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koban [17]
3 years ago
12

In an experiment, one of the forces exerted on a proton is F⃗ =−αx2i^, where α=12N/m2. What is the potential-energy function for

F⃗ ? Let U=0 when x=0.
Physics
1 answer:
Blizzard [7]3 years ago
7 0

Answer:

The potential energy is -4x^3

Explanation:

Given that,

Force F=-\alpha x^2 i

We need to calculate the potential energy

Using formula of work done

\Delta U=F(x) dx

Put the value of F

\Delta U=-\alpha x^2\ dx

On integration

U=-\alpha \dfrac{x^3}{3}+C

U=-\dfrac{\alpha x^3}{3}+C...(I)

U = 0, x = 0 so C = 0

Put the value of c and α in equation (I)

U=-\dfrac{12x^3}{3}+0

U=-4x^3

Hence, The potential energy is -4x^3

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What is the relationship between the poles of a magnet? which direction does the magnet field move?
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A 35 N force makes a 10 degree angle with the positive x-axis. What is the magnitude of the vertical component of the force?
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Answer:

6.07 N

Explanation:

Given that,

Force, F = 35 N

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We need to find the magnitude of the vertical component of the force. It can be given by :

F_y=F\sin\theta\\\\=35\times \sin(10)\\\\=6.07\ N

So, the magnitude of the vertical component of the force is 6.07 N.

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Read 2 more answers
For the different values given for the radius of curvature RRR and speed vvv, rank the magnitude of the force of the roller-coas
nirvana33 [79]

Explanation:

The force of the roller-coaster track on the cart at the bottom is given by :

F=\dfrac{mv^2}{R}, m is mass of roller coaster

Case 1.

R = 60 m v = 16 m/s

F=\dfrac{(16)^2m}{60}=4.26m\ N

Case 2.

R = 15 m v = 8 m/s

F=\dfrac{(8)^2m}{15}=4.26m\ N

Case 3.

R = 30 m v = 4 m/s

F=\dfrac{(4)^2m}{30}=0.54m\ N

Case 4.

R = 45 m v = 4 m/s

F=\dfrac{(4)^2m}{45}=0.36m\ N

Case 5.

R = 30 m v = 16 m/s

F=\dfrac{(16)^2m}{30}=8.54m\ N

Case 6.

R = 15 m v =12 m/s

F=\dfrac{(12)^2m}{15}=9.6m\ N

Ranking from largest to smallest is given by :

F>E>A=B>C>D

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3 years ago
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Answer:

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