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mart [117]
3 years ago
13

The gravitational force of a star on an orbiting planet 1 is F1. Planet 2, which is twice as massive as planet 1 and orbits at t

hree times larger distance from the star, experiences gravitational force F2.
What is the ratio F2F1? You can ignore the gravitational force between the two planets.
Physics
1 answer:
Andrej [43]3 years ago
8 0

Answer:

ratio = 1 : 4.5

Explanation:

If m₁ is the mass of the star and m₂ the mass of the planet, the force of gravity F₁ for planet 1 is given by:

F_1=\frac{Gm_1m_2}{r^2}

The force F₂:

F_2=\frac{Gm_1(2m_2)}{(3r)^2}

The ratio:

\frac{F_2}{F_1}=\frac{2}{9}

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A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi
nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

W = F * d

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

d=h*N, for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

V_f^2-V_i^2 = 2a\Delta X

Where,

V_f = Final velocity

V_i = Initial Velocity

a = Acceleration

\Delta X = Displacement

PART A) For the particular case of work we know then that,

W = F*d

W = m*g*(h*N)

W = 50*9.8*(0.3*30)

W = 4.41kJ

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

V_f^2-V_i^2 = 2a\Delta X

Here,

V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow3 steps in one second

v_f = 0

Replacing,

V_f^2-V_i^2 = 2a\Delta X

0-0.9^2=2a(30*0.3)

Re-arrange for a,

a = -\frac{0.9^2}{2*30*0.3}

a = -45*10^{-3}m/s^2

At this point we can calculate the time, which is,

t = \frac{\Delta V}{a}

t = \frac{0-0.9}{-45*10^{-3}}

t = 20s

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

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In the parallelogram shown, AE = t + 2, CE = 3t − 14, and DE = 2t + 8.
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The answer would be 48
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In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin
alexandr402 [8]

Answer:

a) v=0.999124c

b) E=7.566*10^{22}

c) E_a=760 times\ larger

Explanation:

From the question we are told that

Distance to Betelgeuse d_b=430ly

Mass of Rocket M_r=20000

Total Time in years traveled T_d=36years

Total energy used by the United States in the year 2000 E_{2000}=1.0*10^20

Generally the equation of speed of rocket v mathematically given by

v=\frac{2d}{\triangle t}

v=860ly/ \triangle t

where

\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}

\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}

\triangle t=\sqrt{(860)^2+(36)^2}

\triangle t=860.7532

Therefore

v=\frac{860ly}{ 860.7532}

v=0.999124c

b)

Generally the equation of the energy E required to attain prior speed mathematically given by

E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2

E=7.566*10^{22}

c)Generally the equation of the energy E_a required to accelerate the rocket mathematically given by

E_a=\frac{E}{E_{2000}}

E_a=\frac{7.566*10^{22}}{1.0*10^{20}}

E_a=760 times\ larger

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3 years ago
A force, 10 N drags a mass 10 kg on a horizontal table with an acceleration of 0.2
kifflom [539]

Answer:

20&£+)##&843&()-_££-()&_2+0&&-£_!)

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