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luda_lava [24]
3 years ago
14

A police car is located 50 feet to the side of a straight road. A red car is driving along the road in the direction of the poli

ce car and is 140 feet up the road from the location of the police car. The police radar reads that the distance between the police car and the red car is decreasing at a rate of 70 feet per second. How fast is the red car actually traveling along the road?

Physics
1 answer:
NNADVOKAT [17]3 years ago
6 0

Answer:74.33 feet per sec

Explanation:

Given

Police car is 50 feet side of road

Red car is 140 feet up the road

Distance between them is decreasing at the rate of 70 feet per sec

From figure

x^2+y^2=z^2

z=\sqrt{140^2+50^2}

z=148.66 feet

as the red car is moving

therefore its velocity magnitude is given by

\frac{\mathrm{d} x}{\mathrm{d} t}

2x\times \frac{\mathrm{d} x}{\mathrm{d} t}+0=2z\frac{\mathrm{d} z}{\mathrm{d} t}

x\frac{\mathrm{d} x}{\mathrm{d} t}=z\frac{\mathrm{d} z}{\mathrm{d} t}

140\times \frac{\mathrm{d} x}{\mathrm{d} t}=148.66\times 70

\frac{\mathrm{d} x}{\mathrm{d} t}=74.33 feet\ per\ sec

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Answer:

No. Twice as much work will give the ball twice as much kinetic energy. But since KE is proportional to the speed squared, the speed will be sqrt{2} times larger.

Explanation:

The work done on the ball is equal to the kinetic energy gained by the ball:

W=K

So when the work done doubles, the kinetic energy doubles as well:

2W = 2 K

However, the kinetic energy is given by

K=\frac{1}{2}mv^2

where

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We see that the kinetic energy is proportional to the square of the speed, v^2. We can rewrite the last equation as

v=\sqrt{\frac{2K}{m}}

which also means

v=\sqrt{\frac{2W}{m}}

If the work is doubled,

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So the new speed is

v'=\sqrt{\frac{2(2W)}{m}}=\sqrt{2}\sqrt{\frac{2W}{m}}=\sqrt{2} v

So, the speed is \sqrt{2} times larger.

5 0
3 years ago
A thin disk of mass 2.2 kg and radius 61.2 cm is suspended by a horizonal axis perpendicular to the disk through its rim. The di
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Answer: The period of the subsequent simple harmonic motion is 1.004 sec.

Explanation:

The given data is as follows.

  Mass of disk (m) = 2.2 kg,     radius of the disk (r) = 61.2 cm,

Formula to calculate the moment of inertia around the center of mass is as follows.

          I_{cm} = \frac{1}{2}mr^{2}  

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Also,

Distance between center of mass and axis of rotation (d) = r = 0.612 m

Moment of inertia about the axis of rotation (I)

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Now, we will calculate the time period as follows.

        T = 2\pi sqrt{\frac{I}{mgd})

        T = 2 \times 3.14 sqrt{(\frac{0.339}{2.2 \times 9.8 \times 0.612}

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Thus, we can conclude that the period of the subsequent simple harmonic motion is 1.004 sec.

7 0
2 years ago
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