Explanation:
When water is collected from vapour form to liquid form then this process is known as condensation.
For example, water evaporated from the atmosphere reaches to the clouds and it gets converted into liquid form and again reaches to Earth in the form of rain.
Thus, we can conclude that process in which water vapor turns to liquid is called condensation.
PLZZ
Answer:
speaker
Explanation:
speaker runs by the help of electrical energy and the magnetic energy is used and finally it is converted into sound energy
Answer:
the motor in today's standard power drills. the motor in today's standard power saws. the motor in an electric tooth brushes.
Explanation:
More axamples could be that both devices work because of electromagnetic induction, which is when a voltage is induced by a changing magnetic field.
Using the formula q = mcΔT, and substituting the values for mass, specific heat capacity of iron, and temperature change, it is calculated that it takes approximately 3.058 KJ to warm 125 g of iron from 23.5 °C to 78.0 °C.
To calculate the amount of heatneeded to warm 125 g of iron from 23.5 °C to 78.0 °C, we use the formula q = mcΔT, where 'm' is the mass in kilograms, 'c' is the specific heat capacity, and 'ΔT' is the temperature change. In this case, the mass 'm' is 0.125 kg (since 1 g = 10^-3 kg), the specific heat capacity 'c' of iron is 0.449 J/g°C (or 449 J/kg°C), and 'ΔT' is 78.0 °C - 23.5 °C = 54.5 °C.
Substituting these values into the formula, we get q = (0.125 kg) * (449 J/kg°C) * (54.5 °C), which gives a result of approximately 3.058 KJ.
Therefore, it would take approximately 3,058 KJ to warm 125 g of iron from 23.5 °C to 78.0 °C.
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To warm 125 g of iron from 23.5 °C to 78.0 °C, it requires approximately 3.93 kilojoules of energy.
To calculate the number of kilojoules required to warm 125 g of iron from 23.5 °C to 78.0 °C, we can use the formula:
q = m * c * ΔT
Where:
Using the given values:
Substituting the values into the formula:
q = 125 g * 0.450 J/g°C * (78.0 °C - 23.5 °C)
Simplifying the equation:
q = 125 * 0.450 * (78.0 - 23.5)
q ≈ 3933.75 J ≈ 3.93 kJ
Therefore, it requires approximately 3.93 kilojoules of energy to warm 125 grams of iron from 23.5 °C to 78.0 °C.
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c. 4
b. 3
d. 5 Please select the best answer from the choices provided A B C D
How many valence electrons are in an atom of phosphorus?
a. 2
c. 4
b. 3
d. 5 Please select the best answer from the choices provided A B C D
There are five valence electrons are in an atom of phosphorus. The answer is letter D.
to move?
Select one:
0 The weight of the box is less than the force of friction.
O
The weight of the box is greater than the force of friction.
The applied force is greater than the force of friction.
O
The applied force is less than the force of friction
Answer:
the applied force is less than the force of friction.
Explanation: