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4 years ago

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A car of mass 1000kg is travelling in an eastward ( x) direction with a constant speed of 20m/s. It then hits a stationary truck

. The truck is twice as heavy as the car. After the collision, the car travels in a direction 300 North of East and the truck travels 600 South of East. (a) What are the final speeds of the car and the truck after collision

1 answer:

Aleonysh [2.5K]4 years ago

8 0

Answer:

4.33 m/s, 5 m/s

Explanation:

mass of car, m = 1000 kg

initial velocity of car, u = 20 m/s along east

mass of truck, M = 2 x mass of car = 2000 kg

initial velocity of truck, U = 0 m/s

After collision, the velocity of car is v and the direction is 30° North of east.

The velocity of truck is V and the direction is 60° South of east.

Use conservation of momentum along X axis.

m x u = m x v x Cos 30 + M x V x Cos 60

1000 x 20 = 1000 x v x 0.866 + 2000 x V x 0.5

20 = 0.866 v + V .... (1)

Use conservation of momentum along y axis

0 = m x v x Sin 30 - M x V x Sin 60

0 = 1000 x v x 0.5 - 2000 x V x 0.866

v = 3.464 V ..... (2)

Substitute the value of v in equation (1)

20 = 0.866 x 3.464 V + V

V = 5 m/s

v = 0.866 x 5 = 4.33 m/s

Thus, the final speed of car is 4.33 m/s and the final speed of truck is 5 m/s after the collision.

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3 years ago

In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo

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Explanation:

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Substitute this value into the Snell's Law equation:

$\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}$ .

Rearrange to obtain an expression for the speed of light in the first medium:

$v_{1} = v_{\text{air}} \cdot \sin(\theta_{1})$ .

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For $0 < \theta < 90^{\circ}$ , $\sin(\theta)$ is monotonically increasing with respect to $\theta$ . In other words, for $\!\theta$ in that range, the value of $\sin(\theta)\!$ increases as the value of $\theta\!$ increases.

Therefore, compared to the medium in this question with $\theta_{c} = 27^{\circ}$ , the medium with the larger critical angle $\theta_{c} = 32^{\circ}$ would have a larger $\sin(\theta_{c})$ . such that light would travel faster in that medium.

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Read 2 more answers

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A stationary police car emits a sound of frequen

bixtya [17]

Answer:

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b) The frequency is 1404.08 Hz

Explanation:

If the police car is a stationary source, the frequency is:

$f_{a} =(\frac{v+v_{c} }{v} )f_{s}$ (eq. 1)

fs = frequency of police car = 1200 Hz

fa = frequency of moving car as listener

v = speed of sound of air

vc = speed of moving car

If the police car is a stationary observer, the frequency is:

$f_{L} =f_{a} (\frac{v}{v-v_{c} } )=(\frac{v+v_{c} }{v-v_{c} } )f_{s}$ (eq. 2)

Now,

fL = frequecy police car receives

fs = frequency police car as observer

a) The velocity of car is from eq. 2:

$1250=1200(\frac{v+v_{c} }{v-v_{c} } )\\1250(v-v_{c} )=1200(v+v_{c} )\\v_{c} =\frac{50*344}{2450} =7.02m/s$

b) Substitute eq. 1 in eq. 2:

$f_{L} =(\frac{v+v_{p} }{v-v_{c} } )(\frac{v+v_{c} }{v-v_{p} } )f_{s} =(\frac{344+20}{344-7.02} )(\frac{344+7.02}{344-20} )*1200=1404.08Hz$

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