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masya89 [10]
3 years ago
5

. A 2.00 kg ball is attached to a ceiling by a string. The distance from the ceiling to the center of the ball is 1.00 m, and th

e height of the room is 3.00 m. What is the gravitational potential energy associated with the ball relative to each of the following? (a) the ceiling (b) the floor (c) a point at the same elevation as the ball
Physics
1 answer:
cupoosta [38]3 years ago
4 0
<h2>a) Potential energy of ball relative to ceiling is 19.62 J</h2><h2>b) Potential energy of ball relative to floor is 39.24 J</h2><h2>c) Potential energy of ball relative to same elevation is 0 J</h2>

Explanation:

Mass of ball, m = 2 kg

Acceleration due to gravity, g = 9.81 m/s²

Height of ball from ground = 3 - 1 = 2 m

Potential energy, PE = mgh

Potential energy of ball, PE = 2 x 9.81 x 2 = 39.24 J

a) Potential energy of ball at ceiling = 2 x 9.81 x 3 = 58.86 J

Potential energy of ball relative to ceiling = 58.86 - 39.24 = 19.62 J

b) Potential energy of ball at floor = 2 x 9.81 x 0 = 0 J

Potential energy of ball relative to floor = 39.24 - 0 = 39.24 J

c) Potential energy of ball at same elevation = 2 x 9.81 x 2 = 39.24 J

Potential energy of ball relative to same elevation = 39.24 - 39.24 = 0 J

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Answer: a) 3.85 days

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Explanation:-

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

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t = time taken for decomposition  = 3 days

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{58}{100}\times 100=58g

First we have to calculate the rate constant, we use the formula :

Now put all the given values in above equation, we get

k=\frac{2.303}{3}\log\frac{100}{58}

k=0.18days^{-1}

a) Half-life of radon-222:

t_{\frac{1}{2}}=\frac{0.693}{k}

t_{\frac{1}{2}}=\frac{0.693}{0.18}=3.85days

Thus half-life of radon-222 is 3.85 days.

b) Time taken for the sample to decay to 15% of its original amount:

where,

k = rate constant  = 0.18days^{-1}

t = time taken for decomposition  = ?

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{15}{100}\times 100=15g

t=\frac{2.303}{0.18}\log\frac{100}{15}

t=10.54days

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Answer:

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Explanation:

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Answer:

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Explanation:

Given that:

<u><em>initially:</em></u>

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volume of gas, = V_1

<em><u>finally:</u></em>

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