Answer:
Explanation:
Capacitor of 0.75μF, charged to 70V and connect in series with 55Ω and 140 Ω to discharge.
Energy dissipates in 55Ω resistor is given by V²/R
Since the 55ohms and 140ohms l discharge the capacitor fully, the voltage will be zero volts and this voltage will be shared by the resistor in ratio.
So for 55ohms, using voltage divider rule
V=R1/(R1+R2) ×Vt
V=55/(55+140) ×70
V=19.74Volts is across the 55ohms resistor.
Then, energy loss will be
E=V²/R
E=19.74²/55
E=7.09J
7.09J of heat is dissipated by the 55ohms resistor
Answer:
50 N
Explanation:
Let the natural length of the spring = L
so
100 = k(40 - L) (1)
200 = k(60 - L) (2)
(2)/(1): 2 = (60 - L)/(40 - L)
60 - L = 2(40 - L)
60 - L = 80 - 2L
2L - L = 80 - 60
L = 20
Sub it into (1):
100 = k(40 - 20) = 20k
k = 100/20 = 5 N/in
Now
X = k(30 - L) = 5(30 - 20) = 50 N
Oil, grease and dry lubricants
Answer:
v = 0.489 m/s
Explanation:
It is given that,
Mass of a box, m = 1.5 kg
The compression in the spring, x = 6.5 cm = 0.065 m
Let the spring constant of the spring is 85 N/m
We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.
So, the speed of the box is 0.489 m/s.
