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3 years ago

A heavy fuel oil has a specific gravity of 0.918. How much will 100 gallons(300 liters) of this oil weigh? INCLUDE UNITS!

Physics

1 answer:

nalin [4]3 years ago

6 0

Answer:

275.4 kg

Explanation:

specific gravity = 0.918

Density = 0.918 g/cm^3 = 918 kg/m^3

Volume = 100 gallons = 300 litres = 300 x 10^-3 m^3 = 0.3 m^3

Mass = volume x density = 0.3 x 918 = 275.4 kg

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An insulating sphere is 8.00 cm in diameter and carries a 6.50 ÂµC charge uniformly distributed throughout its interior volume.

Kobotan [32]

Explanation:

(a) Formula to calculate the density is as follows.

$\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

= $2.42 \times 10^{-2} C/m^{3}$

Now, calculate the charge as follows.

$q_{in} = \rho(\frac{4}{3} \pi r^{3})$

= $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

= $10.106 \times 10^{-8}$ C

or, = 101.06 nC

(b) For r = 6.50 cm, the value of charge will be calculated as follows.

$q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

= 7.454 $\mu C$

7 0

3 years ago

620 J of heat is added to the cylinder of an engine, which causes the gas inside to expand. As a result, the piston of the engin

Maru [420]

Answer:

400 J work is done BY the engine.

The internal energy of the gas is 620 J

Explanation:

The given information are;

The heat added to the cylinder = 620 J

The force applied by the piston of the engine = 8.0 kN = 8,000 N

The distance over which the force moves (the piston) = 5.0 cm = 0.05 m

The work done (by the engine) = Force × Distance = 8,000 N × 0.05 m = 400 J

The internal energy is the sum of the kinetic and potential energy of the system

Therefore, given that the internal energy, U, is the sum total of the energy in the system

∴ U = The heat supplied to the system = 620 J

Which gives;

400 J work is done BY the engine.

The internal energy of the gas is 620 J.

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3 years ago

What is the chief benefit of anaerobic exercise?

ra1l [238]

Answer:

c isthe answer if not see my profile m e. e. t me I will explain u

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3 years ago

PLEASE ANSWER ASAP!!!!!!!

Viefleur [7K]

Answer:

B. Mechanical energy= 50J+30J=80J

4 0

3 years ago

The current that flows from and back to the power supply in a parallel circuit is called __________ current.

blagie [28]

Answer:

mainline current

Explanation:

Current that flows from and back to the power supply in a parallel circuit. Fuse. A type of circuit protection device.

6 0

3 years ago