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tia_tia [17]
3 years ago
5

Pls Help

Physics
1 answer:
QveST [7]3 years ago
8 0

Answer:

Use the formula

g = GM/R^2

where

g = acceleration of gravity on moon

G = universal gravitational constant

= 6.67 x 10^-11 m^3/kg-s^2

M = mass of the moon

R = radius of the moon

solving for R, we get

R = 1.74 x 10^6 kg

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An ant is crawling up a leaf that is hangin vertically. it lifts 0:007 kg crumb up 0:12 meters. to do this, the ant exerts 0.069
Afina-wow [57]

Answer:

.000828 j

Explanation:

Work = F * d

          .0069 N * .12 m = .000828 j

5 0
2 years ago
Three batteries are connected in series so that the total voltage is 54 volts. The voltage of the first battery is twice the vol
Alex_Xolod [135]

Answer:

v_1 = 12 volts

v_2 = 6 volts

v_3 = 36 volts

Explanation:

As we know that all the batteries are in series

so the net voltage of all three batteries is given as

V = v_1 + v_2 + v_3

now we know that

v_1 = 2v_2

v_1 = \frac{1}{3}v_3

now plug in all the values in it

54 = v_1 + \frac{v_1}{2} + 3v_1

54 = 4.5 v_1

v_1 = 12 volts

now we have

v_2 = 6 volts

v_3 = 36 volts

3 0
3 years ago
Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its
ruslelena [56]

Answer:

a

The orbital speed is v= 2.6*10^{3} m/s

b

The escape velocity of the rocket is  v_e= 3.72 *10^3 m/s

Explanation:

Generally angular velocity is mathematically represented as

            w = \frac{2 \pi}{T}

Where T is the period which is given as 1.6 days = 1.6 *24 *60*60 = 138240 sec

       Substituting the value

         w = \frac{2 \pi}{138240}

             = 4.54*10^ {-5} rad /sec

At the point when the rocket is on a circular orbit  

   The gravitational force =  centripetal force and this can be mathematically represented as

              \frac{GMm}{r^2} = mr w^2

Where  G is the universal gravitational constant with a value  G = 6.67*10^{-11}

            M is the mass of the earth with a constant value of M = 5.98*10^{24}kg

            r is the distance between earth and circular orbit where the rocke is found

               Making r the subject

                     r = \sqrt[3]{\frac{GM}{w^2} }

                        = \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }

                        = 5.78 *10^7 m

The orbital speed is represented mathematically as

                   v=wr

Substituting value

                  v= (5.78*10^7)(4.54*10^{-5})

                     v= 2.6*10^{3} m/s    

The escape velocity is mathematically represented as

                            v_e = \sqrt{\frac{2GM}{r} }

Substituting values

                             = \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }

                             v_e= 3.72 *10^3 m/s

7 0
4 years ago
An atom of an element is electrically neutral because the
lora16 [44]

Answer:

Number of protons equals the number of neutrons

6 0
3 years ago
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It takes 50 seconds for Tyler to do 450 Joules of work. How much power did Tyler use?​
Bogdan [553]
Power=work/time
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Power =450/50=9watts
8 0
3 years ago
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