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Oksi-84 [34.3K]
3 years ago
10

A container of rocks is brought back from the Moon's surface where the acceleration due to gravity is 162 meters per second if t

he total mass of the container of rocks is 650 kilograms, what is its weight on
Earth's surface
Physics
1 answer:
spin [16.1K]3 years ago
6 0

Answer:

6318 N

Explanation:

From the question given above, the following data were obtained:

Acceleration due to gravity of the moon (gₘ) = 1.62 m/s²

Mass (m) of container = 650 kg

Weight (W) of container on the earth =.?

Next, we shall determine the acceleration due to gravity of the earth. This can be obtained as follow:

Acceleration due to gravity of the moon (gₘ) = 1.62 m/s²

Acceleration due to gravity of the earth (gₑ) =.?

gₘ = 1/6 × gₑ

1.62 = 1/6 × gₑ

1.62 = gₑ /6

Cross multiply

gₑ = 1.62 × 6

gₑ = 9.72 m/s²

Finally, we shall determine the weight of the container on the earth as follow:

Mass (m) of container = 650 kg

Acceleration due to gravity of the earth (gₑ) = 9.72 m/s²

Weight (W) of container on the earth =.?

W = m × gₑ

W = 650 × 9.72

W = 6318 N

Therefore, the weight of the container on earth is 6318 N

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1.62 m/s²

Explanation:

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A speed skater just finished a race. After she crossed the finish line, she coasted to a complete stop. If her initial speed was
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2 years ago
8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s
allsm [11]

Answer:

a) \omega = 50\,\frac{rad}{s}, b) \omega = 0\,\frac{rad}{s}

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

\tau = F \cdot r

\tau = m\cdot a \cdot r

\tau = m \cdot \alpha \cdot r^{2}

Where \alpha is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}

\alpha = -3\,\frac{rad}{s^{2}}

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)

\omega = 50\,\frac{rad}{s}

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}

t = 66.667\,s

Since t > 66.667\,s, then the angular velocity is equal to zero. Therefore:

\omega = 0\,\frac{rad}{s}

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4 years ago
If a box is pushed across a floor with a force of 130N. The frictional force acting between the box and the floor is 30N, over a
Sidana [21]
The force acting in the front direction is the 130N.
The frictional force is acting backwards          30N.

1) The net force is 130N - 30N    =  100N

2)  s  = ut + (1/2)at^2             u = 0,  Start from rest,  s = 25m t =5.

25 = 0*5  +  (1/2)* a * 5^2.

25 = 0  +  25/2  * a.

25  =    (25/2)a.      Divide 25 from both sides.

1 =  (1/2)* a.          Cross multiply.

2 = a.

a = 2 m/s^2.

3) Mass of the box
Net Force,  F = ma
                   100 =  m*2.        Divide both sides by 2.
                    
                     100/2  =  m
                       50       =  m.
                        m  = 50 kg.

4)  Final velocity ,   v = u + at.
                                   v  =  0  + 2*5 = 10 m/s.
                                  
   Kinetic Energy,  K  =  (1/2) * mv^2.
                                    =    1/2  * 50 * 10 * 10.
                                    =      2500 J.
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