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oksian1 [2.3K]
3 years ago
10

A 1.5-m length of straight wire experiences a maximum force of 1.6 N when in a uniform magnetic field that is 1.8 T. 1) What cur

rent must be passing through it? (Express your answer to two significant figures.)
Physics
1 answer:
Olenka [21]3 years ago
4 0

Answer:

0.6 A

Explanation:

length, l  = 1.5 m

Maximum force, F = 1.6 N

Magnetic field, B = 1.8 T

Let the current is i.

the force on a current carrying conductor placed in a magnetic field is given by

F = B x i x l x sinθ

Where, θ be the angle between the magnetic field and the length element.

for maximum force, θ = 90°

So,

1.6 = 1.8 x i x 1.5 x 1

i = 0.6 A

Thus, the current in the conductor is 0.6 A.

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Let car A's starting position be the origin, so that its position at time <em>t</em> is

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and car B has position at time <em>t</em> of

B: <em>x</em> = 100 m - (60 m/s) <em>t</em>

<em />

They meet when their positions are equal:

(40 m/s) <em>t</em> = 100 m - (60 m/s) <em>t</em>

(100 m/s) <em>t</em> = 100 m

<em>t</em> = (100 m) / (100 m/s) = 1 s

so the cars meet 1 second after they start moving.

They are 100 m apart when the difference in their positions is equal to 100 m:

(40 m/s) <em>t</em> - (100 m - (60 m/s) <em>t</em>) = 100 m

(subtract car B's position from car A's position because we take car A's direction to be positive)

(100 m/s) <em>t</em> = 200 m

<em>t</em> = (200 m) / (100 m/s) = 2 s

so the cars are 100 m apart after 2 seconds.

3 0
3 years ago
Can someone help please and thank you:)
Ray Of Light [21]

Answer:

The answer is C.

Explanation:

4 0
3 years ago
A river flows at a velocity of 3 km/h relative to the riverbank. A boat moves downstream at a velocity of 15 km/h relative to th
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15+3=18km/hour
Think about it like this. The boat is going 15 faster than the river, and the river is going 3 faster than the bank, so the boat is going 18 faster than the river bank
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The swim bladder of a fish helps keep the fish from sinking by_______.
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C. Increasing its buoyancy

6 0
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Read 2 more answers
Two identical trucks have mass 5100 kg when empty, and the maximum permissible load for each is 8000 kg. the first truck, carryi
Oksanka [162]
<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg. The key here is the conservation of momentum. For the first truck, the momentum is 0(5100 + 4300) The second truck has a starting momentum of 60(5100 + x) And finally, after the collision, the momentum of the whole system is 42(5100 + 4300 + 5100 + x) So let's set the equations for before and after the collision equal to each other. 0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x) And solve for x, first by adding the constant terms 0(5100 + 4300) + 60(5100 + x) = 42(14500 + x) Getting rid of the zero term 60(5100 + x) = 42(14500 + x) Distribute the 60 and the 42. 60*5100 + 60x = 42*14500 + 42x 306000 + 60x = 609000 + 42x Subtract 42x from both sides 306000 + 18x = 609000 Subtract 306000 from both sides 18x = 303000 And divide both sides by 18 x = 16833.33 So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums. 60(5100 + 16833.33) = 60(21933.33) = 1316000 42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000 They match. The 2nd truck was definitely over loaded.</span>
6 0
4 years ago
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