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dimulka [17.4K]
3 years ago
5

How is the mass of an object affected when the object changes state?

Physics
1 answer:
MrMuchimi3 years ago
3 0

From the transitions of ice to water, water to steam, steam to water,
or water to ice, mass doesn't change.


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a roller coaster has a mass of 275kg. it sits at the top of a hill with a height 85m if it drops from this hill, how fast is it
prohojiy [21]
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3 0
4 years ago
A string of length L, mass per unit length \mu, and tension T is vibrating at its fundamental frequency. What effect will the fo
viva [34]

The fundamental frequency on a vibrating string is given by:

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where

L is the length of the string

T is the tension

\mu is the mass per unit length of the string

Keeping this equation in mind, we can now answer the various parts of the question:

(a) The fundamental frequency will halve

In this case, the length of the string is doubled:

L' = 2L

Substituting into the expression of the fundamental frequency, we find the new frequency:

f'=\frac{1}{2(2L)}\sqrt{\frac{T}{\mu}}=\frac{1}{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{2}

So, the fundamental frequency will halve.

(b) the fundamental frequency will decrease by a factor \sqrt{2}

In this case, the mass per unit length is doubled:

\mu'=2\mu

Substituting into the expression of the fundamental frequency, we find the new frequency:

f'=\frac{1}{2L}\sqrt{\frac{T}{2 \mu}}=\frac{1}{\sqrt{2}}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{\sqrt{2}}

So, the fundamental frequency will decrease by a factor \sqrt{2}.

(c) the fundamental frequency will increase by a factor \sqrt{2}

In this case, the tension is doubled:

T'=2T

Substituting into the expression of the fundamental frequency, we find the new frequency:

f'=\frac{1}{2L}\sqrt{\frac{2T}{\mu}}=\sqrt{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\sqrt{2}f

So, the fundamental frequency will increase by a factor \sqrt{2}.

8 0
4 years ago
When a ball player throws a ball straight up, by how much does the speed of the ball decrease each second while ascending?
Alexxandr [17]
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Hope it helps
6 0
3 years ago
Calculate the entropy change that occurs when 1.0kg of water at 20.00 C is mixed with 2.0kg of water at 80.00 C
SOVA2 [1]

Answer:

The change in entropy ΔS = 0.0011 kJ/(kg·K)

Explanation:

The given information are;

The mass of water at 20.0°C = 1.0 kg

The mass of water at 80.0°C = 2.0 kg

The heat content per kg of each of the mass of water is given as follows;

The heat content of the mass of water at 20.0°C = h₁ = 83.92 KJ/kg

The heat content of the mass of water at 80.0°C = h₂ = 334.949 KJ/kg

Therefore, the total heat of the the two bodies = 83.92 + 2*334.949 = 753.818 kJ/kg

The heat energy of the mixture =

1 × 4200 × (T - 20) = 2 × 4200 × (80 - T)

∴ T = 60°C

The heat content, of the water at 60° = 251.154 kJ/kg

Therefore, the heat content of water in the 3 kg of the mixture = 3 × 251.154 = 753.462

The change in entropy ΔS = ΔH/T = (753.818 - 753.462)/(60 + 273.15) = 0.0011 kJ/(kg·K).

8 0
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4 0
4 years ago
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