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Nikolay [14]
3 years ago
15

Mary pushes a crate by applying force of 18 newtons. Unable to push it alone, she gets help from her friend, Anne. Together they

apply a force of 43 newtons, and the crate just starts moving. If the coefficient of static friction is 0.11, what is the value of the normal force?
Physics
2 answers:
dangina [55]3 years ago
8 0
Answer is 3.9 x 10^2
maw [93]3 years ago
7 0
 <span>This problem is relatively simple, in order to solve this problem the only formula you need to know is the formula for friction, which is: 

Ff = UsN 

where Us is the coefficient of static friction and N is the normal force. 

In order to get the crate moving you must first apply enough force to overcome the static friction: 

Fapplied = Ff 

Since Fapplied = 43 Newtons: 

Fapplied = Ff = 43 = UsN 

and it was given that Us = 0.11, so all you have to do is isolate N by dividing both sides by 0.11 

43/0.11 = N = 390.9 which is approximately 391 or C. 3.9x10^2</span>
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A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mas
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Answer:

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

Explanation:

The expression for conservation of the angular momentum (L) is

L_{i} = L_{f}  I_{i}\times\omega_{i} = I_{f}\times\omega_{f}

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I_{i}\ and \ \omega_{i} initial moment of inertia and angular velocity

I_{f}\ and \ \omega_{f} is the final moment of inertia and angular velocity

The expression of moment of inertia of the satellite (a solid sphere) is

I_{i} = \frac{2}{5}m_{s}r^{2}

Where m_{s} is the satellite mass

r is the  radus of the sphere

Substititute 1900kg for m and 4.6m for r

I_{i} = \frac{2}{5}m_{s}r^{2}\\\\ = \frac{2}{5}\times1900 kg\times (4.6 m)^{2} \\\\= 1.61 \cdot 10^{4} kgm^{2}

The final moment of inertia of the satellite about the centre of mass

I_{f} = I_{i} + 2\timesI_{x} \\\\= 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}m_{x}l^{2}

Where m_{x} is the antenna's mass and

I is the length of the antenna

I_{f} = 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}150.0 kg\times(6.6 m)^{2} \\\\= 2.05 \cdot 10^{4} kgm^{2}

So, the Final rotation rate of the satellite is:

I_{i}\times\omega_{i} = I_{f}\times\omega_{f} \\\\\omega_{f} = \frac{I_{i}\times\omega_{i}}{I_{f}} \\\\= \frac{1.61 \cdot 10^{4} kgm^{2}\times8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kgm^{2}} \\\\= 6.3 rev/s

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

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aalyn [17]

Answer:

Wg is positive and WT negative.

(Letters in options are all wrongly written).

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5 0
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Answer: 4.4 cm.

Explanation:

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Using Jurin's law;

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8.8 cm × r = h2 × 2r

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Divide both sides by 2r, we then have;

8.8cm r/ 2r = h2

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