Answer:
In a circuit ,<u> VOLTAGE </u>can be said to be the "source" or the "push of electrons". This push then creates what is known as a <u> CURRENT , </u>which is the flow of electric charge through the circuit. This flow can the slowed down or restricted by <u>RESISTOR </u>, and this is also what can be harnessed in order to use electric <u>ENERGY </u>.
Explanation:
Voltage:
It is the 'push' that causes charges to move in a wire or other electrical conductor, also it is a Source input to the electric circuit.
Measured in Volts.
Current:
An electric current is the rate of flow of electric charge from a point or through a region.
Measured in Ampere.
Resistor:
Resistor is used to resist the flow of charge or to resist the current called as Resistance.
Measured in Ohms.
Electric Energy:
Electrical energy is a form of energy resulting from the flow of electric charge.
Measured in Joules.
In a circuit , voltage can be said to be the "source" or the "push of electrons". This push then creates what is known as a current, which is the flow of electric charge through the circuit. This flow can the slowed down or restricted by resistor, and this is also what can be harnessed in order to use electric energy.
Answer:
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Answer: 757m/s
Explanation:
Given the following :
Mole of neon gas = 1.00 mol
Temperature = 465k
Mass = 0.0202kg
Using the ideal gas equation. For calculating the average kinetic energy molecule :
0.5(mv^2) = 3/2 nRt
Where ;
M = mass, V = volume. R = gas constant(8.31 jK-1 mol-1, t = temperature in Kelvin, n = number of moles
Plugging our values
0.5(0.0202 × v^2) = 3/2 (1 × 8.31 × 465)
0.0101 v^2 = 5796.225
v^2 = 5796.225 / 0.0101
v^2 = 573883.66
v = √573883.66
v = 757.55109m/s
v = 757m/s
I assume the block plows into the bank of sand with a velocity of 6 m/s and comes to a stop in 2 s.
Answer:
His third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A.
