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3 years ago

A 0N B 6N C 10 N D 12 N

Physics

1 answer:

umka21 [38]3 years ago

3 0

Answer:

The net force acting on the object is 0 N

Explanation:

Newton's Second Law of Forces

The net force acting on a body is proportional to the mass of the object and its acceleration.

The net force can be calculated as the sum of all the force vectors in each rectangular coordinate separately.

The image shows a free body diagram where four forces are acting: two in the vertical direction and two in the horizontal direction.

Note the forces in the vertical direction have the same magnitude and opposite directions, thus the net force is zero in that direction.

Since we are given the acceleration a =0, the net force is also 0, thus the horizontal forces should be in equilibrium.

The applied force of Fapp=10 N is compensated by the friction force whose value is, necessarily Fr=10 N in the opposite direction.

The net force acting on the object is 0 N

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3 years ago

If atmospheric pressure suddenly changes from 1.00 atm to 0.896 atm at 298 k, how much oxygen will be released from 3.30 l of wa

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At a temperature of 298 K, the Henry's law constant is 0.00130 M/atm for oxygen. The solubility of oxygen in water 1.00 atm would be calculated as follows:

S = (H) (Pgas) = 0.00130 M / atm x 0.21 atm = 0.000273 M

At 0.890 atm,
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If atmospheric pressure would suddenly change from 1.00 atm to 0.890 atm at the same temperature, the amount of oxygen that will be released from 3.30 L of water in an unsealed container would be as follows

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4 years ago

A cylinder of mass mm is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the

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Answer:

$y = \frac{-f +/- \sqrt{f^{2} +2kmg}}{k}$

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Let y₀ be the initial position of the cylinder when the spring is attached and y its position when it is momentarily at rest.From work-kinetic energy principles, The work done by the spring force + work done by friction + work done by gravity = kinetic energy change of the cylinder

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So ¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = ¹/₂m(v₁² - v₀²)

Since the cylinder starts at rest, v₀ = 0. Also, when it is momentarily at rest, v₁ = 0

¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = ¹/₂m(0² - 0²)

¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = 0

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¹/₂ky₀² + fy₀ - mgy₀ = ¹/₂ky² + fy - mgy

Let y₀ = 0, then the left hand side of the equation equals zero. So,

0 = ¹/₂ky² + fy - mgy

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