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Elden [556K]
3 years ago
11

A 2.5 m -long wire carries a current of 8.0 A and is immersed within a uniform magnetic field B⃗ . When this wire lies along the

+x axis, a magnetic force F⃗ =(−2.3j)N acts on the wire, and when it lies on the +y axis, the force is F⃗ =(2.3i−5.6k)N.
Part A - Find the x-component of B⃗ .

Part B - Find the y-component of B⃗ .

Part C - Find the z-component of B⃗ .
Physics
1 answer:
leva [86]3 years ago
8 0

Answer:

Explanation:

Let the magnetic field be B = B₁i + B₂j + B₃k

Force = I ( L x B )  , I is current , L is length and B is magnetic field .

In the first case

force = - 2.3 j N

L = 2.5 i

puting the values in the equation above

- 2.3 j = 8 [ 2.5 i x ( B₁i + B₂j + B₃k )]

= - 20 B₃ j + 20 B₂ k

comparing LHS and RHS ,

20B₃ = 2.3

B₃ = .115

B₂ = 0

In the second case

L = 2.5 j

Force = I ( L x B )

2.3i−5.6k = 8 ( 2.5 j x (B₁i + B₂j + B₃k )

=  - 20 B₁ k + 20B₃ i

2.3i−5.6k = - 20 B₁ k + 20B₃ i

B₃ = .115

B₁ = .28

So magnetic field B = .28 i + .115 B₃

Part A

x component of B = .28 T

Part B

y component of B = 0

Part C

z component of B = .115 T .

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75kg man climbs a mountain 1000m high in 3hrs and uses 4100 joulse/min. (a) calculate the power consumption in watt, (b) what is
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Answer:

Explanation:

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3 years ago
Some of the highest tides in the world occur in the Bay of Fundy on the Atlantic Coast of Canada. At Hopewell Cape the water dep
Nady [450]

Answer:

(a) 1.939 m/h

(b) 0.926 m/h

(c) -0.315 m/h

(d) -1.21 m/h

Explanation:

Here, we have the water depth given by the function of time;

D(t) = 7 + 5·cos[0.503(t-6.75)]

Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;

D'(t) = \frac{d(7 + 5\cdot cos[0.503(t-6.75)])}{dt}

= 5×(-sin(0.503(t-6.75))×0.503

= -2.515×(-sin(0.503(t-6.75))

= -2.515×(-sin(0.503×t-3.395))

Therefore we have;

(a) at 5:00 AM = 5 -  0:00 = 5

D'(5) =  -2.515×(-sin(0.503×5-3.395)) = 1.939 m/h

(b) at 6:00 AM = 6 -  0:00 = 6

D'(5) =  -2.515×(-sin(0.503×6-3.395)) = 0.926 m/h

(c) at 7:00 AM = 7 -  0:00 = 7

D'(5) =  -2.515×(-sin(0.503×7-3.395)) = -0.315 m/h

(d) at Noon 12:00 PM = 12 -  0:00 = 12

D'(5) =  -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.

4 0
2 years ago
An enormous thunderstorm covers Dallas-Ft. Worth. Your best friend Clark is a storm chaser and heads to the center of the storm
Rus_ich [418]

Answer:

t = 0.437 s

Explanation:

The speed of sound is a constant that is worth v = 343 m / s

           v = d / t

            t = d / v

the time it takes for the sound to reach Clark at d = 150 m is

           t = 150/343

           t = 0.437 s

This same sound takes much longer to reach you

          t₂ = 127 10³/343

          t₂ = 370 s

6 0
3 years ago
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