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nataly862011 [7]
3 years ago
10

A metal bar, pivoted at one end, oscillates freely in the absence of a magnetic field. But when it oscillates between the poles

of a magnet, its oscillations are quickly damped. Why? (Such magnetic damping is used in a number of practical devices.)
Physics
1 answer:
miv72 [106K]3 years ago
8 0

Explanation:

When the metal bar oscillates between the poles of a magnet, it experience a change in the magnetic flux ( no of magnetic field lines passing through the metal bar) as it enter or leaves the magnetic field of the poles. As we know that the change in magnetic field induces the electric current in the metal bar (conductor). By considering the lenz law which states that the direction of induced current in the conductor will be such that as to oppose the initial magnetic field that is producing it. This opposing force acting on the metal bar will damp the oscillations of the bar between the poles of a magnet.

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the correct answers are d, e, & f

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What is the difference between a neutralization reaction and a double-displacement reaction? (1 point) Double-displacement react
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Answer:

Neutralization reactions always produce water and a salt.

Explanation:

There are different types of chemical reactions that occurs. Two of them are neutralization and double-displacement reaction. Neutralization reaction is the reaction between an acid and a base to give an acid salt and water as products. For example:

2NaOH (base) + H2SO4 (acid) → Na2SO4 (salt) + 2H2O (water)

On the other hand, Double displacement reaction is the reaction that involves an ion exchange between the reactants to form new products. This ionic exchange occurs in such a way that two reactants form two new products. For example:

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3 years ago
A rocket is launched with a velocity of 30 m/s at 40° to the horizontal. How far will it
steposvetlana [31]

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A. Rocket A will travel farther horizontally than rocket B.

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What is a tiltmeter?
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A car is being tested for safety by colliding it with a brick wall.The 1300 kg car is initially driving towards the wall at a sp
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Answer:

44200 N

Explanation:

To calculate the average force exerted on the car, we will use the following equation

\begin{gathered} Ft=\Delta p \\ Ft=m(v_f-v_i) \end{gathered}

Where F is the average force, t is the time, m is the mass, vf is the final velocity and vi is the initial velocity of the car.

Replacing t = 0.5s, m = 1300 kg, vf = -2 m/s, and vi = 15 m/s and solving for F, we get

\begin{gathered} F(0.5s)=(1300\text{ kg\rparen\lparen-2 m/s - 15 m/s\rparen} \\ F(0.5s)=(1300\text{ kg\rparen\lparen-17 m/s\rparen} \\ F(0.5s)=-22100\text{ kg m/s} \\ F=\frac{-22100\text{ kg m/s}}{0.5\text{ s}} \\ F=-44200\text{ N} \end{gathered}

Therefore, the average force exerted on the car by the wall was 44200 N

4 0
1 year ago
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