Answer:
heat energy is used in boiling water and to make steam at power stations
Explanation:
Answer:
E = q V B describes the electric field induced
E Proportional to V B
while the magnet is pushed into the coil the induced field (B) will increase (consider 1 turn of the coil)
If V is constant the E-field will increase due to increasing B and the galvanometer will deflect accordingly
When V drops to zero the deflection must again be zero
So one would see a blip due to the deflection of the galvanometer
Note that as V increases the galvanometer will deflect one way and then as V drops to zero the deflection will be opposite (drop to zero when V is zero)
B always increases to a constant value because of the properties of the magnet.
Answer:
216 m
Explanation:
Assuming a straight line:
Δx = vt
Δx = (12 m/s) (18 s)
Δx = 216 m
Answer:
(a) A+B = 2i-3j
(B) A-B = 4i + j
Explanation:
We have given two vectors A = 3i-j and B = -1-2j
We have to find the two vectors that is A+B and A-B
(A) In first art we have calculate A+B for this we have to add simply vector A and v ector B
So A+B = 3i-j-i-2j = 2i-3j
(B) In this part we have to find A-B for this we have to simply subtract B from A so A-B = 3i-j-(-i-2j) =3i-j+i+2j =4i+j
Answer:
total work is = 52450 J
Explanation:
given data
mass = 5000-lb
density = 10 lb/ft
height = 50 ft
solution
as we will treat here cable and ball are separate
and
here work need to lift cable is
w = (10Δy )(9.8 y ) j
and
now summing all segment of cable
so passing limit Δy to 0
so total work need
= 
= ![[49 y^2]^{50}_0](https://tex.z-dn.net/?f=%5B49%20y%5E2%5D%5E%7B50%7D_0)
= 2450J
so lifting 5000 lb wrcking 50 m required additional 5000 + 2450
so total work is = 52450 J
