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blsea [12.9K]
3 years ago
6

The weight of an astronout is 60kg on earth, find the wight of the same object on the planet where the gravitational attraction

has been reduced to 1/10 of the earths pull
Physics
1 answer:
Marianna [84]3 years ago
3 0

The astronaut's weight is not 60 kg anywhere, because kg is a unit of mass, not weight.

If the astronaut's mass is 60 kg, then his weight is (60 kg)x(acceleration of gravity).

That's 588 Newtons on Earth, and 58.8 Newtons on a planet with 1/10 Earth's gravity.

The astronaut's mass of 60 kg goes with her, and doesn't depend on where she is.

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The value of the second charge is 1.2 nC.

<h3>Electric potential</h3>

The work done in moving the charge from infinity to the given position is calculated as follows;

W = Eq₂

E = W/q₂

<h3>Magnitude of second charge</h3>

The magnitude of the second charge is determined by applying Coulomb's law.

E = \frac{kq_2}{r^2} \\\\\frac{kq_2}{r^2} = \frac{W}{q_2} \\\\kq_2^2 = Wr^2\\\\q_2^2 = \frac{Wr^2}{k} \\\\q_2 = \sqrt{\frac{Wr^2}{k} } \\\\q_2 =  \sqrt{\frac{(8.1 \times 10^{-6}) \times (0.04)^2}{9\times 10^9} } \\\\q_2 = 1.2 \times 10^{-9} \ C\\\\q_2 = 1.2 \ nC

Thus, the  value of the second charge is 1.2 nC.

Learn more about electric potential here: brainly.com/question/14306881

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An ideal spring hangs from the ceiling. A 1.95 kg mass is hung from the spring, stretching the spring a distance d=0.0865 m from
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kinetic energy = 0.1168 J

Explanation:

From Hooke's law, we know that ;

F = kx

k = F/x

We are given ;

Mass; m = 1.95 kg

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So, Force = mg = 1.95 × 9.81

k = 1.95 × 9.81/0.0865 = 221.15 N/m

Now, initial energy is;

E1 = mgL + ½k(x - L)²

Also, final energy; E2 = ½kx² + ½mv²

From conservation of energy, E1 = E2

Thus;

mgL + ½k(x - L)² = ½kx² + ½mv²

Making the kinetic energy ½mv² the subject, we have;

½mv² = mgL + ½k(x - L)² - ½kx²

We are given L=0.0325 m

Plugging other relevant values, we have ;

½mv² = (1.95 × 9.81 × 0.0325) + (½ × 221.15(0.0865 - 0.0325)² - ½(221.15 × 0.0865²)

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3 years ago
Three children are riding on the edge of a merry-go-round that is 100 kg, has a 1.60-m radius, and is spinning at 20.0 rpm. The
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Answer:

25.33 rpm

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(1/2 M + m1 + m2 + m3) x 2 x π x f = (1/2 M + m1 + m3) x 2 x π x f'

( 1/2 x 100 + 22 + 28 + 33) x 20 = (1/2 x 100 + 22 + 33) x f'

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