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sattari [20]
2 years ago
6

Assuming the acceleration due to gravity on the moon is exactly one-sixth of the acceleration due to gravity on Earth, what is t

he weight of the object on the moon?
Physics
1 answer:
uysha [10]2 years ago
5 0

<u>I have assumed a weight of 120 N on Earth.</u>

Answer:

<em>The object weighs 20 N on the moon</em>

Explanation:

Weight

The weight of an object depends on the mass m of the object and the acceleration of gravity g of the place they are in.

The formula to calculate the weight is:

W = m.g

If g_e is the acceleration of gravity on Earth, and g_m is the acceleration of gravity on the moon, we know:

g_m=1/6 g_e

Dividing by ge:

g_m/g_e = 1/6

An object of weight We=120 N on planet Earth has a mass of:

m = 120 / g_e

Multiplying by gm:

m.g_m=120 g_m/g_e

Substituting the ratio of accelerations of gravity:

m.g_m=120 * 1/6

Since m.gm is the weight on the Moon Wm:

W_m=20~N

The object weighs 20 N on the moon

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The acceleration due to gravity on the moon is about 1/6 of the acceleration due to gravity on the earth. A net force F acts hor
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Answer:

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An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 60 cm long and has a mass of 3.8 kg, with the center of
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Answer:

(a) τ = 26.58 Nm

(b) τ = 18.79 Nm

Explanation:

(a)

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

d₁ = perpendicular distance between ball and shoulder = 60 cm = 0.6 m

τ₁ = (29.4 N)(0.6 m)

τ₁ = 17.64 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of 60 cm = (0.4)(60 cm) = 24 cm = 0.24 m

τ₂ = (37.24 N)(0.24 m)

τ₂ = 8.94 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 17.64 Nm + 8.94 Nm

<u>τ = 26.58 Nm</u>

<u></u>

(b)

Now, the arm is at 45° below horizontal line.

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

42.42 cm = 0.4242 m

τ₁ = (29.4 N)(0.4242 m)

τ₁ = 12.47 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of (60 cm)(Cos 45°) = (0.4)(42.42 cm) = 16.96 cm = 0.1696 m

τ₂ = (37.24 N)(0.1696 m)

τ₂ = 6.32 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 12.47 Nm + 6.32 Nm

<u>τ = 18.79 Nm</u>

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