Answer:
A) ΔU = 3.9 × 10^(10) J
B) v = 8420.75 m/s
Explanation:
We are given;
Potential Difference; V = 1.3 × 10^(9) V
Charge; Q = 30 C
A) Formula for change in energy of transferred charge is given as;
ΔU = QV
Plugging in the relevant values gives;
ΔU = 30 × 1.3 × 10^(9)
ΔU = 3.9 × 10^(10) J
B) We are told that this energy gotten above is used to accelerate a 1100 kg car from rest.
This means that the initial potential energy will be equal to the final kinetic energy since all the potential energy will be converted to kinetic energy.
Thus;
P.E = K.E
ΔU = ½mv²
Where v is final velocity.
Plugging in the relevant values;
3.9 × 10^(10) = ½ × 1100 × v²
v² = [7.8 × 10^(8)]/11
v² = 70909090.9090909
v = √70909090.9090909
v = 8420.75 m/s
E = ½KA^2 is the mechanical energy of any oscillator. It is the sum of elastic potential energy and
kinetic energy. When amplitude A
decreases by 3%, then
(E2-E1)/E1 = {½K(A2^2/A1^2) }/ ½K(A1^2)
= {(A2^2 – A1^2) / (A1^2)}
= 97^2 – 100^2/100^2
= 5.91% of the mechanical energy is lost each cycle.
Acceleration is the change of velocity, and velocity is the change of distance. The opposite of finding change, or differentiation, is integration.
Acceleration = 1.3 m/s²
Velocity: ∫ 1.3 dx = 1.3x + c m/s
Distance: ∫ 1.3x dx = 1.3x²/2 + c m
Distance run: 1.3*3²/2 = 5.85 m
<em>What</em><em> </em><em>bad</em><em> </em><em>thing</em><em> </em><em>happened</em><em>?</em>
Since the current is inversely propotional to its resistance, when the voltage is doubled the current will be one-half