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2 years ago

about how much more energy is released in a 6.5 richter magnitude earthquake than in one with magnitude 5.5?

Physics

1 answer:

OverLord2011 [107]2 years ago

8 0

Answer:

For example, an earthquake of magnitude 5.5 releases about 32 times as much energy as an earthquake measuring 4.5. Another way to look at this is that it takes about 900 magnitude 4.5 earthquakes to equal the energy released in a single 6.5 earthquake.

Explanation:

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When the current in a toroidal solenoid is changing at a rate of 0.0240 A/s , the magnitude of the induced emf is 12.4 mV . When

Gemiola [76]

Answer:

The number of turns in the solenoid is 230.

Explanation:

Given that,

Rate of change of current, $\dfrac{dI}{dt}=0.0240\ A/s$

Induced emf, $\epsilon=12.4\ mV=12.4\times 10^{-3}\ V$

Current, I = 1.5 A

Magnetic flux, $\phi=0.00338\ Wb$

The induced emf through the solenoid is given by :

$\epsilon=L\dfrac{dI}{dt}$

$L=\dfrac{\epsilon}{(di/dt)}$ ........(1)

The self inductance of the solenoid is given by :

$L=\dfrac{N\phi}{I}$ .........(2)

From equation (1) and (2) we get :

$\dfrac{\epsilon}{(di/dt)}=\dfrac{N\phi}{I}$

N is the number of turns in the solenoid

$N=\dfrac{\epsilon I}{\phi (dI/dt)}$

$N=\dfrac{12.4\times 10^{-3}\times 1.5}{0.00338 \times 0.024}$

N = 229.28 turns

N = 230 turns

So, the number of turns in the solenoid is 230. Hence, this is the required solution.

3 0

3 years ago

A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa

Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

$Q=P\tan\phi$ ...(I)

We need to calculate the $\phi$

Using formula of $\phi$

$\phi=\cos^{-1}(Power\ factor)$

Put the value into the formula

$\phi=\cos^{-1}(0.6)$

$\phi=53.13^{\circ}$

Put the value of Φ in equation (I)

$Q=600\tan(53.13)$

$Q=799.99\ kVAR$

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

$Q'=600\tan(0.95)$

$Q'=9.94\ KVAR$

We need to calculate the difference between Q and Q'

$Q''=Q-Q'$

Put the value into the formula

$Q''=799.99-9.94$

$Q''=790.05\ KVAR$

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

8 0

3 years ago

If a radio wave carries information by changing the amplitude of the transmitted wave, what kind of broadcast might it possibly

mariarad [96]

Option A because those are the correct ones

4 0

2 years ago

Select the correct answer.

BaLLatris [955]

It is due to the excess stress.

6 0

3 years ago

A wire of length L is made up of two sections of two different materials connected in series. The first section of length L1 = 1

ololo11 [35]

Answer:

Explanation:

velocity of wave in a tense wire is given by the expression

$v= \sqrt{\frac{T}{m} }$

v is velocity . T is tension and m is mass per unit length .

for steel wire

m = π r² ρ where r is radius and ρ is density

= 3.14 x (5.3 x 10⁻⁴)²x7.75 x 10³

= 683.57 x 10⁻⁵ kg/m

v = $\sqrt{\frac{148}{683.57\times 10^{-5}} }$

= 1.47 x 10² m /s

= 147 m /s

for iron wire

m = π r² ρ where r is radius and ρ is density

= 3.14 x (5.3 x 10⁻⁴)²x7.86 x 10³

= 693.27 x 10⁻⁵ kg/m

$v = \sqrt{\frac{148}{693.27\times 10^{-5}} }$

= 146 m /s

Time taken to move from one end to another

= 17.7 / 147 + 28.5 / 146

= .12 + .195

= .315 s .

4 0

3 years ago