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timurjin [86]
3 years ago
5

A 2500-kg car falls over a cliff and converts 1000,000J of PE into KE.

Physics
1 answer:
MArishka [77]3 years ago
7 0

B is correct makes more sense

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A cart is loaded with a brick and pulled at constant speed along an inclined plane to the height of a seat-top. If the mass of t
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Answer:

13.23J

Explanation:

PE = m*g*h

PE = (3 kg ) * (9.8 m/s/s) * (0.45 m)

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Which of the following is the best name for NO3?
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<span>Nitrogen trioxide is the best name for NO3. </span>
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Lightning a. occurs when a positively charged cloud base induces a negative charge on the Earth's surface. b. travels most often
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Answer:

c. may be accompanied by the sound of explosively expanding hot air, called thunder.

Explanation:

Lightning is a discharge which is due to the reaction between oppositely charged charges in the clouds, or between clouds base and the Earth surface.

The motion of the cloud causes charging of clouds by friction, thus the reaction between opposite charges (jumping of charges from one cloud to another) in the cloud can lead to lightning. Also, oftentimes the bottom of a cloud is negatively charged so that this is attracted to the positive charge on the earth surface. Thus leading to a discharge called lightning.

Thus in the given question, the appropriate option is C. This implies that, lightning may be accompanied by the sound of explosively expanding hot air, called thunder.

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A solid non-conducting sphere of radius R carries a charge Q1 distributed uniformly. The sphere is surrounded by a concentric sp
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Answer:

E = k Q₁ / r²

Explanation:

For this exercise that asks us for the electric field between the sphere and the spherical shell, we can use Gauss's law

           Ф = ∫ E .dA = q_{int} / ε₀

where Ф the electric flow, qint is the charge inside the surface

To solve these problems we must create a Gaussian surface that takes advantage of the symmetry of the problem, in this almost our surface is a sphere of radius r, that this is the sphere of and the shell, bone

           R <r <R_a

for this surface the electric field lines are radial and the radius of the sphere are also, therefore the two are parallel, which reduces the scalar product to the algebraic product.

         E A = q_{int} /ε₀

The charge inside the surface is Q₁, since the other charge Q₂ is outside the Gaussian surface, therefore it does not contribute to the electric field

          q_{int} = Q₁

The surface area is

          A = 4π r²

we substitute

          E 4π r² = Q₁ /ε₀

          E = 1 / 4πε₀ Q₁ / r²

          k = 1/4πε₀

 

          E = k Q₁ / r²

6 0
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