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Readme [11.4K]
3 years ago
7

47. Calculate the final steady temperature obtained when

Physics
1 answer:
Rashid [163]3 years ago
4 0

Answer:

the final steady temperature of the mixture is 46.67 ⁰C

Explanation:

Given;

mass of first water,m₁ = 6 g = 0.006 kg

initial temperature, t₁ = 50 ⁰C

mass of the second water, m₂ = 3.0 g = 0.003 kg

initial temperature of the second water, t₂ = 40 ⁰C

specific heat capacity of water, C = 4,200 J/kg.K

Let the final temperature of the mixture = T

Based on Law of conservation of energy, the temperature of the final mixture is calculated as follows;

m₁c(t₁ - T) = m₂c(T - t₂)

0.006 x 4200 x (50 - T) = 0.003 x 4200 x (T - 40)

1,260 - 25.2T = 12.6T - 504

1,260 + 504 = 12.6T + 25.2T

1764 = 37.8T

T = 1764/37.8

T = 46.67 ⁰C

Therefore, the final steady temperature of the mixture is 46.67 ⁰C

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Answer:

11.78meters

Explanation:

Given data

Mass m = 100kg

Length of cord= 10m

Spring constant k= 35N/m

At the greatest vertical distance, the spring potential energy is equal to the gravitational potential energy

That is

Us=Ug

Us= 1/2kx^2

Ug= mgh

1/2kx^2= mgh

0.5*35*10^2= 100*9.81*h

0.5*35*100=981h

1750=981h

h= 1750/981

h= 1.78

Hence the bungee jumper will reach 1.78+10= 11.78meters below the surface of the bridge

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diamong [38]

The answer & explanation for this question is given in the attachment below.

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Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.500 A when R2 is removed,
KATRIN_1 [288]

Answer:

a)   R₁ = 14.1 Ω,   b)  R₂ =  19.9 Ω

Explanation:

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all resistors connected

           V = i (R₁ + R₂)

with R₁ connected

           V = (i + 0.5) R₁

with R₂ connected

           V = (i + 0.25) R₂

We have a system of three equations with three unknowns for which we can solve it

We substitute the last two equations in the first

           V = i ( \frac{V}{ i+0.5} + \frac{V}{i+0.25} )

           1 = i ( \frac{1}{i+0.5} + \frac{1}{i+0.25} )

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with the second equation we look for R1

          R₁ = \frac{V}{i+0.5}

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with the third equation we look for R2

          R₂ = \frac{V}{i+0.25}

          R₂ =\frac{12}{0.35355+0.25}

          R₂ =  19.9 Ω

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