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il63 [147K]
3 years ago
8

20.What factor is more important in determining electrical force

Physics
1 answer:
Agata [3.3K]3 years ago
6 0

Answer:

20.The first factor is the amount of charge on each object. The greater the charge, the greater the electric force. The second factor is the distance between the charges. The closer together the charges are, the greater the electric force is

Explanation:

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Three carts of masses 4.0 kg, 10kg, and 3.0 kg move on a frictionless track with speeds of v1 = 5.0m/s, v2=3.0m/s, and v3=-3.6 m
gogolik [260]

2.24 m/s is the calculated velocity.

Initial velocity (u) squared plus two times the acceleration (a) times the displacement equals final velocity (v) squared (s). Final velocity (v) is equal to the square root of initial velocity (u) squared plus two times the acceleration (a) times displacement when v is the variable being solved for (s).

The cart's masses and speeds are known.

M1 = 4.00 kg, M2 = 10.0 kg, M3 = 3.00 kg, etc.

v1 = 5.00 m/s = 5.00 m/s, v2 = 3.00 m/s = 3.00 m/s, v3 = -4.00 m/s = 4.00 m/s, and m1v1+m2v2+m3v3 = (m1+m2+m3) v=d frac m 1v 1+m 2v 2+m 3v 3, where (m1+m2 + m3) is the product of (v1 v 1+m2v2+m3v3).

"m 1+m 2+m 3" is equivalent to "m 1+m2+m3/m1v 1+m2v2 +m3v3"

the three carts' final velocities are calculated as follows: v=d frac

{4.00kg\sdot5.00m/s+10.0kg\sdot3.00m/s-3.00kg\sdot4.00m/s} 4.24m/s = 4.50kg+10.0kg+3.00kg vs. 4.50kg+10.0kg+3.00kg

5.00m/s/4.00kg/5.00m/s+10.0m/s/3.00m/s/4.00m/s =2.24m/s.

2.24 m/s is the calculated final velocity.

Learn more about velocity here-

brainly.com/question/18084516

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2 years ago
Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 18.0 cm and carries a
Helen [10]

Answer:

Explanation:

The wires are in circular shape . They have common center .

magnetic field due to circular wire is given by the formula

B = \frac{\mu_0\times I }{2r}

 where I is current , r is radius of the coil .

magnetic field due to inner wire

= \frac{\mu_0\times 10 }{2\times.09}

magnetic field due to outer wire

= \frac{\mu_0\times I }{2\times.15}

These should be equal  and opposite so that by cancelling each other , they create zero field.

\frac{\mu_0\times 10 }{2\times.09}  = \frac{\mu_0\times I }{2\times.15}

I = 16.66  A

Direction of current should be in opposite direction ie anticlockwise when looking from above.

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4 years ago
What kind of model is shown below?
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3 years ago
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[3 points] Question: Consider a pendulum made from a uniform, solid rod of mass M and length L attached to a hoop of mass M and
aliina [53]

Answer:

I=\frac{4}{3}ML^2+2MR^2+2MRL

Explanation:

We are given that

Mass of rod=M

Length of rod=L

Mass of hoop=M

Radius of hoop=R

We have to find the moment of inertia I of the pendulum about pivot depicted at the left end of the slid rod.

Moment of inertia of rod about center of mass=\frac{1}{12}ML^2

Moment of inertia of hoop about center of mass=MR^2

Moment of inertia of the pendulum about the pivot left end,I=\frac{1}{12}ML^2+M(\frac{L}{2})^2+MR^2+M(L+R)^2

Moment of inertia of the pendulum about the pivot left end,I=\frac{1}{12}ML^2+\frac{1}{4}ML^2+MR^2+MR^2+ML^2+2MRL

Moment of inertia of the pendulum about the pivot left end,I=\frac{1+3+12}{12}ML^2+2MR^2+2MLR

Moment of inertia  of the pendulum about the pivot left end,I=\frac{16}{12}ML^2+2MR^2+2MRL

Moment of inertia of the pendulum about the pivot left end,I=\frac{4}{3}ML^2+2MR^2+2MRL

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