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Natali5045456 [20]
3 years ago
15

Water has a very high specific heat capacity when compared to most other common materials. In fact, ethyl alcohol has a specific

heat that is only about that of water, whereas the specific heat of lead is about that of water. Suppose that you have equal‑mass samples of each material and that each sample is at the same initial temperature. You then carefully transfer the same amount of heat into each sample and measure the resulting final temperature of each. Rank the final temperature of each sample from highest to lowest.
Physics
1 answer:
AveGali [126]3 years ago
5 0

Answer:

Lead, Ethyl alcohol and water.

Explanation:

Specific heat capacity of a substance can be define as the quantity of heat that is absorbed by a substance needed to change the temperature of a unit mass of one kilogram of the substance by one kelvin

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Assume this is an acceleration graph, where the X axis represents time in seconds and the Y axis represent velocity in m/s. Whic
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D) Acceleration is positive and increasing.

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Acceleration is defined as the rate of change of velocity per unit time; in formulas:

a=\frac{\Delta v}{\Delta t}

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The graph shows the velocity vs the time of a moving object. We can see that \Delta v is the increment on the y-axis, while \Delta t is the increment on the x axis: therefore, the ratio \frac{\Delta v}{\Delta t} is the slope of the curve. In fact, in a velocity-time graph, the slope of the curve corresponds to the acceleration of the object.

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repel

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Which statement forms the basis of Faraday’s law of induction?
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A 4g bullet, travelling at 589m/s embeds itself in a 2.3kg block of wood that is initially at rest, and together they travel at
VARVARA [1.3K]

Answer:

The  percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %

Explanation:

Given;

mass of bullet, m₁ = 4g = 0.004kg

initial velocity of bullet, u₁ = 589 m/s

mass of block of wood, m₂ = 2.3 kg

initial velocity of the block of wood, u₂ = 0

let the final velocity of the system after collision = v

Apply the principle of conservation of linear momentum

m₁u₁ + m₂u₂ = v(m₁+m₂)

0.004(589) + 2.3(0) = v(0.004 + 2.3)

2.356 = 2.304v

v = 2.356 / 2.304

v = 1.0226 m/s

Initial kinetic energy of the system

K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E₁ = ¹/₂(0.004)(589)² = 693.842 J

Final kinetic energy of the system

K.E₂ = ¹/₂v²(m₁ + m₂)

K.E₂ = ¹/₂ x 1.0226² x (0.004 + 2.3)

K.E₂ = 1.209 J

The kinetic energy left in the system = final kinetic energy of the system

The percentage of the kinetic energy that is left in the system after collision to that before = (K.E₂ / K.E₁) x 100%

                       = (1.209 / 693.842) x 100%

                        = 0.174 %

Therefore, the  percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %

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3 years ago
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