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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:

x=2.4365\ m

and

x=-1.4365\ m

Explanation:

Given:

  • first charge, q_1=5\times 10^{-3}\ C
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Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

E_1=E_2

  • since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}

\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}

3(r^2+1+2r)=5r^2

2r^2-6r-3=0

r=3.4365 \&\ r=-0.4365

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

x=-1+3.4365=2.4365\ m

and

x=-1-0.4365=-1.4365\ m

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