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gizmo_the_mogwai [7]
3 years ago
7

When the current through an electromagnet ceases, what generally happens to the magnetic field?

Physics
1 answer:
kati45 [8]3 years ago
3 0
When the current flow ceases, the magnetic flow also decreases
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A flat surface is in a uniform magnetic field. Given only the area of the surface and the magnetic flux through the surface, it
Tasya [4]

Answer:

Given the area A of a flat surface and the magnetic flux through the surface \Phi it is possible to calculate the magnitude \frac{\Phi}{A}=B\ cos \theta.

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux \Phi is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is (m^{2}). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with \theta being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

                                                     \Phi=B\ A\ cos\theta

We are told the values of \Phi and B, then we can calculate the magnitude

                                                      \frac{\Phi}{A}=B\ cos\theta

3 0
2 years ago
A tank of volume 0.25 ft is designed to contain 50 standard cubic feet of air. The temperature is 80° F.Calculate the pressure i
slavikrds [6]

Answer:

The inside Pressure of the tank is 4499.12 lb/ft^{2}

Solution:

As per the question:

Volume of tank, V = 0.25 ft^{3}

The capacity of tank, V' = 50ft^{3}

Temperature, T' = 80^{\circ}F = 299.8 K

Temperature, T = 59^{\circ}F = 288.2 K

Now, from the eqn:

PV = nRT                      (1)

Volume of the gas in the container is constant.

V = V'

Similarly,

P'V' = n'RT'                       (2)

Also,

The amount of gas is double of the first case in the cylinder then:

n' = 2n

\]frac{n'}{n} = 2

where

n and n' are the no. of moles

Now, from eqn (1) and (2):

\frac{PV}{P'V'} = \frac{nRT}{n'RT'}

P' = 2P\frac{T'}{T} = 2\times 2116\times \frac{299.8}{288.2} = 4499.12 lb/ft^{2}

7 0
3 years ago
When making maps of the large-scale universe, astronomers estimate distances to the vast majority of galaxies by using:
Vesnalui [34]

Answer:

<em>The comoving distance and the proper distance scale</em>

<em></em>

Explanation:

The comoving distance scale removes the effects of the expansion of the universe, which leaves us with a distance that does not change in time due to the expansion of space (since space is constantly expanding). The comoving distance and proper distance are defined to be equal at the present time; therefore, the ratio of proper distance to comoving distance now is 1. The scale factor is sometimes not equal to 1. The distance between masses in the universe may change due to other, local factors like the motion of a galaxy within a cluster.  Finally, we note that the expansion of the Universe results in the proper distance changing, but the comoving distance is unchanged by an expanding universe.

4 0
3 years ago
A fish tank is a cube of size L × L × L, where L = 1 m, filled with water. Find
Murljashka [212]

At the bottom of the tank :

P = ρgH

P = (1000 kg/m³)(10 m/s²)(1 m)

P = 10000 N/m²

F = P • A

F = (10000 N/m²)(1 m²)

F = 10000 N

At the side of the tank :

Pav = ½ρgH

Pav = ½(1000 kg/m³)(10 m/s²)(1 m)

Pav = 5000 N/m²

F = P • A

F = (5000 N/m²)(1 m²)

F = 5000 N

3 0
2 years ago
Which formula can be used to solve problems related to the first law of thermodynamics?
docker41 [41]

Answer: The formula used to solve the problems related to first law of thermodynamics is \Delta U=Q+W

Explanation:

First law of thermodynamics states that the total energy of the system remains conserved. Energy can neither be destroyed, nor be created but it can only be transformed into one form to another.

Its implication is any change in the internal energy will be either due to heat energy or work energy.

Mathematically,

\Delta U=Q+W

where, Q = heat energy

W = work energy

\Delta U = Change in internal energy

Sign convention for these energies:

For Q: Heat absorbed will be positive and heat released will be negative.

For W: Work done by the system is negative and work done on the system is positive.

For \Delta U: When negative, internal energy is decreasing and when positive, internal energy is increasing.

Hence, the formula used to solve the problems related to first law of thermodynamics is \Delta U=Q+W

8 0
3 years ago
Read 2 more answers
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