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andreyandreev [35.5K]
2 years ago
10

2 Jupiter orbits the sun in a nearly circular path with radius 7.8x10^11m. The orbital period of Jupiter is 12 years.

Physics
1 answer:
bazaltina [42]2 years ago
4 0

Mass of Jupiter=1.9×10

27

㎏=M

1

Mass of Sun=1.99×10

30

㎏=M

2

Mean distance of Jupiter from Sun=7.8×10

11

m=r

G=6.67×10

−11

N㎡㎏

−2

Gravitational Force, F=

r

2

GM

1

M

2

F=

(7.8×10

11

)

2

6.67×10

−11

×1.9×10

27

×1.99×10

30

F=4.16×10

23

N

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nydimaria [60]
J.J. Thompson is the scientist who recieved credit for discovering them.
7 0
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Indicate whether the statement is true or false. an astronaut weighs the same on earth as in space.
tino4ka555 [31]
Weight is different (but mass is the same)
6 0
2 years ago
Read 2 more answers
In an RC circuit, what fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for
4vir4ik [10]

Answer:

The  fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is  

      k  = 0.903

Explanation:

From the question we are told that

     The time  constant  \tau  =  3

The potential across the capacitor can be mathematically represented as

     V  =  V_o  (1 -  e^{- \tau})

Where V_o is the voltage of the capacitor when it is fully charged

    So   at  \tau  =  3

     V  =  V_o  (1 -  e^{- 3})

     V  =  0.950213 V_o

   Generally energy stored in a capacitor is mathematically represented as

             E = \frac{1}{2 } * C  * V ^2

In this equation the energy stored is directly proportional to the the square of the potential across the capacitor

Now  since capacitance is  constant  at  \tau  =  3

        The  energy stored can be evaluated at as

         V^2 =  (0.950213 V_o )^2

       V^2 =  0.903  V_o ^2

Hence the fraction of the energy stored in an initially uncharged capacitor is  

      k  = 0.903

4 0
3 years ago
When you and a friend move a couch to another room you exert a force of 75 N over 5m how much work will you do
sesenic [268]

The work done is 375 J

Explanation:

The work done by a force in moving an object is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and the displacement

In this problem,

F = 75 N is the force applied to the couch

d = 5 m is the displacement

Assuming the force applied to the couch is parallel to the motion, \theta=0

And so, the work done is

W=(75)(5)(cos 0)=375 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

6 0
3 years ago
Two resistors, R1=3.85 Ω and R2=6.47 Ω , are connected in series to a battery with an EMF of 24.0 V and negligible internal resi
Bond [772]

Answer:

(a) 2.33 A

(b) 15.075 V

Explanation:

From the question,

The total resistance (Rt) = R1+R2 = 3.85+6.47

R(t) = 10.32 ohms.

Applying ohm's law,

V = IR(t)..........equation 1

Where V = Emf of the battery, I = current flowing through the circuit, R(t) = combined resistance of both resistors.

Note: Since both resistors are connected in series, the current flowing through them is the same.

Therefore,

I = V/R(t)............. Equation 2

Given: V = 24 V, R(t) = 10.32 ohms

Substitute these values into equation 2

I = 24/10.32

I = 2.33 A.

Hence the current through R1 = 2.33 A.

V2 = IR2.............. Equation 3

V2 = 2.33(6.47)

V2 = 15.075 V

7 0
3 years ago
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