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Marrrta [24]
2 years ago
5

A circular loop of radius 0.0400 m is

Physics
1 answer:
Alenkinab [10]2 years ago
4 0

Let's see

Find enclosed area of loop

\\ \rm\dashrightarrow \pi r^2

\\ \rm\dashrightarrow 3.14(0.04)^2

\\ \rm\dashrightarrow 0.005m^2

Now

\\ \rm\dashrightarrow \phi=BAcos\theta

  • We need B

\\ \rm\dashrightarrow 9.59\times 10^{-7}=B(0.005)cos75

\\ \rm\dashrightarrow 9.59\times 10^{-7}=0.0046B

\\ \rm\dashrightarrow B=2084.78\times 10^{-7}

\\ \rm\dashrightarrow B=2.08\times 10^{-4}T

\\ \rm\dashrightarrow B=0.208mT

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miv72 [106K]

Answer:

Option A

You need a Angle C congruent to angle F

Explanation:

EX) Side angle Side = sas

6 0
3 years ago
The distance between the earth and sun is 1.5 x 108 kilometers and the speed of light is 3.00 x 108 meters per second. Calculate
butalik [34]

Answer:

time = 8.3333 minutes.

Explanation:

distance between earth and sun = 1.5 * 10^{8}km

speed of light = 3* 10^{8}m/s

convert the distance unit from km to m so we can have uniform units.

distance between earth and sun = 1.5 *10^{8}*1000m

distance between earth and sun = 1.5 * 10^{11}m

speed = distance /time

time = distance / speed

time = \frac{1.5*10^{11} }{3*10^{8} }

= 0.5*10^{3}

time =500 sec

time = 500/60 minutes

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3 0
3 years ago
What is the eccentricity of an ellipse with a foci distance of 50,000,000 km and
inysia [295]

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25,000,000 Km ;)

Explanation:

5 0
3 years ago
n the Bohr model of the hydrogen atom (see Section 39.3), in the lowest energy state the electron orbits the proton at a speed o
Ivahew [28]

Answer:

(a) T=1.5*10^{-6}s

(b) I=1.1*10^{-3}A

(c) \mu=9.71*10^{-24}A\cdot m^2

Explanation:

(a) The orbital period is the time that the electron spend to travel the orbit of the atom. Thus, it is given by the length of the circular orbit divided by its velocity:

T=\frac{2\pi r}{v}\\T=\frac{2\pi(5.3*10^{-11}m)}{2.2*10^{6}\frac{m}{s}}\\T=1.5*10^{-6}s

(b) Current means charge over time, So, in this case is charge over period:

I=\frac{q}{t}\\I=\frac{e}{T}\\I=\frac{1.6*10^{-19}C}{1.5*10^{-6}s}\\\\I=1.1*10^{-3}A

(c) Magnetic moment is given by:

\mu=IA

Here A is the area of the orbit.

\mu=I\pi r^2\\\mu=(1.1*10^{-3}A)\pi(5.3*10^{-11}m)^2\\\mu=9.71*10^{-24}A\cdot m^2

4 0
3 years ago
What happens to the gravitational attraction between 2 objects
amm1812

Answer:

It decreases.

Explanation:

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2 years ago
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