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3 years ago

What is magnetisim? What things are magnets?

1 answer:

5 0

A physical phenomenon produced by the motion of electric charge, resulting in attractive and repulsive forces between objects.

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A transformer has a primary coil with 106 turns and a secondary coil of 340 turns. The AC voltage across the primary coil has a

UkoKoshka [18]

To solve this problem it is necessary to apply the concepts related to transformers, that is to say passive electrical device that transfers electrical energy from one electrical circuit to one or more circuits.

From the mathematical definition we have that the relationship between the voltage of the first coil and the second coil is proportional to the number of loops of the first and second loop, that is:

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$

Where

$V_p =$ input voltage on the primary coil.

$V_s=$ input voltage on the secondary coil.

$N_p=$ number of turns of wire on the primary coil.

$N_s =$ number of turns of wire on the secondary coil.

Replacing our values we have:

$V_p = 128V$

$N_p = 106$

$N_s = 340$

Replacing,

$\frac{V_s}{128} = \frac{340}{106}$

$V_s = 410.56V$

From the same relations of number of turns and the voltage of the first and second coil we also have the relation of electricity and voltage whereby:

$V_s I_s = V_p I_p$

Where

$I_p$ = Current Primary Coil

$I_s$ = Current secundary Coil

Therefore:

$I_s = \frac{V_p I_p}{V_s}$

$I_s = \frac{(128)(6)}{410.56}$

$I_s = 1.87 A$

Therefore the maximum values for the secondary coil of the voltage is 410.56V and Current is 1.87A

5 0

3 years ago

Problem: The frequency of an FM radio station is 89.3 MHz. Calculate its period. Part B: From the Library, select the general eq

vekshin1

Answer:

Time period, $T=1.11\times 10^{-8}\ s$

Explanation:

We have,

The frequency of an FM radio station is 89.3 MHz.

It is required to find the period of the wave.

The reciprocal of frequency is called time period of a wave. It can be given by :

$T=\dfrac{1}{f}\\\\T=\dfrac{1}{89.3\times 10^{6}\ Hz}\\\\T=1.11\times 10^{-8}\ s$

So, the period of the wave is $1.11\times 10^{-8}\ s$ .

5 0

2 years ago

An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.

Anna35 [415]

Answer

Given,

Energy absorbed, $Q_H = 1.69\ kJ$

Energy expels, $Q_C = 1.25\ kJ$

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

$\eta = \dfrac{Q_H - Q_C}{Q_H}\times 100$

$\eta = \dfrac{1.69 - 1.25}{1.69}\times 100$

$\eta =26.03 %$

b) Work done by the engine

$W = Q_H- Q_C$

$W =1.69 - 1.25$

$W = 0.44\ kJ$

c) Power output

t = 0.296 s

$P = \dfrac{W}{t}$

$P = \dfrac{0.44}{0.296}$

$P = 1.486\ kW$

8 0

3 years ago

A power plant produces 1000 MW to supply a city 40 km away. Current flows from the power plant on a single wire of resistance 0.

nikitadnepr [17]

Answer:

Current = 8696 A

Fraction of power lost = $\dfrac{80}{529}$ = 0.151

Explanation:

Electric power is given by

$P=IV$

where I is the current and V is the voltage.

$I=\dfrac{P}{V}$

Using values from the question,

$I=\dfrac{1000\times10^6 \text{ W}}{115\times10^3\text{ V}} = 8696 \text{ A}$

The power loss is given by

$P_\text{loss} = I^2R$

where R is the resistance of the wire. From the question, the wire has a resistance of $0.050\Omega$ per km. Since resistance is proportional to length, the resistance of the wire is