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2 years ago

For two objects in space with very different masses, the gravitational force causes:

1 answer:

7 0

The answer is B. The less massive object to orbit the more massive object.
Explanation: The more massive the object, the larger its gravitational pull.
Example: the moon orbits the earth because the moon’s mass is smaller than earth, earth orbits the sun because the earth’s mass is smaller than the sun..

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Suppose a single light bulb burns out. How do you think this will affect lights that are strung

Gnom [1K]

The other lights wouldn't be affected?

:/

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4 years ago

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3) Ęplain why muddy water is<br> Heterogeneous<br> mixture

nataly862011 [7]

Answer:

muddy water is a heterogeneous mixture, which is Suspension.

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3 years ago

On a horizontal frictionless surface a mass M is attached to two light elastic strings both having length l and both made of the

EastWind [94]

Answer:

ω = √(2T / (mL))

Explanation:

(a) Draw a free body diagram of the mass. There are two tension forces, one pulling down and left, the other pulling down and right.

The x-components of the tension forces cancel each other out, so the net force is in the y direction:

∑F = -2T sin θ, where θ is the angle from the horizontal.

For small angles, sin θ ≈ tan θ.

∑F = -2T tan θ

∑F = -2T (Δy / L)

(b) For a spring, the restoring force is F = -kx, and the frequency is ω = √(k/m). (This is derived by solving a second order differential equation.)

In this case, k = 2T/L, so the frequency is:

ω = √((2T/L) / m)

ω = √(2T / (mL))

6 0

3 years ago

Identify the characteristics of protists. Check all that apply.

romanna [79]

Answer:

There are no examples

Explanation:

3 1

3 years ago

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Four identical masses of 2.5 kg each are located at the corners of a square with 1.0-m sides. What is the net force on any one o

Ghella [55]

Answer:

F=8.0*10^{-10}N

Explanation:

See the attached file for the masses distributions

The force between two masses at distance r is expressed as

$F=\frac{Gm_{1}m_{2} }{r^{2} }\\ G=Gravitional constant \\$

since the masses are of the same value, the above formula can be reduce to

$F=\frac{Gm^{2}}{r^{2} }\\$

using vector notation,Let use consider the force on the lower left corner of the mass due to the upper left side of the mass is

$F_{12} =\frac{Gm^{2}}{r^{2} }j\\$

The force on the lower left corner of the mass due to the lower right side of the mass is

$F_{14} =\frac{Gm^{2}}{r^{2} }i\\$

The force on the lower left corner of the mass due to the upper right side of the mass is

$F_{13} =\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\$

The net force can be express as

$F=\frac{Gm^{2}}{r^{2} }j +\frac{Gm^{2}}{r^{2} }i +\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\\\F=Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}cos\alpha }]i + Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}sin\alpha }]j\\\alpha=45^{0}, G=6.67*10^{-11}Nmkg^{-2}$

if we insert values we arrive at

$F=6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}cos45 }]i + 6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}sin45}]j\\F=5.643*10^{-10}i+5.643*10^{-10}j$

if we solve for the magnitude, we arrive at

$F=5.643*10^{-10}i+5.643*10^{-10}j \\F=\sqrt{(5.643*10^{-10})^{2} +(5.643*10^{-10})}^{2} \\F=8.0*10^{-10}$

Hence the net force on one of the masses is

$F=8.0*10^{-10}N$

8 0

3 years ago