What is the energy equivalent of 3 eV? a) 1.6x10-19J b) 3.2x10-19J c) 4.8x10-19J d) 0.53x10-19J
1 answer:
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Answer:
a) v² = v₀² + 2 g h, b) t = v₀/g (1+ √ (1 + 2gh/ v₀²))
Explanation:
a) This is an exercise that we can solve using conservation of energy.
Starting point. High point
Em₀ = K + U = ½ m v₀² + m gh
Final point. Soil
= K = ½ m v²
energy is conserved because there is no friction
Emo = Em_{f}
½ m v₀² + m g h = ½ m v²
v² = v₀² + 2 g h
b) the time it takes to reach the ground can be calculated with kinematics
let's create a reference frame with positive upward direction
v = vo - g t
when it reaches the ground it has a velocity v, the initial velocity is downwards v₀ = -v₀
v = -v₀ - gt
t = - (v + v₀) / g
we substitute the velocity values calculated in the previous part
t = - (√(v₀² + 2 g h) + vo) / g
we will simplify the equation a bit
t = - v₀/g (1+ √ (1 + 2gh/ v₀²))
c) is now thrown vertically upward with the same initial velocity vo.
To find the final velocity we use the conservation of energy where the velocity is squared, so it does not matter if it is positive or negative, therefore in this section the value should be the same as in part a
v = √ (v₀² + 2gh)
d) for this part if there is change since the speed is not squared
v₀ = v₀
v = v₀ - gt
t = (v₀ - v) / g
t = (v₀ - √(v₀² + 2 g h)) / g
t = v₀/g (1 - √(1 + 2gh / v₀²))
Answer:
Final volume, V2 = 24.62 L
Explanation:
Given the following data;
Initial volume = 40 L
Initial pressure = 80 Pa
Final pressure = 130 Pa
To find the final volume V2, we would use Boyles' law.
Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.
Mathematically, Boyles law is given by;
Substituting into the equation, we have;
Final volume, V2 = 24.62 Liters
Answer:
The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%
Explanation:
Mechanical energy (Potential energy, PE) of the oscillator is calculated as;
PE = ¹/₂KA²
During the first oscillation;
PE₁ = ¹/₂KA₁²
During the second oscillation;
A₂ = A₁ - 0.0342A₁ = 0.9658A₁
PE₂ = ¹/₂KA₂²
PE₂ = ¹/₂K (0.9658A₁)²
PE₂ = (0.9658²)¹/₂KA₁²
PE₂ = (0.9328)¹/₂KA₁²
PE₂ = 0.9328PE₁
Percentage of the mechanical energy of the oscillator lost in each cycle;
Therefore, the percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%