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Nady [450]
3 years ago
7

Density of water 1000kgper m^3 What will be the volume of 3500kg water​

Physics
1 answer:
skad [1K]3 years ago
5 0

Answer:

<em>The volume of water is 3.5 cubic meter</em>

Explanation:

<u>Density </u>

The density of a substance or material is the mass per unit volume. The density varies with temperature and pressure.

The formula to calculate the density of a substance of mass (m) and volume (V) is:

\displaystyle \rho=\frac{m}{V}

We are given the density of water as \rho=1,000\ kg/m^3.It's required to find the volume of m=3,500 kg of water. Solving for V:

\displaystyle V=\frac{m}{\rho}

\displaystyle V=\frac{3,500}{1,000}

V=3.5\ m^3

The volume of water is 3.5 cubic meter

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Hydrogen gas (H2) can be found in trace amounts in Earth’s atmosphere. Which of these statements describes a physical property o
notka56 [123]
<span>Hydrogen gas (H2) can be found in trace amounts in Earth’s atmosphere.

a. Hydrogen is found in acids.
b. Hydrogen gas is highly flammable.
c. Hydrogen reacts with oxygen to form water.
These are true, but they all describe chemical properties
of Hydrogen, not physical properties.

d. Hydrogen gas is less dense than oxygen gas.
Yes !  This is a physical property of Hydrogen.
</span>
7 0
3 years ago
A video game includes an asteroid that is programmed to move in a straight line across a 17-inch monitor according to the equati
Ainat [17]

Answer:

The asteroid's acceleration at this point is 2.71\ m/s^2

Explanation:

The equation that governs the trajectory of asteroid is given by :

x=6.5t-2.3t^3

The velocity of asteroid is given by :

v=\dfrac{dx}{dt}\\\\v=\dfrac{d(6.5t-2.3t^3)}{dt}\\\\v=6.5-6.9t^2

At some point during the trip across the screen, the asteroid is at rest. It means, v = 0

So,

6.5-6.9t^2=0\\\\t=0.971\ s                      

Acceleration,

a=\dfrac{dv}{dt}\\\\a=\dfrac{d(6.5-6.9t^2)}{dt}\\\\a=-13.8t                        

Put t = 0.971 s

a=-13.8\times 0.197\\\\a=-2.71\ m/s^2

So, the asteroid's acceleration at this point is 2.71\ m/s^2 and it is decelerating.

6 0
3 years ago
Calculate the amount of heat liberated (in kj) from 411 g of mercury when it cools from 88.0°c to 12.0°c.
Katen [24]
You should note that the melting point of mercury is -38.83°C, while the boiling point is at 356.7°C. Then, that means that there is no latent heat involved here. We only compute for the sensible heat.

ΔH = mCpΔT
The Cp of mercury is 0.14 J/g·°C
Thus,
ΔH = (411 g)(0.14 J/g·°C)(88 - 12°C)
<em>ΔH = 4,373.04 J</em>
5 0
3 years ago
Water enters a baseboard radiator at 180 °F and at a flow rate of 2.0 gpm. Assuming the radiator releases heat into the room at
beks73 [17]

Answer:

Temperature of water leaving the radiator = 160°F

Explanation:

Heat released = (ṁcΔT)

Heat released = 20000 btu/hr = 5861.42 W

ṁ = mass flowrate = density × volumetric flow rate

Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³

ṁ = 1000 × 0.000126 = 0.126 kg/s

c = specific heat capacity for water = 4200 J/kg.K

H = ṁcΔT = 5861.42

ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C

And in change in temperature terms,

10°C= 18°F

11.08°C = 11.08 × 18/10 = 20°F

ΔT = T₁ - T₂

20 = 180 - T₂

T₂ = 160°F

8 0
3 years ago
What is same about all myriapods
seraphim [82]
They all have segmented limbs, a hard exoskeleton, a pair of antennae and a segmented body.
4 0
3 years ago
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