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AlexFokin [52]
3 years ago
8

Future space rockets might propel themselves by firing laser beams, rather than exhaust gases, out the back. The acceleration wo

uld be small, but it could continue for months or years in the vacuum of space. Consider a 1200 kg uncrewed space probe powered by a 15 MW laser. After one year, how far will it have traveled and how fast will it be going
Physics
1 answer:
Nina [5.8K]3 years ago
8 0

Answer:

Acceleration = 0.0282 m/s^2

Distance = 13.98 * 10^12 m

Explanation:

we will apply the energy theorem

work done = ΔK.E ( change in Kinetic energy )  ---- ( 1 )

<em>where :</em>

work done = p * t

                  = 15 * 10^6 watts * ( 1 year ) = 473040000 * 10^6 J

( note : convert 1 year to seconds )

and ΔK.E = 1/2 mVf^2   given ; m = 1200 kg  and initial V = 0

<u>back to equation 1 </u>

473040000 * 10^6  = 1/2 mv^2

Vf^2 = 2(473040000 * 10^6 ) / 1200

∴ Vf = 887918.92 m/s

<u>i) Determine how fast the rocket is ( acceleration of the rocket )</u>

a = Vf / t

  = 887918.92 / ( 1 year )

  = 0.0282 m/s^2

<u>ii) determine distance travelled by rocket </u>

Vf^2 - Vi^2 = 2as

Vi = 0

hence ; Vf^2 = 2as

s ( distance ) =  Vf^2 / ( 2a )

                     = ( 887918.92 )^2 / ( 2 * 0.0282 )

                    = 13.98 * 10^12 m

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) A satellite of mass m has an orbital period T when it is in a circular orbit of radius R around the earth. If the satellite in
Mrrafil [7]

Answer:

A) T.

Explanation:

Kepler's third law states that the orbital period (T) of a satellite is related with the radius (R) and the mass of the object (M) it orbits:

T=\frac{2\pi R^{\frac{3}{2}}}{\sqrt{GM}}  

So the orbital period is independent of the mass of the satellite, that means no matter the mass every satellite at a radius R around the earth have an orbital period A.

4 0
3 years ago
Write the following as powers of ten with one figure before the decimal point:
vodka [1.7K]
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5 0
2 years ago
Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant
irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

a_g = \frac{GM}{R^2}

Here

M = \text{Mass inside the Orbit of the star}

R = \text{Orbital radius}

G = \text{Universal Gravitational Constant}

Mass inside the orbit in terms of Volume and Density is

M =V \rho

Where,

V = Volume

\rho =Density

Now considering the volume of the star as a Sphere we have

V = \frac{4}{3} \pi R^3

Replacing at the previous equation we have,

M = (\frac{4}{3}\pi R^3)\rho

Now replacing the mass at the gravitational acceleration formula we have that

a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho

a_g = \frac{4}{3} G\pi R\rho

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration.  So centripetal acceleration of the star is

a_c = \frac{4}{3} G\pi R\rho

At the same time the general expression for the centripetal acceleration is

a_c = \frac{\Theta^2}{R}

Where \Theta is the orbital velocity

Using this expression in the left hand side of the equation we have that

\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2

\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}

\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R

Considering the constant values we have that

\Theta = \text{Constant} \times R

\Theta \propto R

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0
3 years ago
PLEASEEE HELPPP!!!!
Zigmanuir [339]

Answer: The work is 1863 N*m

Explanation:

We can define work as:

W = F*d

Where F is the force that the mover needs to apply to the refrigerator, and d is the distance that the refrigerator is moved.

To move the refrigerator, the minimal force that the mover needs to do is exactly the friction force (In this case, the refrigerator will move with constant speed).

Then we will have:

F = 230 N

and the distance is 8.1 meters, then the work will be:

W = 230N*8.1 m = 1863 N*m

3 0
2 years ago
Help a girl out ?! Plz ill give brainlest
shepuryov [24]

Answer:

I am pretty sure it is B.

Explanation:

I hope this helped if it didn't I am truly sorry

5 0
2 years ago
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