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Zanzabum
3 years ago
15

The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the

planet, along a line connecting the centers of planet and moon, the net force on an object will be zero. Natividad, Joshua - [email protected] @theexpertta - tracking id: 1F76-1A-38-41-94BC-19489. In accordance with Expert TA's Terms of Service. copying this information to any solutions sharing website is strictly forbidden. Doing so may result in termination of your Expert TA Account. show answer No Attempt 50% Part (a) Derive an expression for x.

Physics
1 answer:
seraphim [82]3 years ago
6 0

Answer:

x=\dfrac{D}{1+\sqrt{\dfrac{m}{M}}}

Explanation:

Mass of moon = m

Mass of planet =M

We know that gravitational force given as

F=G\dfrac{m_1m_2}{d^2}

F'=G\dfrac{m'M}{x^2}

F=G\dfrac{m'm}{(D-x)^2}

Given that force is zero so

F=F'

G\dfrac{m'm}{(D-x)^2}=G\dfrac{m'M}{x^2}

\dfrac{m}{(D-x)^2}=\dfrac{M}{x^2}

\dfrac{x}{D-x}=\sqrt{\dfrac{M}{m}}

\dfrac{D-x}{x}=\sqrt{\dfrac{m}{M}}

x=\dfrac{D}{1+\sqrt{\dfrac{m}{M}}}

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Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and ex
Ahat [919]

The question is incomplete. The complete question is :

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.

What is the frequency of the sound?

Solution :

Given :

The distance between the two loud speakers, d = 1.8 \ m

The speaker are in phase and so the path difference is zero constructive interference occurs.

At the point D, the speakers are out of phase and so the path difference is $=\frac{\lambda}{2}$

Therefore,

$AD-BD = \frac{\lambda}{2}

$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$

$\lambda = 2 \times 0.4985$

$\lambda = 0.99714 \ m$

Thus the frequency is :

$f=\frac{v}{\lambda}$

$f=\frac{340}{0.99714}$

f=340.9744 Hz

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2 years ago
PLSS HELP WILL GIVE POINTS OR WHATEVER JUST PLS HELP I WILL GIVE 98 POINTS
galina1969 [7]

Answer:

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Explanation:

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3 years ago
A swimmer is swimming to the left with a speed of 1.0 m/s when she starts to speed up with constant acceleration. The swimmer re
kolezko [41]

Answer:

Correct answer: t = 2.86 seconds

Explanation:

We first use this formula

V² - V₀² = 2 a d    

where V is the final velocity (speed), V₀ the initial velocity (speed),

a the acceleration and d the distance.

We will calculate the acceleration from this formula

a = (V² - V₀²) / (2 d) = (2.5² - 1²) / (2 · 5) = (6.25 - 1) / 10 = 5.25 / 10

a = 0.525 m/s²

then we use this formula

V = V₀ + a t  => t = (V - V₀) / a = (2.5 - 1) / 0.525 = 1.5 / 0.525 = 2.86 seconds

t = 2.86 seconds

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3 years ago
A tennis ball of mass 0.060 kg travels horizontally at a speed of 25m/s. The ball hits a tennis
stepladder [879]
It would be A because a is perfect
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Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibriu
Citrus2011 [14]

Answer:

0.247 μC

Explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

F_y:  T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N

T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N

T_x = T*sin(50) = 0.0234 N

The electric force is given by the expression:

F = k*\frac{q_1*q_2}{r^2}

In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m

And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.

F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}

q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C

O 0.247 μC

8 0
2 years ago
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