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brilliants [131]
3 years ago
13

A very large magnet applies a repulsive force to a smaller (0.05 Kg) magnet. If the smaller magnet accelerates across a friction

less table at 2m/s2, then what is the magnitude of the repulsive force?
Physics
1 answer:
makkiz [27]3 years ago
5 0

Answer:

0.1 N

Explanation:

using F = ma yields 0.1 N.

net

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How much work w is done by the electrostatic force on the moving point charge?
V125BC [204]

As we know that electrostatic force is a conservative force

so we can say by the condition of conservative force

F_c = -\frac{dU}{dr}

here we can rearrange the above equation as

dU = - F_c.dr

now integrate both sides

\int dU = - \int F_c . dr

Now we know by the definition of work done by a force is given by

W = \int F.dr

now work done by conservative force is given as

W_c = \int F_c . dr

Now from above work done by electric field to move charge from one point to other is given as

W_{EF} = \int F_{e}.dr = - \int dU

so here work done is given as

W_{EF} = U_i - U_f

so change in potential energy is given by work done

3 0
4 years ago
Suppose that you have a spring gun that you use to launch a small metal ball. You try the first two settings of the gun. The fir
Ivan

Answer:

The distance s of how far the ball will go at the highest setting = 2.25m

Explanation:

Let consider x to be the representative of the compression and the distance to be s

Recall that:

\dfrac{1}{2}\times K \times  x^2 = mgs +c

By cross multiplying

K \times  x^2 = 2(mgs +c)

K \times  x^2 = 2\times 9.81(ms) +2c

K \times  x^2 = 19.62(ms) +2c

x^2 = A \times  s+B

Thus, for the first setting

x = 1 , s = 0.25

for the second setting

x = 2,   s = 1

1 = 0.25A + B ---  (1)

4 = A + B    ----- (2)

From (1); let B =  1 - 0.25A  and substitute it into (2)

4 = A + 1 - 0.25 A

4 - 1 = A - 0.25 A

3 = 0.75 A

A = 3/0.75

A = 4

From (2)

4 = A + B

4 = 4 + B

B = 4 - 4

B = 0

Therefore, for the highest setting, where x = 3

Then :

x^2 = A \times  s+B will be:

3² =   4s + 0

9 = 4s

s = 9/4

s = 2.25 m

∴

The distance s of how far the ball will go at the highest setting = 2.25m

6 0
4 years ago
A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m high
Andrews [41]

Answer:

The boat is approaching at 0.9938 m/s to the dock when it is 9 m from the dock.

Explanation:

α=tan⁻¹(1 m / 9 m)

α= 6.34º

V= 1m/s

Vx = V * cos (6.34º)

Vx= 0.9938 m/s

8 0
4 years ago
Each plate of a parallel‑plate capacitor is a square of side 0.0479 m, and the plates are separated by 0.479 × 10 − 3 m. The cap
Anon25 [30]

Explanation:

It is known that the relation between electric field and potential is as follows.

             E = \frac{V}{d}

And, formula to calculate the capacitance is as follows.

           C = \frac{\epsilon_{o} A}{d}

              = \frac{8.854 \times 10^{-12} \times (0.479 m)^{2}}{0.479 \times 10^{-3}}

              = 4.24 \times 10^{-9} F

Hence, energy stored in a capacitor is as follows.

         W = \frac{1}{2}CV^{2}

          V = \sqrt{\frac{2W}{C}}

        E = \sqrt{\frac{2W}{d^{2}C}}

            = \frac{2 \times 8.11 \times 10^{-9} J}{(0.479 \times 10^{-3})^{2} \times 4.24 \times 10^{-9}}

            = 16.687 \times 10^{3} N/C

Thus, we can conclude that electric field strength E inside the capacitor is 16.687 \times 10^{3} N/C.

7 0
3 years ago
A ball at rest starts rolling down a hill with a constant acceleration of 3.2 meters/second2. What is the final velocity of the
leva [86]
Used an app called Mephyso to do the calculation.

8 0
3 years ago
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