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Grace [21]
2 years ago
6

You have just landed on Planet X. You take out a ball of mass 101 g , release it from rest from a height of 16.0 m and measure t

hat it takes a time of 2.91 s to reach the ground. You can ignore any force on the ball from the atmosphere of the planet. How much does the ball weigh on the surface of Planet X?
Physics
1 answer:
IRINA_888 [86]2 years ago
8 0

Answer:

0.3817 N

Explanation:

Remark

One thing is certain: the ball has a mass of 101 grams wherever it is in the universe. That is not true of the force. The force on the moon is a whole lot less than it is on earth, and maybe planet x as well.

Givens

m = 101 g

vi = 0       That's what at rest means.

t = 2.91 s

d = 16 m

F= ?

Formulas

d = vi*t + 1/2*a * t^2

Force = m * a

Solution

16 = 0 + 1/2 a * 2.91^2

16 = 4.234 a                       Divide by 4.234

16/4.234 = a

a = 3.779

F = m * a

a = 3.779

m = 101 g = 1 kg / 1000 grams

m = 0.101 kg

F = 0.101 * 3.779

F = 0.3817N

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A 10.2-kg mass is located at the origin, and a 4.6-kg mass is located at x = 8.1 cm. Assuming g is constant, what is the locatio
goldfiish [28.3K]

Answer:

center of mass of the two masses will lie at x = 2.52 cm

center of gravity of the two masses will lie at x = 2.52 cm

So center of mass is same as center of gravity because value of gravity is constant here

Explanation:

Position of centre of mass is given as

r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}

here we have

m_1 = 10.2 kg

m_2 = 4.6 kg

r_1 = (0, 0)

r_2 = (8.1cm, 0)

now we have

r_{cm} = \frac{10.2 (0,0) + 4.6 (8.1 , 0)}{10.2 + 4.6}

r_{cm} = {(37.26, 0)}{14.8}

r_{cm} = (2.52 cm, 0)

so center of mass of the two masses will lie at x = 2.52 cm

now for center of gravity we can use

r_g_{cm} = \frac{m_1gr_1 + m_2gr_2}{m_1g + m_2g}

here we have

m_1 = 10.2 kg

m_2 = 4.6 kg

r_1 = (0, 0)

r_2 = (8.1cm, 0)

now we have

r_g_{cm} = \frac{10.2(9.8) (0,0) + 4.6(9.8) (8.1 , 0)}{10.2(9.8) + 4.6(9.8)}

r_g_{cm} = {(37.26, 0)}{14.8}

r_g_{cm} = (2.52 cm, 0)

So center of mass is same as center of gravity because value of gravity is constant here

3 0
3 years ago
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