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3 years ago

a high speed train travels with an average speed of 250 km/h. the train travels for 2 hrs. how far does the train travel

Physics

2 answers:

Afina-wow [57]3 years ago

7 0

Answer:

The answer would be 454 miles.

Explanation:

You multiply the average speed by time travelled. Hope this helps!

steposvetlana [31]3 years ago

5 0

Answer:

<h3>The answer is 500 km </h3>

Explanation:

The distance covered by an object given it's velocity and time taken can be found by using the formula

<h3>distance = average velocity × time</h3>

From the question

average speed = 250 km/h

time = 2 hrs

We have

distance = 250 × 2

We have the final answer as

Hope this helps you

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The answer is <span>50 km/hr. </span>

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3 years ago

The potential (relative to infinity) at the midpoint of a square is 3.0 V when a point charge of +Q is located at one of the cor

Sveta_85 [38]

Answer:

d) 12 V

Explanation:

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We can apply the superposition principle (as the potential is linear with the charge) and calculating the total potential due to the 4 charges, just adding the potential due to any of them:

V = V(Q₁) + V(Q₂) +V(Q₃) + V(Q₄) = 4* 3.0 V = 12. 0 V

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3 years ago

A 6 kg block is released from rest at the top of an incline, as shown above, and slides to the bottom. The incline is 1.97 m lon

viva [34]

The net force on the block acting perpendicular to the incline is

∑ <em>F</em> = <em>n</em> - <em>w</em> cos(29.4°) = 0

where <em>n</em> is the magnitude of the normal force and <em>w</em> = <em>m g</em> is the weight of the block.

The equation itself comes from splitting up the forces acting on the block into components pointing parallel or perpendicular to the incline. The only forces acting on the block in the perpendicular direction are the normal force and the perpendicular component of the block's weight.

Solve for <em>n</em> :

<em>n</em> = <em>m g</em> cos(29.4°)

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<em>n</em> ≈ 51.2 N

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3 years ago

Two loudspeakers, A and B are driven by the same amplifier

DedPeter [7]

<h3><u>Complete Question </u></h3>

Q)Two loudspeakers, A and B, are driven by the same amplifier as shown and emit sinusoidal waves in phase. Speaker B is 2.00 m to the right of speaker A. Consider point Q along the extension of the line connecting the speakers, 1.00 m to the right of speaker B. Both speakers emit sound waves that travel directly from the speaker to point Q. a) What is the lowest frequency for which constructive interference occurs at point Q? b) What is the lowest frequency for which destructive interference occurs at point Q? The speed of sound wave is 344 m/s.

Answer:

lowest frequency is 172 Hz. for n = 1

lowest frequency is 86 Hz for n = 0

Explanation:

(a) Constructive interference takes place when the path difference is n where n=1,2,3 ...

frequency f = v /λn

= v n / d

= n ( 344 / 2 )

= n ( 172 ) Hz.

lowest frequency is 172 Hz

b)Destructive interference takes place when the path difference is n/2 x where n=1,3,5 ...

λ = 4/2n+1

for n = 0

λ = 4m

f = 344/4

f = 86 Hz.

lowest frequency is 86 Hz

3 0

4 years ago