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LUCKY_DIMON [66]
3 years ago
11

A straight 2.20 m wire carries a typical household current of 1.50 A (in one direction) at a location where the earth's magnetic

field is 0.550 gauss from south to north.. . Find the magnitude of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running from west to east.
Physics
1 answer:
Snowcat [4.5K]3 years ago
7 0
Use F = I*B*L*sinθ 

<span>I is the current, B is the magnitude of the magnetic field (Note: 1 gauss is equal to 1E-4 Tesla, the SI unit), and L is the length of wire. Adjust θ for each separate orientation.

All you need to do is just plug the given values to the equation. You should receive the same answer for parts a and b.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
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A kid on a playground swing makes a complete to-and-fro swing each 2 seconds.
sashaice [31]

Answer:

The frequency of the swing: 1/2 Hertz

The period is: 2 Seconds

Explanation:

The time the kid takes to make a complete to-and-fro swing = 2 seconds

The period, T, is the time it takes to make one complete cycle of an oscillatory motion, therefore, we have;

The frequency, f, is the number of cycles completed each second, therefore, we have;

The time for 1 cycle = 2 seconds

2 seconds = 1 cycle

Dividing both sides by 2 gives;

2/2 seconds = 1/2 cycles

2/2 = 1

In 2/2 = 1 seconds The number of cycles completed = 1/2 cycles

Therefore, the number of cycles completed per (one) second = 1/2 cycles

Therefore the frequency of the swing, f = 1/2 cycle/seconds = 1/2 Hertz

The period, T, is the time it takes to complete one to-and-fro swing which is one cycle which is 2 seconds

Therefore, the period is 2 Seconds.

4 0
3 years ago
A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the mer
son4ous [18]

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, $\omega = \frac{1}{8}$  rev/sec

                             $=\frac{2 \pi \times 7.5}{8}$  rad/sec

                              = 5.89 rad/sec

Therefore, force required,

$F=m.\omega^2.r$

   $$=1640 \times (5.89)^2 \times 7.5  

   = 427126.9 N

Thus, the net work done for the acceleration is given by :

W = F x r

   = 427126.9 x 7.5

   = 3,203,451.75 J

6 0
2 years ago
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Genrish500 [490]

Answer:

Universe

Explanation:

I took the quiz.

3 0
3 years ago
What is the internal energy of 2.00 mol of diatomic hydrogen gas (H2) at 35°C?
djyliett [7]
As you mentioned, we will use <span>Equipartition Theorem.
</span><span>H2 has 5 degrees of freedom; 3 translations and 2 rotation
</span>Therefore:
Internal energy = (5/2) nRT
You just substitute in the equation with the values of R and T and calculate the internal energy as follows:
Internal energy = (5/2) x 2 x <span>8.314 x 308 = 32.0089 x 10^3 J</span>
4 0
3 years ago
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Anna35 [415]

Answer:

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Explanation:\\\

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