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3 years ago

5

on 28 may _______ at Chaghi, Balochistan; Dr A.Q Khan commanded to perform six successful nuclear explsions.

1 answer:

KATRIN_1 [288]3 years ago

8 0

I think the answer would be 1998

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2. A portion of the body receives 0.15 mGy from radiation with a quality factor Q = 6 and 0.22 mGy from radiation with Q = 10. (

DiKsa [7]

Answer with Explanation:

We are given that

$D_1=0.15mGy$

$D_2=0.22mGy$

$Q_1=6$

$Q_2=10$

a.We have to find the total dose

Total dose= $D=D_1+D_2$

Using the formula then, we get

$D=0.15+0.22$

$D=0.37mGy$

b.We have to find the total dose equivalent

Total dose equivalent=H= $D_1Q_1+D_2Q_2$

Using the formula

$H=0.15(6)+0.22(10)$

H=3.1mSv

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3 years ago

Wich advantage of reproduction does the graph shown

inn [45]

Answer:

No photo or graph is there

Explanation:

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2 years ago

Which of these is most likely to lead directly to a black market?

Sloan [31]

The answer is C.rationing.

<span>♡♡Hope I helped!!! :)♡♡
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7 0

3 years ago

Read 2 more answers

What is the definition of acceleration

PtichkaEL [24]

Acceleration is the way the motion is changing.

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3 years ago

A multimeter in an RL circuit records an rms current of 0.600 A and a 50.0-Hz rms generator voltage of 110 V. A wattmeter shows

jeka57 [31]

Answer:

(a) The impedance in the circuit is $Z=183.33\Omega$ .

(b)The resistance is $R=38.89\Omega$ .

(c) The inuctance is 0.57 H.

Explanation:

(a)

The expression for the impedance is as follows:

$Z=\frac{V_rms}{I_rms}$

Here, $V_rms$ is the rms voltage and $I_rms$ is the rms current.

Put $V_rms=110 V$ and $I_rms=0.600 A$ .

$Z=\frac{110}{0.600}$

$Z=183.33\Omega$

Therefore, the impedance in the circuit is $Z=183.33\Omega$ .

(b)

The expression for the average power is as follows;

$P_{a}=I_{rms}^{2}R$

Here, $P_{a}$ is the average power and R is the resistance.

Calculate the resistance by rearranging the above expression.

$R=\frac{P_{a}}{I_{rms}^{2}}$

Put $P_{a}=14W$ and

$R=\frac{14}{{0.600}^{2}}$

$R=38.89\Omega$

Therefore, the resistance is $R=38.89\Omega$ .

(c)

The expression for the impedance is as follows;

$Z^{2}=R^{2}+X_{L}^{2}$

Here, $X_{L}$ is the inductive reactance.

Put $Z=183.33\Omega$ and $R=38.89\Omega$ .

$(183.33)^{2}=(38.89)^{2}+X_{L}^{2}$

$X_{L}=179.16\Omega$

The expression for the inductive reactance in terms of frequency is as follows;

$X_{L}=2\pi fL$

Here, L is the inductance.

Calculate the inductance by rearranging the above expression.

$L=\frac{X_{L}}{2\pi f}$

Put $X_{L}=179.16\Omega$ and f=50Hz.

$L=\frac{179.16}{2\pi (50)}$

L=0.57 H

Therefore, the inuctance is 0.57 H.

4 0

3 years ago