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3 years ago

Which of the following is an example of balanced forces?

Physics

2 answers:

Sergio [31]3 years ago

6 0

A seesaw remains stationary when two students of equal weight sit on the ends

Leya [2.2K]3 years ago

5 0

Answer:

the answer is c.

Explanation:

a balanced force is a constant motion, or when the object does not move. because the children have the same weight, they can manage the seesaw to stay still.

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Nikitich [7]

Answer:

Protons and Neutrons are found in the nucleus and the electrons are found on the outside of the nucleus

Explanation:

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7 0

3 years ago

A common physics demonstration is to drop a small magnet down a long, vertical aluminum pipe. Describe the motion of the magnet

Rzqust [24]

Answer and Explanation:

This experiment is known as Lenz's tube.

The Lenz tube is an experiment that shows how you can brake a magnetic dipole that goes down a tube that conducts electric current. The magnet, when falling, along with its magnetic field, will generate variations in the magnetic field flux within the tube. These variations create an emf induced according to Faraday's Law:

$\varepsilon =-\frac{d\phi_B}{dt}$

This emf induced on the surface of the tube generates a current within it according to Ohm's Law:

$V=IR$

This emf and current oppose the flux change, therefore a field will be produced in such a direction that the magnet is repelled from below and is attracted from above. The magnitude of the flux at the bottom of the magnet increases from the point of view of the tube, and at the top it decreases. Therefore, two "magnets" are generated under and above the dipole, which repel it below and attract above. Finally, the dipole feels a force in the opposite direction to the direction of fall, therefore it falls with less speed.

7 0

3 years ago

Read 2 more answers

PLEASE HURRY

VLD [36.1K]

Answer:

Wilhelm Conrad Roentgen (1845-1923)

Antoine Henri Becquerel (1852-1908)

Pierre (1859-1906) and Marie (1867-1934) Curie

Explanation:

Wilhelm Conrad Roentgen (1845-1923)

Contribution: Discovery of x-rays in 1901.

Antoine Henri Becquerel (1852-1908)

Contribution: He discovered that radioactivity is the separation of x-rays and document and the difference between two.

Pierre (1859-1906) and Marie (1867-1934) Curie

Contribution: She discovered Polonium and Radium in1911

5 0

3 years ago

The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0. 10. The mass of the object

lilavasa [31]

The force of friction is 78.4 N

What is force of friction?

Frictional force is the force created when two surfaces come into contact and slide against each other. it is given by the equation:

$F_{f} =$ μ * $F_{n}$

where,

$F_{f}$ = force of friction

μ = coefficient of friction

$F_{n}$ = normal force

Given,

μ $= 0.10, m = 8.0 kg$

We know,

$F_{n} = mg$

$F_{n} = 8.0 * 9.8\\\\ F_{n} = 78.4 N$

Subsituing the values in equation:

$F_{f} =$ μ * $F_{n}$

$F_{f} = 0.10 * 78.4 \\\\F_{f} = 7.84 N$

The force of friction is 78.4 N.

To know more about force of friction, check out:

brainly.com/question/24386803

#SPJ4

4 0

1 year ago

A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa

andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

$\omega=\frac{v}{r}$

where

$\omega$ is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s$

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

$\omega=\frac{v}{r}$

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s$

(c) 5550 m

The maximum playing time of the CD is

$t =74.0 min \cdot 60 s/min = 4,440 s$

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

$d=vt=(1.25 m/s)(4,440 s)=5,550 m$

(d) $-6.4\cdot 10^{-3} rad/s^2$

The angular acceleration of the CD is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 21.6 rad/s$ is the final angular speed (when the CD is scanned at the outermost part)

$\omega_i = 50.0 rad/s$ is the initial angular speed (when the CD is scanned at the innermost part)

$t=4440 s$ is the time elapsed

Substituting into the equation, we find

$\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2$

5 0

3 years ago

Read 2 more answers