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s344n2d4d5 [400]

2 years ago

14

A lump of steel of mass 10kg at 627 degree Celsius is dropped in 100kg oil at 30 degree Celsius . the specific heat of steel And

oil are 0.5kj/kg.k and 3.5kj/kg.k calculate the entropy change in steel,oil and in the universe.

1 answer:

Naily [24]2 years ago

8 0

Answer:

700J

Explanation:

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Why does it take significantly stronger magnetic and electric field strengths to move the beam of alpha particles compared with

wlad13 [49]

It takes significantly stronger magnetic and electric field strengths to move a beam of alpha particles compared with the beam of electrons(betaparticles) because the charge of an alpha particle is twice stronger than a beta particle. Therefore, more energy is needed to move the alpha particle.

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3 years ago

Eugene describes the physical property of a material as “sweet and floral.” What physical property of the material is Eugene mos

shusha [124]

The correct answer would be odor. Because it's sweet. Boiling shape and hardness have nothing to do with sweet and floral :)

7 0

2 years ago

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The magnitude of each force is 208 N the force on the right is applied at an angle 36° and the mass of the block is 17 kg the co

djyliett [7]

Answer:

<em>11.06m/s²</em>

Explanation:

According to Newtons second law of motion

$\sm F_x = ma_x\\F_m - F_f = ma_x\\mgsin \theta - \mu R mgcos \theta = ma_x\\$

Given

Mass m = 17kg

Fm = 208N

theta = 36 degrees

g = 9.8m/s²

a is the acceleration

Substitute

208 - 0.148(17)(9.8)cos 36 = 17a

208 - 24.6568cos36 = 17a

208 - 19.9478 = 17a

188.05 = 17a

a = 188.05/17

a = 11.06m/s²

<em>Hence the the magnitude of the resulting acceleration is 11.06m/s²</em>

6 0

2 years ago

Help not to sure with this one need help plz asap

DiKsa [7]

A is the answer for the problem

4 0

2 years ago

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Ten identical steel wires have equal lengths L and equal "spring constants" k. The wires are connected end to end so that the re

lapo4ka [179]

Answer:

$K_{system} = \frac{k}{10}$

Explanation:

When the springs are connected end to end, it means they are connected in series. When the springs are connected in series, the stress applied to the system gets applied to each of the springs without any change in magnitude while the strain of the system is the sum total of strains of each spring. The spring constant of the resultant system is given as,

$\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{n}})$

Here, n = 10

Spring constant of each spring = k

Thus,

$\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{10}})$

$\frac{1}{K_{system}} = (\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})$

$\frac{1}{K_{system}} = \frac{10}{k}$

$K_{system} = \frac{k}{10}$

7 0

3 years ago

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