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Semmy [17]
3 years ago
10

Consider the expression below. Assume m is an integer. 6m(2m + 18) Enter an expression in the box that uses the variable m and m

akes the equation true. (Simplify your answer completely. If no expression exists, enter DNE.)
Physics
1 answer:
Anika [276]3 years ago
8 0
6x2=12m
6x18=108
12m+108
Simplified: m+9 bc 12/12 and 108/12
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Two thin wires rings each having a radius R are placed at a distance d apart with their axes coinciding. The charges on the two
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Answer:

Q/4πε0 [1/R - 1/√R2+d2]

Explanation:

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What happens at the end of silence of the lambs?
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3 years ago
A single-turn current loop, carrying a current of 4.03 A, is in the shape of a right triangle with sides 68.1, 151, and 166 cm.
SashulF [63]

Answer:

Part a)

F = 0

Part b)

F = 0.25 N

Part c)

F = 0.25 N

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

F_{net} = 0

Explanation:

As we know that force on a current carrying wire is given as

\vec F = i(\vec L \times \vec B)

now we have

Part a)

current in side 166 cm and magnetic field is parallel

so we have

F = i(\vec L \times \vec B)

here we know that L and B is parallel to each other so

F = 0

Part b)

For 68.1 cm length wire we have

F = iLB sin\theta

here we know that

cos\theta = \frac{68.1}{166}

\theta = 65.8

so we have

F = (4.03)(0.681)(99.3 \times 10^{-3})sin65.8

F = 0.25 N

Part c)

For 151 cm length wire we have

F = iLB sin\phi

here we know that

cos\phi = \frac{151}{166}

\theta = 24.5

so we have

F = (4.03)(1.51)(99.3 \times 10^{-3})sin24.5

F = 0.25 N

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

F_{net} = 0

4 0
3 years ago
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