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arsen [322]
3 years ago
9

A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?

Physics
1 answer:
stiv31 [10]3 years ago
4 0

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, m_1 = 300\ g = 0.3\ kg

Initial speed of the cart, u_1=1.2\ m/s

Mass of the larger cart, m_2 = 2\ kg

Initial speed of the larger cart, u_2=0

After the collision,

Final speed of the smaller cart, v_1=-0.81\ m/s (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let v_2 is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

m_1u_1+m_2u_2-m_1v_1=m_2v_2

v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}

v_2=0.301\ m/s

So, the speed of the large cart after collision is 0.301 m/s.

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The rectangular plates in a parallel-plate capacitor are 0.063 m x 5.4 m. A distance of 3.5 x 10^-5m separate the plates. The pl
Aleks04 [339]

The capacitance of the capacitor is 1.8\cdot 10^{-9}F

Explanation:

The capacitance of a parallel-plate capacitor is given by the equation

C=\frac{k\epsilon_0 A}{d}

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Therefore, the capacitance of the  capacitor is

C=\frac{(2.1)(8.85\cdot 10^{-12})(0.0034)}{3.5\cdot 10^{-5}}=1.8\cdot 10^{-9}F

Learn more about capacitors:

brainly.com/question/10427437

brainly.com/question/8892837

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B is the answer that I know of.
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