Explanation:
The work done equals the change in energy.
W = ΔKE
W = 0 − ½mv²
W = -½ (0.270 kg) (-7.50 m/s)²
W = -7.59 J
Work is force times displacement.
W = Fd
-7.59 J = F (-0.150 m)
F = 50.6 N
Answer:
The displacement of the spring due to weight is 0.043 m
Explanation:
Given :
Mass 
Kg
Spring constant 
According to the hooke's law,
Where 
force, 
displacement
Here,
( 
)
N
Now for finding displacement,
Here minus sign only represent the direction so we take magnitude of it.
m
Therefore, the displacement of the spring due to weight is 0.043 m
Force , F = ma
F = m(v - u)/t
Where m = mass in kg, v= final velocity in m/s, u = initial velocity in m/s
t = time, Force is in Newton.
m= 1.2*10³ kg, u = 10 m/s, v = 20 m/s, t = 5s
F = 1.2*10³(20 - 10)/5
F = 2.4*10³ N = 2400 N
Answer:
The time constant and its uncertainty is t ± Δt = 0.526 ± 0.057 s
Explanation:
If we make a comparison we have to:
y = A*(1-e^-(C*x)) + B
If the time remains constant we have to:
t = R*C = 1/C
In this way we calculate the time constant and its uncertainty. this will be equal to:
t ± Δt = (1/1.901) ± (0.2051/1.901)*(1/1.901) = 0.526 ± 0.057 s
In Millikan oil drop experiment, when the switch is opened and by altering supply the charge of electron is determined.
Explanation:
Millikan's oil drop experiment is held to determine the terminal velocity and charge of the oil drop.
Firstly without any supply of voltage when an oil drop is sprinkled and these droplets gather electrons together and gives negative charge as they pass through air.
By applying and altering voltage applied on the plates, drop can be suspended in air. Millikan observed one drop after another, varying the voltage and noting the effect. After many repetitions he concluded that charge could assume only certain fixed values.
After conducting many times he concluded 1.602176487 ×10−19 C as the charge of an electron.
