You push a coin across a table. The coin stops. How does this motion relate to balanced and unbalanced forces?

1 answer:

7 0

If you are pushing the coin across the table at a constant rate, the friction of the table and the horizontal force of your hand pushing are equal, and the coin itself moves at a constant rate. If you push a coin and let it go, there is no horizontal force keeping the coin going. Friction slows the coin to a stop. In both cases, the gravitational downward pull of Earth is equally but oppositely resisted by the upward push of table on the coin.

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A simple electric motor consists of a 220-turn coil, 4.4 cm in diameter, mounted between the poles of a magnet that produces a 9

Verdich [7]

Answer:

$M = 5.01\ A.m^2$

Explanation:

It is given that,

Number of turns in the coil, N = 220

Diameter of the coil, d = 4.4 cm

Radius of the coil, r = 2.2 cm = 0.022 m

Magnetic field produced by the poles of magnet, $B=96\ mT=96\times 10^{-3}\ T$

Current flowing in the coil, I = 15 A

Let M is the coil's magnetic dipole moment. Its formula is given by :

$M=N\times I\times A$

$M=220\times 15\times \pi (0.022)^2$

$M = 5.01\ A.m^2$

So, the coil's magnetic dipole moment is $5.01\ A.m^2$ . Hence, this is the required solution.

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Suppose two children push horizontally, but in exactly opposite directions, on a third child in a sled. The first child exerts a

anyanavicka [17]

Answer:

Please, read the anser below

Explanation:

1. In order to calculate the acceleration of the children you use the Newton second law for the summation of the implied forces:

$F_2-F_1-F_f=Ma$ (1)

Where is has been used that the motion is in the direction of the applied force by the second child

F2: force of the second child = 92N

F1: force of the first child = 79N

Ff: friction force = 5.5N

M: mass of the third child = 24kg

a: acceleration of the third child = ?

You solve the equation (1) for a, and you replace the values of the other parameters:

$a=\frac{F_2-F_1.F_f}{M}=\frac{96N-79N-5.5N}{24kg}=0.48\frac{m}{s^2}$

The acceleration is 0.48m/s^2

2. The system of interest is the same as before, the acceleration calculated is about the motion of the third child.

3. An image with the diagram forces is attached below.

4. If the friction would be 150N, the acceleration would be zero, because the friction force is higher than the higher force between children, which is 92N.

Then, the acceleration is zero