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s2008m [1.1K]
3 years ago
10

Why does the gravitational force between earth and moon predominate over electrical forces?

Physics
1 answer:
nadezda [96]3 years ago
4 0

Answer:

Because the Earth and the Moon are electrically neutral

Explanation:

The gravitational force between the Earth and the Moon is given by

F=G\frac{mM}{r^2}

where G is the gravitational constant, m is the mass of the Moon, M is the mass of the Earth, r is the distance between the Moon and the Earth. Since both the Earth and the Moon have large masses, m and M are big, so the gravitational force between the two objects is large.

On the contrary, the electrostatic force between the Earth and the Moon is given by

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

where k is the Coulomb's constant, q1 is the charge of the Moon, q2 is the charge of the Earth, r is the distance between the Moon and the Earth. However, big objects (and planets as well) are electrically neutral, which means that they have zero charge (because if they had charge, they would tend to attract charge of the opposite sign, becoming neutral again). Therefore, the values of q1 and q2 are approximately zero, so the electrostatic force is zero as well.

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C and d are the right answers
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Explanation:

Distance covered by the satellite in 24 hours

s=2πr

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Therefore speed of satellite,

v=

time taken

distance travelled

=

24×60×60

265464.58

=3.07 km s

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5 0
2 years ago
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Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.
Artist 52 [7]

Answer:

a) C.M =(\bar x, \bar y)=(0.767,0.7)m

b) (x_4,y_4)=(-1.917,-1.75)m

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}

\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}

Where M represent the sum of all the masses on the system.

And the center of mass C.M =(\bar x, \bar y)

Part a

m_1= 3 kg, m_2=5kg,m_3=7kg represent the masses.

(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5) represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m

C.M =(\bar x, \bar y)=(0.767,0.7)m

Part b

For this case we have an additional mass m_4=6kg and we know that the resulting new center of mass it at the origin C.M =(\bar x, \bar y)=(0,0)m and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

(x_4,y_4)=(-1.917,-1.75)m

5 0
3 years ago
a child riding a bicycle at 15 meters per second accelerates at -3,0 meters per second? for 4.0 seconds. What is the child's spe
Vika [28.1K]

Answer:

<h2> 27m/s</h2>

Explanation:

Given data

initital velocity u=15m/s

deceleration a=3m/s^2

time t= 4 seconds

final velocity v= ?

Applying the expression

v=u+at------1

substituting our data into the expression we have

v=15+3*4

v=15+12

v=27m/s

The velocity after 4 seconds is 27m/s

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Answer:

F=5.7×10⁻⁶

Explanation:

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My work is in the attachment, where I double checked the units too, comment with any questions.

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