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GREYUIT [131]
3 years ago
5

a 1.2 m wire carries a current of 10.0 A in a uniform magnetic field of 0.050T. Find the magnitude of the magnetic force on the

wire.
Physics
1 answer:
rodikova [14]3 years ago
8 0
To calculate we use the formula for a magnetic force in a current-carrying wire expressed as the product of the current, magnetic field and the length of the wire.
 F = I x L x B

where F is the force on the wire, I is the current flowing on the wire, L is the length of the wire and B is the magnetic field.

F = 10.0 A x 1.2 m x 0.050 T
F = 0.60 N
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A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the
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Answer:

The minimum coefficient of friction required is 0.35.  

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

-F_{f} + F = 0      

-F_{f} + ma = 0      

\mu mg = ma

\mu = \frac{a}{g}

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

v_{f}^{2} = v_{0}^{2} + 2ad

a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}

Where:  

d: is the distance traveled = 46.1 m

v_{f}: is the final speed of the truck = 0 (it stops)      

v_{0}: is the initial speed of the truck = 17.9 m/s

a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}        

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.  

\mu = \frac{a}{g}  

\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}

\mu = 0.35

Therefore, the minimum coefficient of friction required is 0.35.  

I hope it helps you!

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If a train is 100 kilometers away, how much sooner would you hear the train coming by listening to the rails (iron) as opposed t
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From tables, the speed of sound at 0°C is approximately
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V₃ = 5130 m/s (in iron)

Distance traveled is
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Time required to travel in air is
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The difference in time is
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