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3 years ago

As you increase voltage, what happens to the strength of an electromagnet?

Physics

2 answers:

Neporo4naja [7]3 years ago

6 0

As you increase voltage, the electromagnet would become stronger.

blsea [12.9K]3 years ago

5 0

When the electric current moves through a wire, it makes a magnetic field. ... The strength of an electromagnet can be increased by increasing the number of loops of wire around the iron core and by increasing the current or voltage.

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A spherical balloon is made from a material whose mass is 3.00 kg. The thickness of the material is negligible compared to the 1

vesna_86 [32]

To solve the problem it is necessary to apply the definition of Newton's second Law and the definition of density.

Density means the relationship between volume and mass:

$\rho = \frac{m}{V}$

While Newton's second law expresses that force is given by

F = ma

Where,

m = mass

a= acceleration (gravity at this case)

In the case of the given data we have to,

$m_b = 3Kg$

$r = 1.5m\\V = \frac{4}{3}\pi r^3 \\V = \frac{4}{3} \pi 1.5^3\\V = 14.13m^3$

In equilibrium, the entire system is equal to zero, therefore

$\sum F = 0$

$F_g +F_h-F_b = 0$

Where,

$F_g =$ Weight of balloon

$F_h =$ Weight of helium gas

$F_b =$ Bouyant force

Then we have,

$mg+V\rho g -V\rho_a g = 0$

$\rho = \rho_0-\frac{m}{V}$

Replacing the values we have that

$\rho = 1.19kg/m^3 -\frac{3Kg}{14.13m^3}$

$\rho = 0.978kg/m^3$

Now by ideal gas law we have that

$PV=nRT$

$P\frac{\rho}{m} = nRT$

$P = \rho \frac{n}{m}RT$

But the relation \frac{n}{m} is equal to the inverse of molar mass, that is

$P = \frac{\rho}{M_0} RT$

$P = \frac{0.978kg/m^3}{0.04kg/mol}*8.314J/K.Mol * 305K$

$P = 619995.7Pa$

Therefore the pressure of the helium gas assuming it is ideal is 0.61Mpa

5 0

3 years ago

What is field in physics

Troyanec [42]

Where everything is effected by a force

6 0

3 years ago

A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 μs. (a) Ho

Crank

Answer:

a)N = 3.125 * 10¹¹

b) I(avg) = 2.5 × 10⁻⁵A

c)P(avg) = 1250W

d)P = 2.5 × 10⁷W

Explanation:

Given that,

pulse current is 0.50 A

duration of pulse Δt = 0.1 × 10⁻⁶s

a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles

N = Δq/e

charge is given by Δq = IΔt

so,

N = IΔt / e

$N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1$

N = 3.125 * 10¹¹

b) Q = nqt

where q is the charge of 1puse

n = number of pulse

the average current is given as I(avg) = Q/t

I(avg) = nq

I(avg) = nIΔt

= (500)(0.5)(0.1 × 10⁻⁶)

= 2.5 × 10⁻⁵A

C) If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,

eV = K

V = K/e

the power is given by

P = IV

P(avg) = I(avg)K / e

$P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}$

= 1250W

d) Final peak=

P= Ik/e

= $= P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W$

P = 2.5 × 10⁷W

5 0

3 years ago

Read 2 more answers

Use the formula h = −16t2 + v0t. (if an answer does not exist, enter dne.) a ball is thrown straight upward at an initial speed

makkiz [27]

Using the given formula with v0=56 ft/s and h=40 ft
h = -16t2 + v0t
40 = -16t2 + 56t
16t2 - 56t + 40 = 0
Solving the quadratic equation:
t= (-b+/-(b^2-4ac)^1/2)/2a = (56+/-((-56)^2-4*16*40)^1/2)/2*16 = (56 +/- 24) / 32
We have two possible solutions
t1 = (56+24)/32 = 2.5
t2 = (56-24)/32 = 1
So initially the ball reach a height of 40 ft in 1 second.

3 0

4 years ago

Question 4

zmey [24]

I think the answer is B but not too sure.

8 0

3 years ago