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Mekhanik [1.2K]

2 years ago

5

The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole huma

n bodies has a field strength of 8.36 T, and the current in the solenoid is 3.34 × 102 A. What is the number of turns per meter of length of the solenoid?

1 answer:

gayaneshka [121]2 years ago

6 0

Explanation:

Below is an attachment containing the solution.

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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

7 0

2 years ago

Marie Curie and her husband Pierre were Henri Becquerel's graduate students when Becquerel

LenKa [72]

Answer:

It is C on edge.

Explanation:

Because I just figured it out and got it right and because it says so in the link provided from the question.

7 0

2 years ago

Read 2 more answers

If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2

kaheart [24]

Answer:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

So then the difference of temperature across the material would be $\Delta T = 375 K$

Explanation:

For this case we can use the Fourier Law of heat conduction given by the following equation:

$Q = -kA \frac{\Delta T}{\Delta x}$ (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 represent the cross sectional area

Q= 3KW represent the rate of heat transfer

$\Delta T$ is the temperature of difference that we want to find

$\Delta x=2.5 cm =0.025 m$ represent the thickness of the material

If we solve $\Delta T$ in absolute value from the equation (1) we got:

$\Delta T =\frac{Q \Delta x}{Ak}$

First we convert 3KW to W and we got:

$Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W$

And we have everything to replace and we got:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

So then the difference of temperature across the material would be $\Delta T = 375 K$

5 0

3 years ago

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume tha

belka [17]

Given that,

Energy $H=2.7\times10^{31}\ W$

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

$H=Ae\sigma T^4$

$A=\dfrac{H}{e\sigma T^4}$

$4\pi R^2=\dfrac{H}{e\sigma T^4}$

$R^2=\dfrac{H}{e\sigma T^4\times4\pi}$

Put the value into the formula

$R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}$

$R=5.0\times10^{10}\ m$

(b). Given that,

Radiates energy $H=2.1\times10^{23}\ W$

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

$R^2=\dfrac{H}{e\sigma T^4\times4\pi}$

Put the value into the formula

$R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}$

$R=5.42\times10^{6}\ m$

Hence, (a). The radius of the star is $5.0\times10^{10}\ m$

(b). The radius of the star is $5.42\times10^{6}\ m$

8 0

2 years ago

If you have a block of wood that is 4cm wide ,6cm long and 10cm high and it weighs 100 grams what would happen to the volume if

n200080 [17]

It would make both half 4cm to 2cm wide, and 6cm to 3cm long, 10cm to 5cm hight it would split in half

4 0

2 years ago

Read 2 more answers