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3 years ago

5

Plz di all plz i will give brainest and thanks to best answer do it right

1 answer:

Inga [223]3 years ago

7 0

Answer:

area 4

Explanation:

area 4 has low pressure

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A) A spaceship passes you at a speed of 0.800c. You measure its length to be 31.2 m .How long would it be when at rest?

rosijanka [135]

Answer:

a

$l_o =52 \ m$

b

$l = 37.13 \ LY$

Explanation:

From the question we are told that

The speed of the spaceship is $v = 0.800c$

Here c is the speed of light with value $c = 3.0*10^{8} \ m/s$

The length is $l = 31.2 \ m$

The distance of the star for earth is $d = 145 \ light \ years$

The speed is $v_s = 2.90 *10^{8}$

Generally the from the length contraction equation we have that

$l = l_o \sqrt{1 -[\frac{v}{c } ]}$

Now the when at rest the length is $l_o$

So

$l_o =\frac{l}{\sqrt{ 1 - \frac{v^2}{c^2 } } }$

$l_o =\frac{ 31.2 }{ \sqrt{1 - \frac{(0.800c ) ^2}{c^2} } }$

$l_o=52 \ m$

Considering b

Applying above equation

$l =l_o \sqrt{1 - [\frac{v}{c } ]}$

Here $l_o =145 \ LY(light \ years )$

So

$l=145 * \sqrt{1 - \frac{v_s^2}{c^2 } }$

$l =145 * \sqrt{ 1 - \frac{2.9 *10^{8}}{3.0*10^{8}} }$

$l = 37.13 \ LY$

4 0

3 years ago

9. Pick the best example of Newton's Second Law in action.

madreJ [45]

Answer:

A rocket taking off from earth which pushes gasses in one direction and the rocket in

the other

7 0

3 years ago

Determine how many times per second it would move back and forth across a 6.0-m-long room on the average, assuming it made very

timofeeve [1]

Answer:

The right solution is "24.39 per sec".

Explanation:

According to the question,

⇒ $v=\frac{502.1}{\sqrt{3} }$

$=289.9 \ m/s$

The time will be:

⇒ $t=\frac{d}{v}$

$=\frac{2\times 6}{289.9}$

$=\frac{12}{289.9}$

$=0.041 \ sec$

hence,

⇒ $N=\frac{1}{t}$

$=\frac{1}{0.041}$

$=24.39 \ per \ sec$

4 0

3 years ago

The magnitude of a force vector is 89.6 newtons (N). The x component of this vector is directed along the +x axis and has a magn

insens350 [35]

Answer:

(a) θ = 33.86°

(b) Ay = 49.92 N

Explanation:

You have that the magnitude of a vector is A = 89.6 N

The x component of such a vector is Ax = 74.4 N

(a) To find the angle between the vector and the x axis you use the following formula for the calculation of the x component of a vector:

$A_x=Acos\theta$ (1)

Ax: x component of vector A

A: magnitude of vector A

θ: angle between vector A and the x axis

You solve the equation (1) for θ, by using the inverse of cosine function:

$\theta=cos^{-1}(\frac{A_x}{A})=cos{-1}(\frac{74.4N}{89.6N})\\\\\theta=33.86\°$

the angle between the A vector and the x axis is 33.86°

(b) The y component of the vector is given by:

$A_y=Asin\theta\\\\A_y=(89.6N)sin(33.86\°)=49.92N$

the y comonent of the vecor is Ay = 49.92 N

3 0

3 years ago

When a +0.00235 C charge

irakobra [83]

Answer:177.87

Explanation:

4 0

3 years ago