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nata0808 [166]
3 years ago
14

A rectangular painting measures 1.0 m tall along the y' axis and 3.0 m wide along the

Physics
1 answer:
Gennadij [26K]3 years ago
8 0

Answer:

a) 1 m tall, 3 m wide

b) 1 m tall, 1.31 m wide

Explanation:

According to the captain of the spaceship, the dimensions of the picture is the same i.e 1.0 m tall along the y' axis and 3.0 m wide along the x' axis.

b) The dimensions of the picture as seen by an observer on the Earth along the y axis will remain the same, 1.0 m tall, for the direction of the y axis is perpendicular to the spaceship movement.

The dimensions of the picture as seen by an observer on the Earth along the x axis will reduce if we are to go by the Lorentz contraction:

L(x) = L(x)' * √[1 - (v²/c²)]

where

L(x)' = the dimensions of the picture along the x axis on the spaceship,

v² = the speed of the spaceship and c² = the speed of light in the vacuum.

On substituting, we have

L(x) = 3 * √[1 - (0.81c²/c²)]

L(x) = 1.31 m

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Answer:

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Explanation:

As we know that if the object is placed on the inclined plane then the force of friction on the object is counterbalanced by the component of the weight of the object along the inclined plane.

So we can say

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As we know that the limiting value or the maximum value of friction force at the static condition is given by

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F_f = \mu (mg cos\theta)

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3 0
3 years ago
Help
Blizzard [7]
We can approach this in another way.
We know that sin(∅) = height / hypotenuse.

Thus, for x, height is 1 and hypotenuse is 3. Using Pythagoras theorem,
3² = 1² + b²
b = √8
cos(x) = b/hypotenuse
cos(x) = √8 / 3

Now, lets consider y:
sec(y) = 1 / cos(y) = 1 / base / hypotenuse = hypotenuse / base
The hypotenuse is 25 and the base is 24. We again apply Pythagoras theorem to find the third side, which works out to be:
height = 7
sin(y) = height / hypotenuse
sin(y) = 7/25

Now, sin(x + y) =
sin(x)cos(y) + sin(y)cos(x)
= (1/3)(24/25) + (√8 / 3)(7/25)
= 8/25 + 7√8/75
= (24 + 14√2) / 75
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3 years ago
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A plane starting from rest accelerates at 3m/s2 for 25s. Calculate the increase in velocity after:
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Time not needed now

Case-1:-

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\\ \rm\hookrightarrow v=0+3(1)

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\\ \rm\hookrightarrow v=0+3(3)

\\ \rm\hookrightarrow v=9m/s

Case-3:-

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\\ \rm\hookrightarrow v=0+3(25)

\\ \rm\hookrightarrow v=75m/s

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Answer:

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Answer:

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