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Vsevolod [243]
2 years ago
15

A box is being moved with a velocity v by a force p (parallel to v) along a level horizontal floor. The normal force is FN, the

kinetic firictional force is fk, and the weight of the box is mg. Decide which forces do positive, zero, or megative work. Provide a reason for each of your answers
Physics
1 answer:
Andrej [43]2 years ago
6 0
La neta no se no hablo inglés
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Which of the following statements is true? The square root of the molecular weight of a gas is equal to its rate of diffusion. T
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The diffusion coefficient of the gas is proportional to the average rate of thermal motion of the molecules.

the average velocity is inversely proportional to the square root of the molar mass

so
The gas diffusion rate is inversely proportional to the square root of its molecular weight.
8 0
3 years ago
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A rectangular conducting loop of wire is approximately half-way into a magnetic field B (out of the page) and is free to move. S
Black_prince [1.1K]

Answer:

. The loop is pushed to the right, away from the magnetic field

Explanation

This decrease in magnetic strength causes an opposing force that pushes the loop away from the field

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2 years ago
A physics book slides off a table at 1.25ms and hits the ground after 0.4s
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Answer:

Whats the question here?

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3 years ago
A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls
Mars2501 [29]

Answer:

Therefore % increase in velocity is 18.23 %

Explanation:

we use the equality of mass flow rate and the areas

m_1 = m_2\\p_1v_1 = p_2v_2\\p_1A_1v_1 = p_2A_2v_2\\v_2 = \frac{p_1}{p_2} v_1

The percentage increase in velocity is

Δ v% = \frac{v_2 - v_1}{v_1} \\100%

= \frac{p_1}{p_2} v_1 - v_1.100%

= \frac{\frac{1.2}{1.015} - 1}{1} . 100%

= Therefore % increase in velocity is 18.23 %

5 0
2 years ago
In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr
spin [16.1K]

Answer:

a) v_p=9.35m/s

Explanation:

From the question we are told that:

Open cart of mass   M_o=50.0 kg

Speed of cart   V=5.00m/s

Mass of package   M_p=15.0kg

Speed of package at end of chute V_c=3.00m/s

Angle of inclination   \angle =37

Distance of chute from bottom of cart   d_x=4.00m

a)

Generally the equation for work energy theory is mathematically given by

  \frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2

Therefore

  \frac{1}{2}u^2+gh=\frac{1}{2}v_p^2

  v_p=\sqrt{2(\frac{1}{2}u^2+gh)}

  v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}

  v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}

  v_p=9.35m/s

4 0
2 years ago
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