The concept required to solve this problem is associated with potential energy. Recall that potential energy is defined as the product between mass, gravity, and change in height. Mathematically it can be described as
Here,
= Change in height
m = mass of super heroine
g = Acceleration due to gravity
The change in height will be,
The final position of the heroin is below the ground level,
The initial height will be the zero point of our system of reference,
Replacing all this values we have,
Since the final position of the heroine is located below the ground, there will net loss of gravitational potential energy of 10744.81J
Answer:
Work done, W = 0.0219 J
Explanation:
Given that,
Force constant of the spring, k = 290 N/m
Compression in the spring, x = 12.3 mm = 0.0123 m
We need to find the work done to compress a spring. The work done in this way is given by :
W = 0.0219 J
So, the work done by the spring is 0.0219 joules. Hence, this is the required solution.
Answer:
it will take 36.12 ms to reduce the capacitor's charge to 10 μC
Explanation:
Qi= C×V
then:
Vi = Q/C = 30μ/20μ = 1.5 volts
and:
Vf = Q/C = 10μ/20μ = 0.5 volts
then:
v = v₀e^(–t/τ)
v₀ is the initial voltage on the cap
v is the voltage after time t
R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant
τ = 20µ x 1.5k = 30 ms
v = v₀e^(t/τ)
0.5 = 1.5e^(t/30ms)
e^(t/30ms) = 10/3
t/30ms = 1.20397
t = (30ms)(1.20397) = 36.12 ms
Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.
Answer:
technically yes
Explanation:
with a gun depending on how fast it shoots so when you fire at something you shoot in front of it a little bit so you hit it but a plane that fast you shoot like 100 feet infront of it...
Answer:
The answer to your question is: V2 = 1 l
Explanation:
Data
P1 = 200 kPa
P2 = 300 kPa
V1 = 1.5 l
V2 = ?
Formula
P1V1 = P2V2
V2 = (P1V1) / P2
V2 = (200 x 1.5) / 300
V2 = 1 l
