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Alenkasestr [34]
3 years ago
8

Ellen needs to move a heavy box across the floor and then place it on a shelf that is four feet above the floor. Which use of ma

chines would most likely require the least amount of force in order to get the box on the shelf?
sliding the box up an inclined plane to the shelf
rolling the box up an inclined plane on a cart
sliding the box across the floor and then lifting it using a pulley
rolling the box across the floor in a cart and then using a pulley
Physics
2 answers:
Naily [24]3 years ago
5 0

Option B is correct: Rolling the box up an inclined plane on a cart.

When using the inclined, Ellen needs to overcome frictional force and the weight component of the box along the ramp which will be less than the total weight of the heavy box. Now, in sliding and rolling the weight of the heavy box along the ramp will be same but the difference will come in the type of the frictional force. Since, the rolling friction is less than the sliding friction. Hence, less force will be required when the box is rolled up an inclined plane on a cart.

OleMash [197]3 years ago
3 0

The correct answer is:

B. Rolling the box up an inclined plane on a cart

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Answer:

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A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe
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(a) 2.75 rev/min

The moment of inertia of the rod rotating about its center is:

I_R=\frac{1}{12}ML^2

where

M=3.30\cdot 10^{-2} kg is its mass

L = 0.450 m is its length

Substituting,

I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2

The moment of inertia of the two rings at the beginning is

I_r = 2mr^2

where

m = 0.200 kg is the mass of each ring

r=5.20\cdot 10^{-2} m is their distance from the center of the rod

Substituting,

I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2

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I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2

The initial angular velocity of the system is

\omega_1 = 35.0 rev/min

The angular momentum must be conserved, so we can write:

L=I_1 \omega_1 = I_2 \omega_2 (1)

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r=\frac{0.450 m}{2}=0.225 m

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I_r=2(0.200)(0.225)^2=0.0203 kg m^2

and the total moment of inertia is

I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2

Substituting into (1), we find the final angular speed:

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min

(b) 103.0 rev/min

When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

I_2 = I_R = 5.57\cdot 10^{-4}kg m^2

So using again equation of conservation of the angular momentum:

L=I_1 \omega_1 = I_2 \omega_2

We find the new final angular speed:

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min

7 0
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Why are the alkali metals likely to react with group 17 elements?
Rudiy27

Answer:

Because alkali metals are so reactive, they are found in nature only in combination with other elements. They often combine with group 17 elements, which are very “eager” to gain an electron.

Explanation:

hope this helps you if it does please mark brainliest

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