At a particular instant the magnitude of the gravitational force exerted by a planet on one of its moons is 7 × 1021 N. Collapse
1 answer:
Answer:
Fg = 4.2*10²² N
Explanation:
The gravitational force between any two masses, provided that can be approximated by point masses (comparing their diameters with the distance between them), obeys the Newton's Universal Law of Gravitation, which states that the force (always attractive) is proportional to the product of the masses and inversely proportional to the square of the distance between them (this as a consequence of our Universe being three-dimensional), as follows:
So, if one of the masses increases 6 times, the force between them will be directly 6 times larger, so the new magnitude of the force will be as follows:
Fg₂ = Fg₁*6 = 7*10²¹ N* 6 = 4.2*10²² N
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Answer:
The option (b) is correct.
Explanation:
The expression for the power in terms of work done is as follows;
Here, W is the work done, t is the time taken and P is the power.
According to the given problem, six identical resistors are connected in series with a battery. The number of joules per second supplied by the battery is then determined.
The expression for the equivalent resistance in the series combination is as follows;
R_{eq}=R_{1}+R_{2}......+R_{6}
Put R_{1},R_{2}......,R_{6}=R.
R_{eq}=R+R+R+R+R+R
R_{eq}=6R
It is given in the problem that a seventh resistor is added (in series).
R_{eq}=R_{1}+R_{2}......+R_{7}
Put R_{1},R_{2}......,R_{7}=R.
R_{eq}=R+R+R+R+R+R+R
R_{eq}=7R
The expression for the power in terms of voltage and resistance is as follows;
Here, R is the resistance.
From the above expression, it can be concluded that the power, the number of joules per second supplied by the battery is inversely proportional to the resistance. The equivalent resistance increases if the seventh resistance is connected with a battery.
If a seventh resistor is added (in series) the number of joules per second supplied by the battery decreases.
Therefore, the option (b) is correct.
Answer:
A) 31 kJ
B) 1.92 KJ
C) 40 , 2.48
Explanation:
weight of person ( m ) = 79 kg
height of jump ( h ) = 0.510 m
Compression of joint material ( d ) = 1.30 cm ≈ 0.013 m
A) calculate the force
Fd = mgh
F = mgh / d
W = mg
F(net) = W + F = mg ( 1 +
= 79 * 9.81 ( 1 + (0.51 / 0.013) )
= 774.99 ( 40.231 ) ≈ 31 KJ
B) calculate the force when the stopping distance = 0.345 m
d = 0.345 m
Fd = mgh hence F = mgh / d
F(net) = W + F = mg ( 1 +
= 79 * 9.81 ( 1 + (0.51 / 0.345) )
= 774.99 ( 2.478 ) = 1.92 KJ
C) Ratio of force in part a with weight of person
= 31000 / ( 79 * 9.81 ) = 31000 / 774.99 = 40
Ratio of force in part b with weight of person
= 1920 / 774.99 = 2.48