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aev [14]
3 years ago
5

A 45.5-turn circular coil of radius 4.85 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0

.490 T. If the coil carries a current of 26.7 mA, find the magnitude of the maximum possible torque exerted on the coil.

Physics
2 answers:
il63 [147K]3 years ago
3 0

Answer:

4.399 Nm

Explanation:

The maximum Torque on a coil is given as,

τ = BNIA...................... Equation 1

Where τ = Maximum torque exerted on the coil, B = Magnetic Field, N = Number of turns, I = Current, A = Area.

Given: N = 45.5 Turns, B = 0.49 T, I = 26.7 mA = 0.0267 A,

A = πr², Where r = radius of the coil, r= 4.85 cm = 0.0485 m

A = 3.14(0.0485)²

A = 7.39×10⁻³ m².

Substitute into equation 1

τ  = 45.5×0.49×26.7×7.39×10⁻³

τ = 4.399 Nm

igomit [66]3 years ago
3 0

Explanation:

Below is an attachment containing the solution.

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Answer:

A) 50 seconds

B) 200 m

Explanation:

They are 500 metres apart.

And one of the bike loves at 6 m/s while the other loves at 4 m/s.

A) Let distance of the 6 m/s bike before they meet be x.

Thus, time = x/6

Since time = distance/speed

For the second bike at 4 m/s, his distance covered before they meet will be 500 - x

Thus, time = (500 - x)/4

Now they will meet each other at the same time. Thus;

x/6 = (500 - x)/4

Cross multiply to get;

4x = 3000 - 6x

6x + 4x = 3000

10x = 3000

x = 3000/10

x = 300 m

Thus, time will be;

t = 300/6

t = 50 seconds

B) Distance covered by the slower bike is (500 - x)

Since from a above, x = 300

Thus; distance = 500 - 300 = 200 m

4 0
3 years ago
BRainliest if correct
kolbaska11 [484]

Answer:

C.

hope this helps

have a good day :)

Explanation:

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When light reaches the end of a barrier on the edge of an opening it diffracts, which means...
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3 years ago
(1 point) Find a linearly independent set of vectors that spans the same subspace of R3R3 as that spanned by the vectors ⎡⎣⎢3−1−
Natalka [10]

Answer:

Explanation:

A=\left[\begin{array}{ccc}3&-9&-3\\-1&2&0\\-2&3&-1\end{array}\right] \\\\R_2\rightarrow 3R_2+R_1,R_3\rightarrow 3R_3+2R_1\\\\=\left[\begin{array}{ccc}3&-9&-3\\0&-3&-3\\0&-9&-9\end{array}\right] \\\\R_3\rightarrow 3R_3-9R_2\\\\=\left[\begin{array}{ccc}3&-9&-3\\0&-3&-3\\0&0&0\end{array}\right]

This is the row echelon form of A. This means that only two of the vectors in our set are linearly independent. In other words, the first two vectors alone will span the same subspace of R^4 as all three vectors.

Therefore, the linearly independent spanning set for the subspace is

\left[\begin{array}{ccc}3\\-1\\-2\end{array}\right] \left[\begin{array}{ccc}-9\\2\\3\end{array}\right] \left[\begin{array}{ccc}3\\0\\-1\end{array}\right]

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