What is an electromagnet

Ignoring air resistance and the little friction from the plastic tube, the magnet was a freely-falling object in each trial. If

horsena [70]

Answer:

emf will also be 10 times less as compared to when it has fallen $40 \mathrm{m}$

Explanation:

We know, from faraday's law-

$e m f=-N \frac{\Delta \Phi}{\Delta T}$

and $\Phi=B . A$

So, as the height increases the velocity with which it will cross the ring will also increase. $(v=\sqrt{2 g h})$

Given

$\mathrm{V} 1(\text { Speed at } 40 \mathrm{m})=2 \mathrm{x} \mathrm{V} 2(\text { speed at } 10 \mathrm{m})$

$\sqrt{2 g h_{2}}=2 \times \sqrt{2 g h_{1}}=28.28 \mathrm{m} / \mathrm{s}$

Now, from $40 \mathrm{cm}$

$V_{3}=\sqrt{2 g h_{3}}=\sqrt{2 \times 10 \times 0.4}=2.82 \mathrm{m} / \mathrm{s}$

From equation a and b we see that velocity when dropped from $40 \mathrm{m}$ is 10 times greater when height is 40 $\mathrm{cm}$ so, emf will also be 10 times less as compared to when it has fallen $40 \mathrm{m}$

4 0

3 years ago

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0

3 years ago

Select the correct answer. Brian is repairing an old alarm clock. He needs to replace a device that converts the electric energy

Stella [2.4K]

it shouldbe a buzzer

3 0

3 years ago

Read 2 more answers

A mass hanging from a spring oscillates with a period of 0.35 s. Suppose the mass and spring are swung in a horizontal circle, w

Annette [7]

Answer:

66 rpm

Explanation:

The period of oscillation is given by

$T=2\pi \sqrt{\frac {m}{k}}$

$\frac {k}{m}=\frac {4\pi^{2}}{T^{2}}$ where T is time period of oscillation which is given as 0.35 s, k s spring constant and m is the mass of the object attached to the spring.

Also, net force is given by

Net force= $m\omega^{2} L$

$\omega=\sqrt{\frac {k\triangle L}{mL}}$ where $\triangle L$ is the elongation, L is original length, $\omega$ is the angular velocity

Substituting the equation of $\frac {k}{m}$ into the above we obtain

$\omega=\sqrt {\frac {4\pi^{2}\triangle L}{T^{2} L}}$

$\omega=\sqrt {4\pi^{2}\times 0.15L}{0.35^{2}\times L}}=6.952763\approx 6.95 rad/s$

$6.95\times\frac {60 s}{2\pi rad}\approx 66 rpm$

6 0

3 years ago

What is the net force for this? (please don't forget it's direction and magnitude)

avanturin [10]

Answer:

Explanation:

As the sum of the two right directed forces match exactly the left directed force, the only unbalanced force, and thus the net force, is the upward 25 N force.

3 0

2 years ago

What is an electromagnet ​

What is an electromagnet