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3 years ago

How much will an object weigh on the moon if it weighs 60 lbs on the earth?

Physics

2 answers:

Xelga [282]3 years ago

8 0

Answer:

10lbs on the moon

Explanation:

timofeeve [1]3 years ago

7 0

Your weight on the Moon would only be 33 pounds. Why is your weight on the It's because of the lower gravity on the Moon. Objects on the surface of the Moon experience only 16.5% of the gravity they would experience on Earth.29-Oct-2008

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1. A 1.30 kg ball strikes a wall with a velocity of -10.5 m/s. The ball bounces off with a velocity of 6.50 m/s. If the ball is

bekas [8.4K]

Let F be the magnitude of the force. The impulse of this force while the ball is in contact with the wall is

Ft = F (0.0210 s)

and this impulse is equal to the change in the ball's momentum,

m ∆v = (1.30 kg) (6.50 m/s - (-10.5 m/s)) = (1.30 kg) (17.0 m/s)

Solve for F :

F (0.0210 s) = (1.30 kg) (17.0 m/s)

F = (1.30 kg) (17.0 m/s) / (0.0210 s)

F ≈ 1050 N

4 0

3 years ago

7. Nancy has a mass of 60 kg and sits on the very end of a 3.00 m long plank pivoted in the middle. How much torque must her co-

netineya [11]

Answer:

Torque = 882Nm

Explanation:

Torque = Mg×distance

But plank's is pivoted ,therefore distance=3/2=1.5m

Mass of Nancy=60jg

Acceleration due to gravity, g=9.8m/s^2

Torque= 60×9.8×1.5

Torque= 882Nm

6 0

3 years ago

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

FOR HYDROGEN:

v = √(3KT)/m = 2000 m/s _____ eqn (2)

FOR OXYGEN:

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

K.E of Oxygen = 4000 J

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

v = 2149.24 m/s

8 0

4 years ago

Calculate the angular velocity of a clocks second hand, its minute hand, and its hour hand. State in Rad/s. What is its angular

fenix001 [56]

Second hand:

1 rev per minute = (2π radians/minute) x (1 min/60sec) = π/30 rad/sec

Minute hand:

1 rev per hour = (2π radians/hour) x (1 hr/3600 sec) = π/1800 rad/sec

Hour hand:

1 rev per 12 hours = (2π rad/12 hr) x (1 hr/3600 sec) = π/21,600 rad/sec

As long as the clock is in good working order, and the hands are turning steadily at their normal rate, there is no angular acceleration.

8 0

3 years ago

Read the passage.

nevsk [136]

Answer:

d in my opinion is the most opitmal answer

Explanation:

aka a graph comparing distances traveled by objects thrown on Earth and the moon

7 0

3 years ago

Read 2 more answers