Answer:
Explanation:
- given S = distance from the first = 3.20cm = 0.032m, t = 1.30×10−8 s
- acceleration = 0.032 X 2 /(1.30×10−8)^2
a = 3.79 x 10^14m/s^2
E = ma /q = 9.11 x 10^-31 x 3.79 x 10^14 / 1.6 x 10^-19
E = magnitude of this electric field. = 2156.3N/C
b) Find the speed of the electron when it strikes the second plate ; V^2 = 2as
= 2 X 3.79 x 10^14 X 0.032
= 4.92 X 10^6m/s
Answer:
The fuse must be connected between the device and the power intake source.
Explanation:
A fuse is a protective component of electrical appliances that is designed to be sensitive to a particular range of electric current
The fuse is made of a thing metal strip with a known melting point. Once current abive its carrying capacity flows through it, large heat is generated in the metal strip which melts it and causes the metal strip to cut int two protecting the device from the power spike.
Constant velocity means the netto force = 0, therefore F(gravity) = F(astronaut).
175N divided by 87,5kg = 2.00kg/N
Answer:
work done is -150 kJ
Explanation:
given data
volume v1 = 2 m³
pressure p1 = 100 kPa
pressure p2 = 200 kPa
internal energy = 10 kJ
heat is transferred = 150 kJ
solution
we know from 1st law of thermodynamic is
Q = du +W ............1
put here value and we get
-140 = 10 + W
W = -150 kJ
as here work done is -ve so we can say work is being done on system
Answer:
The speed of the heavier fragment is 0.335c.
Explanation:
Given that,
Mass of the lighter fragment
Mass of the heavier fragment
Speed of lighter fragment = 0.893c
We need to calculate the speed of the heavier fragment
Let v is the speed of the second fragment after decay
Using conservation of relativistic momentum
Hence, The speed of the heavier fragment is 0.335c.