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krok68 [10]
2 years ago
12

a crate is being lifted into a truck. if it is moved with a 2470n force and 3650 j of work is done , then how far is the crate b

eing lifted?
Physics
1 answer:
SVEN [57.7K]2 years ago
4 0

Answer:

The crate was being lifted by a height of 1.48 meters.

Explanation:

In an attempt o move a crate;

Force applied = 2470 N

Work done by the force = 3650 J

We know that the work done is defined as the force used to move an object to a distance.

Given the Force used and the work done by that Force, we need to find out the distance the crate was lifted to.

Work done is defined as:

Work = Force*distance covered in the direction of the force

3650 = 2470*distance

distance = 3650/2470

distance = 1.48 meters

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Answer:

0.0059

Explanation:

According to the question the seismic activity density is given by

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Here,

Number of Earthquakes over a given time span = 424

The land area affected = 71300 mi²

So,

\text{Seismic activity density}=\frac{424}{71300}\\\Rightarrow \text{Seismic activity density}=0.0059

The seismic activity density is 0.0059

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1 year ago
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Identical cannon balls are fired with the same force, one each from four cannons having respective bore lengths of 1.0 meter, 2.
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To answer this question, we must bear in mind the following considerations that are mentioned in the statement:

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Then, the force on the ball increases as the barrel length increases and the impulse depends on the magnitude of the force, then, the cannon that will have the minimum impulse will be the 1 meter one.

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2 years ago
You have two identical pure silver ingots. You place one of them in a glass of water and observe it to sink to the bottom. You p
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Answer: a)

Explanation:

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If this force is larger than the weight of the object (that means that the fluid is denser than the solid), the object floats, which is the case for silver and mercury.

Instead, silver density is larger than water density, which explains why the pure silver ingot sinks.

Finally, as mercury is denser than water, we conclude that for a same object, the buoyant force in mercury is larger than in water (exactly 13.6 times greater).  

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2 years ago
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A 0.300kg glider is moving to the right on a frictionless, ­horizontal air track with a speed of 0.800m/s when it makes a head-o
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Answer:

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

0.010935 J

0.0858675 J

Explanation:

m_1 = Mass of first glider = 0.3 kg

m_2 = Mass of second glider = 0.15 kg

u_1 = Initial Velocity of first glider = 0.8 m/s

u_2 = Initial Velocity of second glider = 0 m/s

v_1 = Final Velocity of first glider

v_2 = Final Velocity of second glider

As momentum and Energy is conserved

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

{\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}

From the two equations we get

v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.3-0.15}{0.3+0.15}\times 0.8+\frac{2\times 0.15}{0.3+0.15}\times 0\\\Rightarrow v_1=0.27\ m/s

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.3}{0.3+0.15}\times 0.8+\frac{0.3-0.15}{0.3+0.15}\times 0\\\Rightarrow v_2=1.067\ m/s

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

Kinetic energy is given by

K=\frac{1}{2}m_1v_1^2\\\Rightarrow K=\frac{1}{2}0.3\times 0.27^2\\\Rightarrow K=0.010935\ J

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Final kinetic energy of second glider is 0.0858675 J

6 0
2 years ago
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