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prohojiy [21]
4 years ago
7

Moon effect. Some people believe that the Moon controls their activities. The Moon moves from being directly on the opposite sid

e of Earth from you to be being directly overhead. Assume that the Earth-Moon (center-to-center) distance is 3.82 108 m and Earth's radius is 6.37 106 m. (a) By what percent does the Moon's gravitational pull on you increase
Physics
1 answer:
Hatshy [7]4 years ago
4 0

Answer:

6.9%

Explanation:

We are given that

Distance between Earth and Moon=d=3.82\times 10^8 m

Radius of Earth=r=6.37\times 10^6 m

a.Initially gravitational pull

F=\frac{GmM}{(r+d)^2}.....(1)

After changing

Gravitational pull

F'=\frac{GmM}{(d-r)^2}.....(2)

Equation (2) divided by equation (1)

\frac{F'}{F}=\frac{(r+d)^2}{(d-r)^2}

\frac{F'}{F}=\frac{(3.82\times 10^8+6.37\times 10^6)^2}{(3.82\times 10^8-6.37\times 10^6)^2}=1.069

Increases in gravitational pull=\frac{F_2}{F_1}-1=1.069-1=0.069\times 100=6.9%

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Explanation:

It is given that,

Weight of the rock in air, W = 110 N

Since, W = mg

m=\dfrac{W}{g}

m=\dfrac{110\ N}{9.8\ m/s^2}

m = 11.22 kg

We need to find the apparent weight of the rock when it is submerged in water. Apparent weight is equal to the weight of liquid displaced i.e.

M=d\times V

d is the density of water, d=1000\ kg/m^3

V is the volume of rock, V=0.00337\ m^3

M=1000\ kg/m^3\times 0.00337\ m^3

M = 3.37 kg

The apparent weight in water, W = m - M

W=7.85\ kg\times 9.8\ m/s^2

W = 76.93 N

So, the apparent weight of the rock is 76.93 N. Hence, this is the required solution.

4 0
3 years ago
A mass on the end of a spring undergoes simple harmonic motion. At the instant when the mass is at its equilibrium position, wha
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Answer:

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2 years ago
Choose the +x-direction to point to the right. • Object 1 has a mass 1.66 kg and is moving to the right at 11.2 m/s. • Object 2
egoroff_w [7]

Answer:

M = 49.4kgm/s (towards the left)

Explanation:

Momentum is the product of mass and velocity of an object

Momentum = mass * velocity

Momentum of Object 1 with mass 1.66 kg moving to the right at 11.2 m/s, is expressed as:

M1 = 1.66 * 11.2

M1 = 18.592kgm/s

Momentum of Object 2 with mass 6.59 kg moving to the left at 10.4 m/s is expressed as:

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M = 18.592kgm/s--68.536kgm/s

M = -49.944kgm/s

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At which location could you place the north pole of a bar magnet so that it would be pushed away from the magnet shown?
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It’s a because b makes more since shown as the magnet
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3 years ago
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In this circuit the battery provides 3 V, the resistance R1 is 7 Ω, and R2 is 5 Ω. What is the current through resistor R2? Give
sveta [45]

Answer:

The current pass the R_2 is  I  = 0.25 A

Explanation:

The diagram for this question is shown on the first uploaded image  

From the question we are told that

    The voltage  is  V =  3V

     The first resistance is  R_1 = 7 \Omega

     The second resistance is  R_2 = 5 \Omega

Since the resistors are connected in series their equivalent resistance is  

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Substituting values

         R_{eq} = 7 + 5

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Since the resistance are connected in serie the current passing through the circuit  is the same current passing through R_2 which is mathematically evaluated as

        I  =  \frac{V}{R_{eq}}

Substituting values  

      I  =  \frac{3}{12}

      I  = 0.25 A

3 0
4 years ago
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