Answer:
b)1 :3
Explanation:
Lets that
The value of a positive charge = q
As we know that electric filed on a point charge given as
Where ,K=Constant
q=Charge ,r=Distance
If the value of the charge gets tripled ,q'= 3 q
Then electric filed E'
E' = 3 E
Therefore we can say that
therefore the answer will be --
b)1 :3
Answer: magnitude of applied force is FA = mg + F
Where F is the resultant force downward that the rope moves with
Explanation:
Force downwards F is,
F = FA - T
T is the upwards tension force on the rope
FA is the actual applied force in pulling the rope down.
Therefore, T = FA - F .....equ. (1)
For the box to move up with force ma ( it's mass times its acceleration upwards) upwards tension on the roap must exceed its own weight mg ( it's mass times acceleration due to gravity 9.8m/s^2)
Therefore, ma = T - mg
T = ma + mg ..... equ. (2)
Equating equ. 1 and 2
T = FA - F = ma + mg
Therefore FA = ma + mg + F
But at constant velocity a = 0
Magnitude of applied force becomes
FA = mg + F
See image below
Answer:
a) 23.51 m/s
b) 1.07 kg
Explanation:
Parameters given:
Kinetic energy, K = 295 J
Momentum, p = 25.1 kgm/s
a) The kinetic energy of a body is given as:
where m = mass of the body and v = speed of the body
We know that momentum is given as:
p = mv
Therefore:
K = 1/2 * pv
=> v = 2K / p
v = (2 * 295) / 25.1 = 23.51 m/s
The velocity of the body at that instant is 23.51 m/s.
b) Momentum is given as:
p = mv
=> m = p / v
m = 25.1 / 23.51 = 1.07 kg
The mass of the body at that instant is 1.07 kg