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3 years ago

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6. A car, 1110 kg, is traveling down a horizontal road at 20.0 m/s when it locks up its brakes. The coefficient of friction betw

een the tires and road is 0.901. How much distance will it take to bring the car to a stop?

1 answer:

USPshnik [31]3 years ago

7 0

Answer:

$x=22.65m$

Explanation:

We have an uniformly accelerated motion, with a negative acceleration. Thus, we use the kinematic equations to calculate the distance will it take to bring the car to a stop:

$v_f^2=v_0^2-2ax\\\frac{0^2-v_0^2}{-2a}=x\\x=\frac{v_0^2}{2a}$

The acceleration can be calculated using Newton's second law:

$\sum F_x:F_f=ma\\\sum F_y:N=mg$

Recall that the maximum force of friction is defined as $F_f=\mu N$ . So, replacing this:

$\mu N=ma\\\mu mg=ma\\a=\mu g\\a=0.901(9.8\frac{m}{s^2})\\a=8.83\frac{m}{s^2}$

Now, we calculate the distance:

$x=\frac{v_0^2}{2a}\\x=\frac{(20\frac{m}{s})^2}{2(8.83\frac{m}{s^2})}\\x=22.65m$

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What would be the speed of an object just before hitting the ground if dropped 91.5 meters

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Answer:

about 42.35 m/s

Explanation:

Use the equation for accelerated motion (g), and with zero initial velocity that doesn't include time:

$v_f^2=v_i^2+2\,a\,\Delta x$

which for our case would reduce to:

$v_f^2=v_i^2+2\,a\,\Delta x\\v_f^2=0+2\,9.8\,(91.5)\\v_f^2= 1793.4\\v_f=\sqrt{1793.4} \\v_f \approx 42.35$

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3 years ago

(33%) Problem 3: Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1100 kg and i

expeople1 [14]

Answer:

The final velocity of the cars is 8.38 m/s at an angle 39.1° south of west.

Explanation:

Given that,

Mass of first car = 1100 kg

Velocity of first car = 8.5 m/s

Mass of second car = 650 kg

Velocity of second car = 17.5 m/s

Suppose we need to find the final velocity of the cars and direction of the cars.

We need to calculate the velocity of the car in west direction

Using conservation of momentum in west direction

$m_{f}v_{f}+m_{s}v_{s}= (m_{f}+m_{s})v_{x}$

$v_{x}=\dfrac{m_{f}v_{f}+m_{s}v_{s}}{(m_{f}+m_{s})}$

Put the value into the formula

$v_{x}=\dfrac{1100\times0+650\times17.5}{1100+650}$

$v_{x}=6.5\ m/s$

We need to calculate the velocity of the car in south direction

Using conservation of momentum in south direction

$m_{f}v_{f}+m_{s}v_{s}= (m_{f}+m_{s})v_{y}$

$v_{y}=\dfrac{m_{f}v_{f}+m_{s}v_{s}}{(m_{f}+m_{s})}$

Put the value into the formula

$v_{y}=\dfrac{1100\times8.5+650\times0}{1100+650}$

$v_{y}=5.3\ m/s$

We need to calculate the final velocity of the cars

Using formula of velocity

$v_{eq}=\sqrt{(6.5)^2+(5.3)^2}$

$v_{eq}=8.38\ m/s$

We need to calculate the direction

Using formula of direction

$\tan\theta=\dfrac{v_{y}}{v_{x}}$

Put the value into the formula

$\tan\theta=\dfrac{5.3}{6.5}$

$\theta=\tan^{-1}(\dfrac{5.3}{6.5})$

$\theta=39.1^{\circ}$

Hence, The final velocity of the cars is 8.38 m/s at an angle 39.1° south of west.

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Answer:

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Explanation:

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$\Delta p = 1.350\,\frac{kg\cdot m}{s}$

b) The change in the magnitude of the ball's momentum:

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3 years ago

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Answer:

Explanation:

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