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3 years ago

In a horse cart system when would there be the smallest amount of acceleration

Physics

1 answer:

Assoli18 [71]3 years ago

3 0

Answer:

well... when the horse stops/rests, or if it is blocked by a surface or anything of solid background.

Explanation:

If it is going up a hill or slope and it just starts to move that would also be considered the smallest amount of acceleration this can go for many things when it just starts to move. but I would go for when it rests amounting to your fitting of the question.

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3 years ago

Light waves travel most slowly through a ____<br><br> A) gas<br> B) solid<br> C) liquid

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Answer:

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3 years ago

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How can you get acceleration when talking about speed?

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Answer:

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Explanation:

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3 years ago

an object is moving in a straight line with a constant acceleration initially it is traveling at 16 meters per second. 3 seconds

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3 years ago

A wire is formed into a circle having a diameter of 10.0cm and is placed in a uniform magnetic field of 3.00mT . The wire carrie

Paul [167]

The range of potential energies of the wire-field system for different orientations of the circle are -

θ U

0° 375 π x $10^{-7}$

90° 0

180° - 375 π x $10^{-7}$

We have current carrying wire in a form of a circle placed in a uniform magnetic field.

We have to the range of potential energies of the wire-field system for different orientations of the circle.

<h3>What is the formula to calculate the Magnetic Potential Energy?</h3>

The formula to calculate the magnetic potential energy is -

U = M.B = MB cos $$\theta$

where -

M is the Dipole Moment.

B is the Magnetic Field Intensity.

According to the question, we have -

U = M.B = MB cos $$\theta$

We can write M = IA (I is current and A is cross sectional Area)

U = IAB cos $$\theta$

U = Iπ $r^{2}$ B cos $$\theta$

For $$\theta$ = 0° →

U(Max) = MB cos(0) = MB = Iπ $r^{2}$ B = 5 × π × $( 0.05 ) ^{2}$ × 3 × $10^{-3}$ =

375 π x $10^{-7}$ .

For $$\theta$ = 90° →

U = MB cos (90) = 0

For $$\theta$ = 180° →

U(Min) = MB cos(0) = - MB = - Iπ $r^{2}$ B = - 5 × π × $( 0.05 ) ^{2}$ × 3 × $10^{-3}$ =

- 375 π x $10^{-7}$ .

Hence, the range of potential energies of the wire-field system for different orientations of the circle are -

θ U

0° 375 π x $10^{-7}$

90° 0

180° - 375 π x $10^{-7}$

To solve more questions on Magnetic potential energy, visit the link below-

brainly.com/question/13708277

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2 years ago