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Zigmanuir [339]
3 years ago
12

What is the magnetic flux density (B-field) at a distance of 0.36 m from a long, straight wire carrying a current of 3.8 A in ai

r? Give your answer in units of tesla.
Physics
1 answer:
olga nikolaevna [1]3 years ago
3 0

Answer:

The magnetic flux density is 2.11\times10^{-6}\ T

Explanation:

Given that,

Distance = 0.36 m

Current = 3.8 A

We need to calculate the magnetic flux density

Using formula of magnetic field

B =\dfrac{\mu_{0}I}{2r}

Where,

r = radius

I = current

Put the value into the formula

B =\dfrac{4\pi\times10^{-7}\times3.8}{2\times\pi\times0.36}

B=2.11\times10^{-6}\ T

Hence, The magnetic flux density is 2.11\times10^{-6}\ T

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I think it's A

Explanation:

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2 years ago
A radio telescope has a circular collecting dish of diameter 5.0 m. It is used to observe two distant
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Find the difference between two masses measured as 123.6 grams and 115.972 grams. Express the answer to the correct number of si
xxTIMURxx [149]

Answer:

The difference is 7.6 grams.

Explanation:

In mathematics the difference of two numbers is express as the subtraction between them:

         

a-b

So to find out the difference between the two measured masses, a will be represented by 123.6 grams since is the bigger number, and b by 115.972 grams.

Therefore, it is get:

123.6grams-115.972grams = 7.6grams

<u>Hence, the difference is 7.6 grams. </u>

The result of 7.628 will be expressed as 7.6 to have the correct number of significant figures.          

 

Notice how that can be express in units of kilograms too since there is 1000 gram in 1 kilogram:

7.6grams . \frac{1Kg}{1000grams} ⇒ 7.6x10^{-3}Kg

8 0
3 years ago
Which of the following are for vector directione?
Inga [223]
Answer:
C. outside 45 degrees
Step-by-step explanation:
C. outside 45 degrees
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8 0
2 years ago
block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8
dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: \sqrt{800} \,\,\,N

Explanation:

First try to write all forces in  vector component form:

The force F1 acting at 45 degrees would have multiplication factors of \frac{\sqrt{2} }{2} on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is    F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N

y-component of F1 is    F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N

As far as force F2, it is given already in x and y components, then:

x-component of F2 = 8 N

y-component of F2 = -6 N (negative meaning pointing down the y-axis)

Force F3 has only component (upwards) in the y-direction

x-component of F3 = 0 N

y-component of F3 =14 N

The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:

x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

Therefore, the acceleration that the mass receives due to this force is given in component form as:

x-component of acceleration: 20 N / 5 kg = 4\,\,\,m/s^2

y-component of acceleration: 20 N / 5 kg = 4\,\,\,m/s^2

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is:     v_x=4\,*\,2=8\,m/s

y-component of the velocity is:     v_y=4\,*\,2=8\,m/s

7 0
3 years ago
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