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Reika [66]
4 years ago
9

An AC generator supplies 230 V peak voltage at 210 Hz frequency. How much power is dissipated in a resistor with R = 225 Ω resis

tance connected to the generator?
Physics
1 answer:
sesenic [268]4 years ago
7 0

Answer:

The power dissipated in a resistor is 117.54 watts.

Explanation:

Given that,

Peak voltage of the Ac generator, V = 230 V

Frequency, f = 210 Hz

Resistance, R = 225 ohms

We need to find the power dissipated in a resistor. The power generated is given by :

P=\dfrac{V^2_{rms}}{R}

V_{rms}=\dfrac{V}{\sqrt{2} }\\\\V_{rms}=\dfrac{230}{\sqrt{2} } \\\\V_{rms}=162.63\ V

So,

P=\dfrac{162.63^2}{225}\\\\P=117.54\ W

So, the power dissipated in a resistor is 117.54 watts. Hence, this is the required solution.

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Explain why mess extinctions occur and why it make sense to measure geologic time between mass extinctions
Nastasia [14]
Mass extinction occur from natural disasters, such as a n asteroid hitting earth or a volcano errupting and spread ash everywhere.

It makes sense to measure geologic time between mass extinctions because after each mass extinction, there is almost no life left and the few left have to repopulate, which may lead way to new mutations and new varieties of plants and animals.
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3 0
3 years ago
A superconducting solenoid has 3300 turns per meter and carries 4. 1 ka. Find the magnetic field strength in the solenoid.
WARRIOR [948]

The magnetic field strength is 17 Weber.

<h3>How can we find the value of magnetic field?</h3>

We know, The magnetic field equation is given by:

B= \mu_{0}  \times n \times I

<h3>Given,</h3>

Where, we know,

\mu_{0} = 4\pi  \times 10^{7} \frac{T. m^{2} }{A}= It is the magnetic constant or the permeability of free space.

n = The turn density which is equivalent to number of turns per unit length = 3300 turns per meter = 3300 m^{-1}

I= the amount of current applied to it = 4. 1 kA = 4 × 10^{3} A.

We have to find the magnetic field strength = B

Substituting the known values into the equation gives the strength of the magnetic field, which is:

B= 4\pi \times 10^{7} \times (3300) \times 4.1 \times 10^{3}

B= 17 T.m

B= 17W  [We know, 1 Tesla.meter = 1 Weber]

So, from the above calculation we can see that, The magnetic field strength is 17 Weber.

Learn more about magnetic field :

brainly.com/question/26257705

#SPJ4

4 0
2 years ago
You throw a ball straight up. The ball has an initial speed of 11.2 m/s when it leaves your hand What is the maximum height the
Jet001 [13]

Answer:

Explanation:

Given

initial velocity u=11.2\ m/s

At maximum height velocity of ball is zero

using

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

(0)^2 -(11.2)^2=2\times (-9.8)\times (s)

s=\frac{11.2^2}{2\times 9.8}

s=6.4

time taken by the ball to reach the maximum height

v=u+at

0=11.2-9.8\times t

t=\frac{11.2}{9.8}

t=1.142\ s

At t=2\ s height of ball is

h=ut+\\frac{1}{2}at^2

h=11.2\times 2-\frac{1}{2}9.8\times (2)^2

h=22.4-19.6

h=2.8\ m

i.e. ball is moving downward

height at v=5\ m/s

v^2-u^2=2as

s=\frac{25-125.4}{2\times (-9.8)}

s=5.12\ m    

8 0
3 years ago
A 2 kg object has a specific heat capacity of 1,700 J/(kg \cdot⋅oC)
Nutka1998 [239]

The amount of heat needed to raise the temperature of a 2kg object from 15°C to 25°C is 34000J.

HOW TO CALCULATE SPECIFIC HEAT CAPACITY:

  • The amount of heat absorbed by an object can be calculated by using the following expression:

  • Q = m.c.∆T

Where;

  1. Q = amount of heat absorbed or released (J)
  2. m = mass of object
  3. c = specific heat capacity (J/g°C)
  4. ∆T = change in temperature (°C)

  • According to this question, 2 kg object has a specific heat capacity of 1,700J/kg°C and was raised from a temperature of 15 Celsius to 25 Celsius. The heat absorbed is calculated as follows:

  • Q = 2 × 1700 × {25 - 15}

  • Q = 3400 × 10

  • Q = 34000J

  • Therefore, the amount of heat needed to raise the temperature of a 2kg object from 15°C to 25°C is 34000J.

Learn more about how to calculate heat absorbed at: brainly.com/question/11194034?referrer=searchResults

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