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Hatshy [7]
3 years ago
14

When a 5.0-kilogram cart moving with a speed of 2.8 meters per second on a horizontal surface collides with a 2.0 kilogram cart

initially at rest, the carts locked together. What is the speed of the combined carts after the collision
Physics
1 answer:
mixer [17]3 years ago
8 0

Here in all such collision type question we can use momentum conservation as we can see that there is no external force on this system

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

as we know that

m_1 = 5 kg

m_2 = 2 kg

v_{1i} = 2.8 m/s

v_{2i} = 0 m/s

now from above equation we have

5(2.8) + 2(0) = m_1v+ m_2v

14 = (5+ 2) v

v = 2 m/s

so the speed of combined system is 2 m/s

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An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.50 mm. A 25
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Answer:

The electric field is 16666.66 V/m.

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Explanation:

Given that,

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\sigma=E\times\epsilon_{0}

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\sigma=16666.66\times8.85\times10^{-12}

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C=\dfrac{\epsilon_{0}A}{d}

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C=\dfrac{8.85\times10^{-12}\times7.60\times10^{-4}}{1.50\times10^{-3}}

C=4.484\times10^{-12}\ F

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q=CV

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q=4.484\times10^{-12}\times25.0

q=112.1\times10^{-12}\ C

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The surface charge density is 1.474\times10^{-7}\ C/m^2

The capacitance is 4.484\times10^{-12}\ F

The charge on each plate is 112.1\times10^{-12}\ C

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