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Alenkinab [10]
2 years ago
10

Two asteroids exert a gravitational force, F, on each other. Some time later, the asteroids are now three times as far from each

other as before. Which of the following represents the gravitational force at this distance? Select one:
a. F/9
b. F/6
c. F/3
d. F/2
Physics
1 answer:
dolphi86 [110]2 years ago
7 0

Answer:

a.F/9

Explanation:

Hello, I can help you wit this

the gravitational Force is described by the next equation

F=G*\frac{m_{1}*m_{2} }{r^{2} } \\

Where, G  is   the universal gravitational constant, M1 and M2 are the masses of the objects, and r is the distance between then.

Step 1

time 1

Let

m1= mass of the asteroid 1

m2= mass of the asteroid 2

distance 1=r1

put the values into the equation to find the force

F_{1}=G*\frac{m_{1}*m_{2} }{r^{2} }\\F_{1}=G*\frac{m_{1}*m_{2} }{r_{1}^{2} }

Step 2

time 2

r2=3r1

F_{2} =G*\frac{m_{1}*m_{2} }{(3*r_{1} )^{2} }\\

Step 3

compare the forces

F_{1}=G*\frac{m_{1}*m_{2} }{r^{2} }\\F_{2} =G*\frac{m_{1}*m_{2} }{9(r_{1}^{2}) }\\F_{2}=\frac{F_{1}}{9}

a.F/9

here it is observed that the force is inversely proportional to the square of the distance, at double distance, it is divided into 4, triple in 9, etc.

Have a great day.

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When operated on a household 110.0 V line, typical hair dryers draw about 1650 W of power. The current can be modeled as a long,
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Explanation:

Given that,

Voltage of household line, V = 110 V

Power of the hairdryer, P = 1650 W

During use, the current is about 1.95 cm from the user's hand.

(a) Power is given by :

P=V\times I\\\\I=\dfrac{P}{V}\\\\I=\dfrac{1650\ W}{110\ V}\\\\I=15\ A

(b) Again the power is given by :

P=\dfrac{V^2}{R}

R is resistance of the dryer

R=\dfrac{V^2}{P}\\\\R=\dfrac{(110)^2}{1650}\\\\R=7.34\ \Omega

(c) The magnetic field produced by the dryer at the user's hand is given by :

B=\dfrac{\mu_o I}{2\pi r}\\\\B=\dfrac{4\pi \times 10^{-7}\times 15}{2\pi \times 1.95\times 10^{-2}}\\\\B=1.53\times 10^{-4}\ T

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2 years ago
The density of atmosphere (measured in kilograms/meter3) on a certain planet is found to decrease as altitude increases (as meas
alexgriva [62]

Answer:

B.  inverse plot, 0.51 kilograms/meter3

Explanation:

First of all, we note that the relationship between the altitude and the atmospheric density is an inverse relationship. In fact, an inverse relationship is a relationship between the x-variable and the y-variable of the form

y \propto \frac{1}{x}

Therefore, as the x increases, the y decreases, and as the x decreases, they increases. This is exactly what occurs with the altitude and the atmospheric density in this plot: as the altitude increases, the density decreases, and vice-versa.

Moreover, we can infer the value of the atmospheric density at an altitude of 1,291 km. This point is located between point A (2550 km) and point B(1000 km), so the density must have a value between 0.30 kg/m^3 and 0.54 kg/m^3, so the correct choice is

B.  inverse plot, 0.51 kilograms/meter3


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2 years ago
An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e
White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}

with this value you can compute the frequency:

a)

f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz

b)

the mass of the block is given by the formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s

Finally, the amplitude is:

x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m

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See Khan Academy.

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Answer:

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Explanation:

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