_{1} a=T_{1} -W_{1} +W_{2} -T_{2} \\m_{1} =\frac{W_{1} }{g} \\m_{2} =\frac{W_{2} }{g} \\Replacing\\T_{2}-T_{1}=W_{2} -W_{1} -(\frac{W_{1} }{g} +\frac{W_{2} }{g} )a$ (eq. 1)

The torque:

$\tau =I\alpha$

Where

I = moment of inertia

α = angular acceleration

If the linear acceleration is

$a=r\alpha \\\alpha =\frac{a}{r} \\I=\frac{1}{2} mr^{2} \\\tau =\frac{mra}{2}$

Torque due the tension is equal:

$\tau =r(T_{2} -T_{1} )$

Substituting torque, mass, in equation 1, the expression respect the acceleration is:

$a=\frac{g*(W_{2}-W_{1})}{W_{1}+W_{2} +\frac{W}{2} }$

Where

W₁ = 75 N

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2 years ago

Please help me with questions 1, 2 and 3. <br> i need a step by step explanation

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Answer:

1) d

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1) Solve equation (ii) for acceleration a and plug the result in equation (i)

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Simplify equation (iv) and use the given values v = 0, x₀ = 0:

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2 years ago

A free falling object has the velocity time graph shown. What is the objects displacement between 0.0 and 6.0s

zloy xaker [14]

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Here acceleration of object is constant as it fall due to gravity so we can use

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