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3 years ago

14

Suppose that a person riding on the top of a freight car shines a searchlight beam in the direction in which the train is travel

ing.
How does the speed of the light beam relative to the ground compare to the speed of beam when the train is at rest?

1 answer:

Snezhnost [94]3 years ago

6 0

Answer:

same

Explanation:

Acc. to Einstien's postulate of special theory of

Relativity , Velocity of the light beam is same in all frames of references

(a) If the freight car is at rest

The frame we can assumed as Non - inertial frame of reference s

In the inertial frame of reference , velocity of the light beam has its own value as : 3 x 10^8 m/s

(b) If the freight car is moving , the frame we can assumed as Non -inertial frame of reference

In thus case also , The velocity of the light beam will also have the same value as ; 3 x 108 m/s

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A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3

Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,

$\text { Electric power }(p)=\frac{V^{2}}{R}$

$\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}$

$\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

For calculating number of turns

$\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

Given that,

$80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }$

$\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}$

$\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .$

We need to find the number of turns in the secondary winding $\left(N_{S}\right)$ to run the bulb at 120W $\left(P_{2}\right)$

Firstly find the secondary voltage in the transformer use, $\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

$\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}$

$V_{2}^{2}=\frac{120^{2} \times 120}{80}$

$V_{2}^{2}=\frac{1728000}{80}$

$V_{2}^{2}=21600$

$V_{2}=\sqrt{21600}$

$V_{2}=146.9 \mathrm{V}=V_{S}$

Now, finding the number of turns in secondary coil. Use, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

$\frac{30}{N_{S}}=\frac{65}{146.9}$

$N_{S}=\frac{30 \times 146.9}{65}$

$N_{S}=\frac{4407}{65}$ $N_{S}=67.8$

The number of turns in the secondary winding are 67.8 turns.

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3 years ago

A box weighing 12,000 N is parked on a 36° slope. What will be the component of the weight parallel to the plane that balances f

sweet-ann [11.9K]

Given :

A box weighing 12,000 N is parked on a 36° slope.

To Find :

What will be the component of the weight parallel to the plane that balances friction.

Solution :

The component of that will be parallel to the plane to balance friction is :

$F \ cos ( 90 - 36)^o\\\\F\ ( sin\ 36^o)$

Therefore, component of force to balance friction is F sin 36° .

Hence, this is the required solution.

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A 1000 kg elevator is rising and its speed is increasing with an acceleration of 3 m/s^2. What is the resulting tension in the v

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Answer:

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Explanation:

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Determine the kinetic of 600 kg roller coaster car that is moving with the speed of 35.2 m/s.

antiseptic1488 [7]

The answer is 10,560 Joules or 1.1*10^4

Explanation:

Step 1: Calculate
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Kinetic energy=.5 times Mass times Velocity²
KE=.5*m*v²

so we plug in our numbers
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This works out to be 10,560 Joules or 1.1*10^4

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Pie

Answer:

b

Explanation:

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3 years ago

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