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NISA [10]
3 years ago
14

A current of 02kA is traveling through a circularly looped wire. The wire makes 35 turns around an induced magnetic field of 8 9

nt, each loop which is of equal size, What is the radius of one of these loops of
Physics
1 answer:
PIT_PIT [208]3 years ago
3 0

Answer:

The radius of the loop is 4.94\times10^{5}\ m

Explanation:

Given that,

Current = 0.2kA = 200 A

Number of turns = 35

Magnetic field = 8.9 nT

We need to calculate the radius of one loop

Using formula of magnetic field

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

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

Where, I = current

N = number of turns

B = magnetic field

r = radius

Put the value into the formula

r=\dfrac{35\times4\pi\times10^{-7}\times200}{2\times8.9\times10^{-9}}

r =494183.11\ m

r=4.94\times10^{5}\ m

Hence, The radius of the loop is 4.94\times10^{5}\ m

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g When the movable mirror of the Michelson interferometer is moved a small distance X while making a measurement, 246 fringes ar
zhenek [66]

Answer:

X = 69.1 x 10⁻⁶ m = 69.1 μm

Explanation:

The relationship between the motion of the moveable mirror and the fringe count of the Michelson's Interferometer is given by the following formula:

d = mλ/2

where,

d = distance moved by the mirror = X = ?

m = No. of Fringes counted = 246

λ = wavelength of light entering interferometer = 562 nm = 5.62 x 10⁻⁷ m

Therefore,

X = (246)(5.62 x 10⁻⁷ m)/2

Therefore,

<u>X = 69.1 x 10⁻⁶ m = 69.1 μm</u>

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3 years ago
98 POINTS FOR ANYONE THAT DOES THIS NOW!!
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In simpler terms, a proton and neutron weigh 1 amu (atomic mass unit) each.

The nucleus has 15 protons and 18 neutrons. Since a proton's and a neutron's weight is only 1 amu, we can simply add the number of protons and neutrons to find the total mass of the nucleus:

15 + 18 = \boxed{33}

The nucleus' mass is 33 amu.

8 0
3 years ago
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Five hundred joules of heat are added to a closed system. The initial internal energy of the system is 87 J, and the final inter
Aleonysh [2.5K]

We can solve the problem by using the first law of thermodynamics:

\Delta U= Q-W

where

\Delta U is the variation of internal energy of the system

Q is the heat added to the system

W is the work done by the system

In this problem, the variation of internal energy of the system is

\Delta U=U_f-U_i=134 J-87 J=47 J

While the heat added to the system is

Q=500 J

therefore, the work done by the system is

W=Q-\Delta U=500 J-47 J=453 J

5 0
3 years ago
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a ball on a string makes 30.0 revolutions in 14.4s, in a circle of radius 0.340m. what is its period.(unit=s)
STALIN [3.7K]

Answer:

0.480 seconds

Explanation:

The period is the time for 1 revolution.  Writing a proportion:

14.4 s / 30.0 rev = t / 1 rev

t = 0.480 s

The period is 0.480 seconds.

7 0
3 years ago
If an object falls and ends with a velocity of 58.8 m/s, how long was the<br><br> object falling?
tensa zangetsu [6.8K]

Answer:

the time taken for the object to fall is 6 s.

Explanation:

Given;

final velocity of the object, v = 58.8 m/s

initial velocity of the object, u = 0

The height of fall of the object is calculated as;

v² = u² + 2gh

v² = 2gh

h = \frac{v^2}{2g} \\\\h = \frac{(58.8)^2}{2(9.8)} \\\\h = 176.4 \ m

The time to fall through the height is calculated as;

h =ut+  \frac{1}{2} gt^2\\\\h = 0 +  \frac{1}{2} gt^2\\\\h =  \frac{1}{2} gt^2\\\\t= \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 176.4}{9.8} } \\\\t = 6 \ s

Therefore, the time taken for the object to fall is 6 s.

6 0
3 years ago
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