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3 years ago

What would we need to do to make an electromagnet strong enough to move cars and trains

Physics

1 answer:

SVEN [57.7K]3 years ago

5 0

Answer:

The combined magnetic force of the magnetized wire coil and iron bar makes an electromagnet very strong. In fact, electromagnets are the strongest magnets made. An electromagnet is stronger if there are more turns in the coil of wire or there is more current flowing through it.

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Which of the following represents an element?

Digiron [165]

H2 is the correct answer

7 0

3 years ago

n isolated charged soap bubble of radius R0=7.45 cmR0=7.45 cm is at a potential of V0=307.0 volts.V0=307.0 volts. If the bubble

Gnesinka [82]

Complete Question

An isolated charged soap bubble of radius R0 = 7.45 cm is at a potential of V0=307.0 volts. V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r

Answer:

The difference is $U_f -U_i = 16 *10^{-7} J$

Explanation:

From the question we are told that

The radius of the soap bubble is $R_o = 7.45 \ cm = \frac{7.45}{100} = 0.0745 \ m$

The potential of the soap bubble is $V_1 =307.0 V$

The new radius of the soap bubble is $R_1 = 0.19 * 7.45=1.4155\ cm = 0.014155 \ m$

The initial electric potential is mathematically represented as

$U_i = \frac{V_1^2 R_o }{2k }$

The final electric potential is mathematically represented as

$U_f = \frac{V_2^2 R_1 }{2k }$

The initial potential is mathematically represented as

$V_1 = \frac{kQ}{R_o}$

The final potential is mathematically represented as

$V_2 = \frac{kQ}{R_1}$

Now

$\frac{V_2}{V_1} = \frac{R_o}{R_1}$

substituting values

$\frac{V_2}{V_1} = \frac{7.45}{1.4155} = \frac{1}{0.19}$

=> $V_2 = \frac{V_1}{0.19}$

$U_f = \frac{V_1^2 R_2 }{2k * 0.19^2}$

Therefore

$U_f -U_i = \frac{V_1^2 R_2 }{2k * 0.19^2} - \frac{V_1^2 R_o }{2k }$

$U_f -U_i = \frac{V_1^2}{2k} [\frac{ R_1 }{ * 0.19^2} - R_o]$

where k is the coulomb's constant with value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$U_f -U_i = \frac{307^2}{9 * 10^{9}} [\frac{ 0.014155 }{ 0.19^2} - 0.0745]$

$U_f -U_i = 16 *10^{-7} J$

8 0

3 years ago

Three fire hoses are connected to a fire hydrant. Each hose has a radius of 0.017 m. Water enters the hydrant through an undergr

alexira [117]

Answer:

2.36 x 10^5 kg

Explanation:

radius of hose, r = 0.017 m

radius of underground pipe, R = 0.088 m

number of hoses, n = 3

velocity of water in underground pipe, V = 2.7 m/s

Let v is the velocity of water in each hose.

According to the equation of continuity

A x V = n x a x v

π R² x V = n x π x r² x v

0.088 x 0.088 x 2.7 = 3 x 0.017 x 0.017 x v

v = 24.12 m/s

(a) Amount of water poured onto a fire in one hour by all the three hoses

= n x a x v x density of water x time

= 3 x 3.14 x 0.017 x 0.017 x 24.12 x 1000 x 3600

= 2.36 x 10^5 kg

Thus, the amount of water poured onto the fire in one hour is 2.36 x 10^5 kg.

6 0

3 years ago

initially at rest. The bullet is traveling 300 m/s when it hits the wood block and sticks inside it. Now the bullet and the wood

madreJ [45]

Needs mass to be reasonably solvable, sorry

6 0

4 years ago

when a temparature of a coin is 75°C, the coin's diameter increases. if the original diameter of a coin is 1.8*10^-2 m and its c

andrey2020 [161]

Answer:

ΔD = 2.29 10⁻⁵ m

Explanation:

This is a problem of thermal expansion, if the temperature changes are not very large we can use the relation

ΔA = 2α A ΔT

the area is

A = π r² = π D² / 4

we substitute

ΔA = 2α π D² ΔT/4

as they do not indicate the initial temperature, we assume that ΔT = 75ºC

α = 1.7 10⁻⁵ ºC⁻¹

we calculate

ΔA = 2 1.7 10⁻⁵ pi (1.8 10⁻²) ² 75/4

ΔA = 6.49 10⁻⁷ m²

by definition

ΔA = A_f- A₀

A_f = ΔA + A₀

A_f = 6.49 10⁻⁷ + π (1.8 10⁻²)² / 4

A_f = 6.49 10⁻⁷ + 2.544 10⁻⁴

A_f = 2,551 10⁻⁴ m²

the area is

A_f = π D_f² / 4

A_f = $\sqrt{4 A_f /\pi }$

D_f = $\sqrt{4 \ 2.551 10^{-4} /\pi }$

D_f = 1.80229 10⁻² m

the change in diameter is

ΔD = D_f - D₀

ΔD = (1.80229 - 1.8) 10⁻² m

ΔD = 0.00229 10⁻² m

ΔD = 2.29 10⁻⁵ m

5 0

3 years ago