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3 years ago

What happens when the wire in an electromagnet is wrapped around a soft iron core?

1 answer:

4 0

The magnetic particles in a soft iron nail will line up with the magnetic field when the current is switched on. ... When you switch on the current, the coil becomes an electromagnet. But also, the soft iron core becomes a magnet. It will add to the strength of the electromagnet.

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A 500-kg elevator is pulled upward with a constant force of 6500 n for a distance of 50.0 m. what is the work done by the 6500 n

umka21 [38]

We know
work=force*distance=6500*50=?
plz tell me the result !
and feel free to ask if u hv any more confusion

3 0

3 years ago

What is Electrostatics

Anna [14]

Answer: Electrostatics is a branch of physics that studies electric charges at rest.

5 0

3 years ago

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1. Two identical bowling balls of mass M and radius R roll side by side at speed v0 along a flat surface. Ball 1 encounters a ra

UNO [17]

Answer:

1/2

Explanation:

We need to make a couple of considerations but basically the problem is solved through the conservation of energy.

I attached a diagram for the two surfaces and begin to make the necessary considerations.

Rough Surface,

We know that force is equal to,

$F_r = mgsin\theta$

$F_r = \mu N$

$F_r = \mu mg cos\theta$

Matching the two equation we have,

$\mu N = \mu mg cos\theta$

$\mu = tan\theta$

Applying energy conservation,

$\frac{1}{2}mv^2_0+\frac{1}{2}I_w^2 = F_r*d+mgh_1$

$\frac{1}{2}mv^2_0+\frac{2}{5}mR^2\frac{V_0^2}{R^2} = F_r*d+mgh_1$

$\frac{1}{2}mv^2_0+\frac{mv_0^2}{5} = mgsin\theta \frac{h_1sin\theta}+mgh_1$

$\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_1+gh_1$

$h_1 = \frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})$

Frictionless surface

$\frac{1}{2}mv_0^2+\frac{1}{2}I\omega^2 = mgh_2$

$\frac{1}{2}m_v^2+\frac{1}{2}\frac{2}{5}mR^2\frac{v_0^2}{R^2} =mgh_2$

$\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_2$

$h_2 = \frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})$

Given the description we apply energy conservation taking into account the inertia of a sphere. Then the relation between $h_1$ and $h_2$ is given by

$\frac{h_1}{h_2} = \frac{\frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})}{\frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})}$

$\frac{h_1}{h_2} = \frac{1}{2}$

8 0

3 years ago

The components of vector A are Ax = +2.2 and Ay = -6.9 , and the components of vector B are given are Bx =

Oksi-84 [34.3K]

For simplicity, let's call vector B-A vector C Then C is
Cx = (-6.1 - 2.2)
Cy = (-2.2 - (-6.9)) Or,
Cx = -8.3 Cy = 4.7
The magnitude is found with the Pythagorean theorem
||C|| = √(-8.3² + 4.7²) = <span>9.538</span>

6 0

3 years ago

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A pitcher throws a ball at 40 m/s. how much time does the ball take to reach the batter 18 m away

Maurinko [17]

18/40=.45
so, .45 of a second

4 0

4 years ago

Read 2 more answers