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3 years ago

Calculate the density of an object with a mass of 25g and a volume of 10ml

Chemistry

1 answer:

Gnoma [55]3 years ago

5 0

Answer:

<h3>The answer is 2.50 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 25 g

volume = 10 mL

We have

$density = \frac{25}{10} = \frac{5}{2} \\$

We have the final answer as

Hope this helps you

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Practice Problem: True Stress and Strain A cylindrical specimen of a metal alloy 49.7 mm long and 9.72 mm in diameter is stresse

amm1812

Answer:

The true stress required = 379 MPa

Explanation:

True Stress is the ratio of the internal resistive force to the instantaneous cross-sectional area of the specimen. True Strain is the natural log to the extended length after which load applied to the original length. The cold working stress – strain curve relation is as follows,

σ(t) = K (ε(t))ⁿ, σ(t) is the true stress, ε(t) is the true strain, K is the strength coefficient and n is the strain hardening exponent

True strain is given by

Epsilon t =㏑ (l/l₀)

Substitute㏑(l/l₀) for ε(t)

σ(t) = K(㏑(l/l₀))ⁿ

Given values l₀ = 49.7mm, l =51.7mm , n =0.2 , σ(t) =379Mpa

379 x 10⁶ = K (㏑(51.7/49.7))^0.2

K = 379 x 10⁶/(㏑(51.7/49.7))^0.2

K = 723.48 MPa

Knowing the constant value would be same as the same material is being used in the second test, we can find out the true stress using the above formula replacing the value of the constant.

σ(t) = K(㏑(l/l₀))ⁿ

l₀ = 49.7mm, l = 51.7mm, n = 0.2, K = 723.48Mpa

σ(t) = 723.48 x 106 x (㏑(51.7/49.7))^0.2

σ(t) = 379 MPa

The true stress necessary to plastically elongate the specimen is 379 MPa.

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3 years ago

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Which of the following substances (with specific heat capacity provided) would show the greatest temperature change upon absorbi

JulijaS [17]

Answer:

Pb is the substance that experiments the greatest temperature change.

Explanation:

The specific heat capacity refers to the amount of heat energy required to raise in 1 degree the temperature of 1 gram of substance. The highest the heat capacity, the more energy it would be required. These variables are related through the equation:

Q = c . m . ΔT

where,

Q is the amount of heat energy provided (J)

c is the specific heat capacity (J/g.°C)

m is the mass of the substance

ΔT is the change in temperature

Since the question is about the change in temperature, we can rearrange the equation like this:

$\Delta T = \frac{Q}{c.m}$

All the substances in the options have the same mass (m=10.0g) and absorb the same amount of heat (Q=100.0J), so the change in temperature depends only on the specific heat capacity. We can see in the last equation that they are inversely proportional; the lower c, the greater ΔT. Since we are looking for the greatest temperature change, It must be the one with the lowest c, namely, Pb with c = 0.128 J/g°C. This makes sense because Pb is a metal and therefore a good conductor of heat.

Its change in temperature is:

$\Delta T = \frac{q}{c.m} = \frac{100.0 J}{0.128 J/g.C . 10.0g } = 78.1$

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3 years ago

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