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Kay [80]
3 years ago
10

A teacher makes the following statement.

Physics
2 answers:
Lostsunrise [7]3 years ago
8 0

Answer: Distillation of mixture

Explanation: Different fractions of crude oil have different

boiling points according molecular weight of hydrocarbons

Papessa [141]3 years ago
6 0
NOT evaporation of a mixture
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Describe how Ubuntu could help to lack of basic services challenges<br>​
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Explanation:

Ubuntu is somewhat a South African concept that involves charity, sympathy, and mainly underlines the concept of universal brotherhood. Hence this concept can help fight social challenges such as racism, crime, violence and many more. It can contribute to maintaining peace and harmony in the country at large

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8 0
3 years ago
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A solid metal ball of radius 1.5 cm bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing a
blsea [12.9K]

We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

E=7*10^{9}N/C

From the question we are told that

A solid metal ball of radius 1.5 cm

bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing

uniformly distributed charge of -7 nC

The distance between the centers of the balls is 9 cm

Generally the equation for the electric field  is mathematically given as

E=\frac{kq_2}{d^2}\\\\E=\frac{(9*10^9)7*10^{-2}}{9*10^{-2}}\\\\

E=7*10^{9}N/C

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brainly.com/question/21811998

4 0
3 years ago
In an experiment Rahul burnt 4 kg fuel and the heat produced was measured to be 160,000 kJ and Raj burnt 2 kg of fuel and heat p
Gnesinka [82]

Answer:

The net calorific value of the fuel 1 is 40000 kilojoules per kilogram.

The net calorific value of the fuel 2 is 50000 kilojoules per kilogram.

Explanation:

The net calorific value is equal to the heat produced (Q), in kilojoules, divided by the mass of the burnt fuel (m). We proceed to calculate the net calorific value of each fuel:

Fuel 1 (m = 4\,kg, Q = 160000\,kJ)

L = \frac{160000\,kJ}{4\,kg}

L = 40000\,\frac{kJ}{kg}

The net calorific value of the fuel 1 is 40000 kilojoules per kilogram.

Fuel 2 (m = 2\,kg, Q = 100000\,kJ)

L = \frac{100000\,kJ}{2\,kg}

L = 50000\,\frac{kJ}{kg}

The net calorific value of the fuel 2 is 50000 kilojoules per kilogram.

6 0
3 years ago
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