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kvasek [131]
3 years ago
9

If the car's speed decreases at a constant rate from 77 mi/h to 50 mi/h in 3.0 s, what is the magnitude of its acceleration, ass

uming that it continues to move in a straight line?
Physics
1 answer:
Lostsunrise [7]3 years ago
3 0

The acceleration of the car,

a = \frac{v-u}{t}

Here, v is final velocity, u is initial velocity and t is time taken by the car.

Given u =77 \ mil/h ,v = 50 mi/h and t = 3.0 s = 3.0 \times \frac{1 \ h}{3600} = 8.3 \times 10^{-4} h

Therefore, from above equation

a =  \frac{50 \ mi/h -77 \ mi/h}{8.33 \times 10^{-4} h} = - \frac{27 \ mi/h }{8.33 \times 10^{-4} h} = - 3.2 \times 10^{4} \ mi/h^2.

Here, negative sign shows deceleration of a car.

Thus the the magnitude of car acceleration is 3.2 \times 10^{4} \ mi/h^2.

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Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

\frac{\Delta l}{l_{o}}=\frac{F}{AY} (1)

T=2 \pi \sqrt{\frac{l_{o}}{g}} (2)

Where:

\Delta l=0.05 mm=5(10)^{-5} m is the length the steel wire streches (taking into account 1mm=0.001 m)

l_{o} is the length of the steel wire before being streched

F=mg=(2 kg)(9.8 m/s^{2})=19.6 N is the force due gravity (the weight) acting on the pendulum with mass m=2 kg

A is the transversal area of the wire

Y=2(10)^{11} Pa is the Young modulus for steel

T is the period of the pendulum

g=9.8 m/s^{2} is the acceleration due gravity

Knowing this, let's begin by finding A:

A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4} (3)

Where d=1.1 mm=0.0011 m is the diameter of the wire

A=\pi \frac{(0.0011 m)^{2}}{4} (4)

A=9.5(10)^{-7}m^{2} (5)

Knowing this area we can isolate l_{o} from (1):

l_{o}=\frac{\Delta l AY}{F} (6)

And substitute l_{o} in (2):

T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}} (7)

T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}} (8)

Finally:

T=1.39 s

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<u>We are given:</u>

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We know the formula:

g = G(mass of planet) / (r)²

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g = (6.87 * 10¹⁵ * 10⁻¹⁴) / 5.15

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Answer:

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Explanation:

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32 N = (1 kg) a

a = 32 m/s²

So the acceleration increases by a factor of 4.

5 0
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