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STatiana [176]
3 years ago
8

When the current through a circular loop is 6.0 A, the magnetic field at its center is 2.0 * 10-4 T. What is the radius of the l

oop?
Physics
1 answer:
Leona [35]3 years ago
5 0

Answer:

ill get back to this question once i get the answer

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Amanda [17]

Answer:

When mass is greater, the force needed must be greater keeping the acceleration of the wheel chair constant

Explanation:

force = mass \times acceleration \\ f \:  \alpha  \: m

force is directly proportional to the mass

6 0
3 years ago
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Can someone please help me ASAP?
Wittaler [7]

Answer:

B

Explanation:

8 0
3 years ago
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A simple pendulum makes 120 complete oscillations in 3.00 min at a location where g 5 9.80 m/s2. Find (a) the period of the pend
choli [55]

Answer:

(a) 1.5 second

(b) 0.56 m

Explanation:

Pendulum makes 120 oscillations in 3 min that means in 180 seconds

time taken by the pendulum to complete one oscillation is called time period.

(a) So, the time period is 180 / 120 = 1.5 second

T = 1.5 second

Thus, the time period of the pendulum is 1.5 second.

(b) g = 9.8 m/s^2

The formula for the time period is given by

T =2\pi \sqrt{\frac{L}{g}}

Where, L be the length of pendulum

1.5 =2\times 3.14 \sqrt{\frac{L}{9.8}}

0.057= {\frac{L}{9.8}}

L = 0.56 m

Thus, the length of the pendulum is 0.56 m .

6 0
3 years ago
Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =
Sloan [31]

Answer:

The current is  I  = 8.9 *10^{-5} \  A

Explanation:

From the question we are told that

     The  radius is r =  3.17 \  mm  =  3.17 *10^{-3} \ m

      The current density is  J =  c\cdot r^2  =  9.00*10^{6}  \ A/m^4 \cdot r^2

      The distance we are considering is  r =  0.5 R  =  0.001585

Generally current density is mathematically represented as

          J  =  \frac{I}{A }

Where A is the cross-sectional area represented as

         A  =  \pi r^2

=>      J  =  \frac{I}{\pi r^2  }

=>    I  =  J  *  (\pi r^2 )

Now the change in current per unit length is mathematically evaluated as

        dI  =  2 J  *  \pi r  dr

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

         I  = 2\pi  \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr

         I  = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr

        I  = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ]  | \left    0.001585} \atop 0}} \right.

        I  = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]

substituting values

        I  = 2 *  3.142  *  9.00 *10^6 *   [ \frac{0.001585^4}{4} ]

        I  = 8.9 *10^{-5} \  A

5 0
4 years ago
Where on this diagram does the ball have the highest point of gravitational potential energy?
mixer [17]
It should be at the very top since it has more space to fall which gives it more potential energy
3 0
3 years ago
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