A fluorine atom has 7 valence electrons in its second shell.

A motor has an internal resistance of 12.1 Ω. The motor is in a circuit with a current of

kotykmax [81]

Answer:

Explanation:

V = I * R

V = 4 * 12.1 = 48.4 v

7 0

2 years ago

Does the voltage of a battery affect the strength of an electromagnet?

swat32

I'm trying to make an electromagnet that's strength is constantly getting incremented by small amounts every second. I need to know, which would have a greater effect on the electromagnet's strength, amps or volts? (I know increasing the turns and/or density of the magnet wire will increase the strength, but I am looking for answers other than that particular one.)

7 0

2 years ago

Your new motorcycle weighs 2450 N.

Snowcat [4.5K]

Answer:

Mass can never be negative. Everything has mass. Just like how they ask you to find area under the graph in maths. If the area is in the 3rd and 4th quadrant, when calculated, you would get negative answer.However, area can not be negative because it is a place/ location. It's exactly the same as mass.

4 0

3 years ago

During which phase of the moon do neap tides occur?

Fynjy0 [20]

Answer:

First Quarter and Third Quarter.

Explanation:

Tides are formed as a consequence of the differentiation of gravity due to the Moon across to the Earth sphere.

Since gravity variates with the distance:

$F = G\frac{m1\cdot m2}{r^{2}}$ (1)

Where m1 and m2 are the masses of the two objects that are interacting and r is the distance between them.

For example, seeing the image below, point A is closer to the Moon than point b, and at the same time the center of mass of the Earth will feel more attracted to the Moon than point B. Therefore, that creates a tidal bulge in point A and point B.

When the Sun and the Moon are alight with respect to the Earth, then the Sun tidal force contributes to the tidal force of the Moon over the Earth. That makes the high tides even higher (spring tides).

However, when the Sun is not in the same line than the Moon (the Moon is at 90° with respect to the Sun), then the low tides are higher and the high tides are lower. That scenario is known as neap tides.

Therefore, that happens when the Moon is at First Quarter and Third Quarter.

4 0

3 years ago

Read 2 more answers

A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the ho

Lady_Fox [76]

Answer:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>

The ball rotates 6.78 revolutions.

Explanation:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>

At the bottom the ball has the following angular speed:

$\omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s$

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

$sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m$

To find the revolutions we need the time, which can be found using the following equation:

$v_{f} = v_{0} + at$

$t = \frac{v_{f} - v_{0}}{a}$ (1)

So first, we need to find the acceleration:

$v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L}$ (2)

By entering equation (2) into (1) we have:

$t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}}$

Since it starts from rest (v₀ = 0):

$t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s$

Finally, we can find the revolutions:

$\theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev$

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!

3 0

3 years ago