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Nataly_w [17]
3 years ago
15

A bag of sand has a density of 45 g/cm3 and a mass of 15 kg. How much space does the sand take up?

Physics
2 answers:
mamaluj [8]3 years ago
6 0

Answer:

Volume of the sand bag, V=333. 33 cm^3

Explanation:

Given that,

Density of sand bag, d=45\ g/cm^3

Mass of the sand bag, m = 15 kg = 15000 kg

We need to find the space taken by the sand bag. The occupied space is called the volume of the sand bag. Density is given by :

d=\dfrac{m}{V}

V=\dfrac{m}{d}

V=\dfrac{15000\ g}{45\ g/cm^3}

V=333.33\ cm^3

or

V=333.33\ cm^3

So, the volume of the sand bag is V=333.33\ cm^3. Hence, this is the required solution.

Aleks [24]3 years ago
3 0

Volume = mass/density

 

Volume = 15000 g/45 g/cm3 ≈ 333.3 cm<span>3</span>

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Which statement is true about the reaction shown by this chemical equation?
KatRina [158]

Answer:

it's C endothermic

Explanation:

5 0
2 years ago
An engineer in India (standard household voltage = 220 volts) is designing a transformer for use on her
Gnom [1K]
He should a step-up transformer with k=220/120=1.83 so output coil must have 240*1.83=440 turns

5 0
3 years ago
An experimenter adds 970 J of heat to 1.75 mol of an ideal gas to heat it from 10.0∘C to 25.0∘C at constant pressure. The gas do
ipn [44]

Answer:

(a) ΔU=747J

(b) γ=1.3

Explanation:

For (a) change in internal energy

According to first law of thermodynamics the change in internal energy is given as

ΔU=Q-W

Substitute the given values

ΔU=970J-223J

ΔU=747J

For(b) γ for the gas.

We can calculate γ by ratio of heat capacities of the gas

γ=Cp/Cv

Where Cp is the molar heat capacity at constant pressure

Cv is the molar heat capacity at constant volume

To calculate γ we first need to find Cp and Cv

So

For Cp

As we know

Q=nCpΔT

Cp=(Q/nΔT)

C_{p}=\frac{970J}{1.75mol*(25^{o}C-10^{o}C )}\\C_{p}=37J/mol.K

From relation of Cv and Cp we know that

Cp=Cv+R

Where R is gas constant equals to 8.314J/mol.K

So

C_{v}=C_{p}-R\\C_{v}=37-8.314\\C_{v}=28.687J/mol.K\\

So

γ=Cp/Cv

γ=[(37J/mol.K) / (28.687J/mol.K)]

γ=1.3

4 0
3 years ago
At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107
dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

K = - V\cdot \frac{dP}{dV}

Where:

K - Bulk module, measured in pascals.

V - Sample volume, measured in cubic meters.

P - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

-\frac{K \,dV}{V} = dP

This resultant expression is solved by definite integration and algebraic handling:

-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP

-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}

\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}

\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }

The final volume is predicted by:

V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }

If V_{o} = 1\,m^{3}, P_{o} - P_{f} = -10132500\,Pa and K = 2.3\times 10^{9}\,Pa, then:

V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }

V_{f} \approx 0.996\,m^{3}

Change in volume due to increasure on pressure is:

\Delta V = V_{o} - V_{f}

\Delta V = 1\,m^{3} - 0.996\,m^{3}

\Delta V = 0.004\,m^{3}

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

8 0
3 years ago
4.
xxTIMURxx [149]
Idk what to say to this
3 0
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