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Tcecarenko [31]
4 years ago
6

Two balls, each with a mass of 0.5 kg, collide on a pool table. Is the law of conservation of momentum satisfied in this collisi

on? Explain why or why not.
Physics
2 answers:
mart [117]4 years ago
6 0
Conservation of momentum<span> is a fundamental law of physics. This law states that the </span>momentum<span> of a system is constant if there are </span>no external forces acting on the system. In a situation in which two balls, each with a mass of 0.5 kg, collide on a pool table<span> the law of conservation of momentum is not satisfied because there are external forces that moved the balls. </span>
swat324 years ago
4 0
Yes, the law of conservation of momentum is satisfied. The total momentum before the collision is 1.5 kg • m/s, and the total momentum after the collision is 
<span>1.5 kg • m/s. The momentum stays the same after the collision.</span>
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What are the types of nuclear radiation ? (Particles let off by nuclear<br> materials.) *
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Answer:

Because nuclear decay releases ionizing radiation (alpha, beta, and gamma radiation) we call such compounds radioactive.

Explanation:

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3 years ago
Determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass.
olga2289 [7]

Answer:

NO_{1.499}

Explanation:

Let assume that 100 kg of the compound is tested. The quantity of kilomoles for each element are, respectively:

n_{N} = \frac{36.86\,kg}{14.006\,\frac{kg}{kmol} }

n_{N} = 2.632\,kmol

n_{O} = \frac{63.14\,kg}{15.999\,\frac{kg}{kmol} }

n_{O} = 3.946\,kmol

Ratio of kilomoles oxygen to kilomole nitrogen is:

n^{*} = \frac{3.946\,kmol}{2.632\,kmol}

n^{*}= 1.499

It means that exists 1.499 kilomole oxygen for each kilomole nitrogen.

The empirical formula for the compound is:

NO_{1.499}

8 0
4 years ago
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A mass with a charge of 4.60 x 10-7 C rests on a frictionless surface. A compressed spring exerts a force on the mass on the lef
stira [4]

Answer:

The compression of the spring is 24.6 cm

Explanation:

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magnitude of the charge on the right, q₂ = 7.5 x 10⁻⁷ C

distance between the two charges, r = 3 cm = 0.03 m

spring constant, k = 14 N/m

The attractive force between the two charges is calculated using Coulomb's law;

F = \frac{kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(4.6\times 10^{-7})(7.5\times 10^{-7})}{(0.03)^2} \\\\F= 3.45 \ N

The extension of the spring is calculated as follows;

F = kx

x = F/k

x = 3.45 / 14

x = 0.246 m

x = 24.6 cm

The compression of the spring is 24.6 cm

4 0
3 years ago
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Answer:

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Explanation:

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3 years ago
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