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Georgia [21]
3 years ago
10

PLEASE HELP!!!!

Physics
1 answer:
Zina [86]3 years ago
4 0
D is the answer
Imagine the magnetic field, it emanates from both

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What is 18000 expressed in scientific notation​
Vikki [24]
1.8x 10^4
This is 18000 expressed in scientific notation.
8 0
3 years ago
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8) A flat circular loop having one turn and radius 5.0 cm is positioned with its plane perpendicular to a uniform 0.60-T magneti
Marrrta [24]

Answer:

EMF induced in the loop is 9.4 V

Explanation:

As we know that initial magnetic flux of the loop is given as

\phi_1 = B.A

\phi_1 = (0.60)(\pi (0.05)^2)

\phi_1 = 4.7 \times 10^{-3} Wb

As soon as the area of the loop becomes zero the final magnetic flux of the loop is ZERO

Now as per faraday's law of electromagnetic induction the EMF is induced due to rate of change in magnetic flux

so we will have

EMF = \frac{\Delta \phi}{\Delta t}

so we will have

EMF = \frac{4.7 \times 10^{-3} - 0}{0.50 \times 10^{-3}}

EMF = 9.4 V

7 0
3 years ago
Hellp.....its physics
blondinia [14]

Answer:

i know it

Explanation:

18.493 its the answer

brainliest pls

4 0
3 years ago
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An asteroid orbits the Sun every 176 years. What is the asteroids average distance from the Sun? P ^ 2 = a ^ 3 where p = period
KatRina [158]

Answer:

The value is  x =  45.99 \  Au

Explanation:

From the question we are told that

   The period of the  asteroid is   T =  176 \ years = 176 * 365 * 24 * 60* 60 = 5.55*10^{9}\ s

Generally the average distance of the asteroid from the sun is mathematically represented as

            R = \sqrt[3]{ \frac{G M * T^2 }{4 \pi} }

Here M is the mass of the sun with a value  

        M  =  1.99*10^{30} \  kg

         G  is the gravitational constant with value  G  =  6.67 *10^{-11}  \  m^3 \cdot kg^{-1} \cdot  s^{-2}

           R = \sqrt[3]{ \frac{6.67 *10^{-11}  * 1.99*10^{30} * [5.55 *10^{9}]^2 }{4 * 3.142 } }

=>       R = 6.88 *10^{12} \  m

Generally

         1.496* 10^{11}  \  m  \to  1 Au (Astronomical \  unit )

So

          R = 6.88 *10^{12} \  m \ \ \ \ \to \ \   x \  Au

=>      x =  \frac{6.88 *10^{12}}{1.496 *10^{11}}

=>       x =  45.99 \  Au

       

7 0
2 years ago
How fast would you have to travel on the surface of earth at the equator to keep up with the sun (that is, so that the sun would
professor190 [17]

Answer: 1037 miles per hour

Explanation: In order to see the sun in the same position in the sky, you would have to travel against the speed of rotation of the earth, because this is what causes the sun to appear in a constantly changing position.

Because of this, we will have to calculate the speed of rotation of the earth. To get started, we must know the circumference of the earth. Assuming the circumference formula for a sphere,

Circumference=2\pi R

Where R is the radius of the earth, we find that the perimeter of the earth is approximately 24881 miles. The equation to calculate speed is given by

v=\frac{Distance}{Time}

Because the earth completes one rotation in 24 hours, we have to find the speed of rotation as the perimeter of the earth divided by 24 hours.

The obtained result is 1037 miles per hour.

You would have to travel at 1037 miles per hour in the direction opposite to the direction the rotation is ocurring in.

5 0
3 years ago
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