Answer:
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.
(a) Determine the position of wire 3.
b) Determine the magnitude and direction of current in wire 3
Explanation:
a) 

position of wire = 50 - 1.2
= 48.8cm
b) 

Direction ⇒ downward
By applying Newton's second law of motion;
ma = mg - T
Where,
m = mass; a = downward accelerations (+ve value) or upward acceleration (-ve value); g = gravitational acceleration; T = tension.
For the current case, the velocity is constant therefore,
a = 0
Then,
0 = mg - T
T = mg = 115*9.81 = 1128.15 N
Tension in the cable is 1128.15 N.
Answer:
E = 1/2 M V^2 = 1/2 P V since P = M V
E2 / E1 = P2 V2 / (P1 V1)
P2 / P1 = E2 V1 / (E1 V2) = V2^2 V1 / (V1^2 V2) = V2 / V1
E2 / E1 = V2^2 / V1^2
V2 / V1 = (E2 / E1)^1/2
V2 / V1 = (.9)1/2 = .95
The linear momentum would have to decrease by 5%
Answer:
841 J/kg.K
Explanation:
The computation of the specific hear of the glass is shown below:
As we know that
E= cmΔt
where
c denotes specific heat
m denotes 0.38 kg
Δt = temperature = 22k
E denotes energy = 7032 J
Now
7032 J = (0.38) (22) (c)
7032 J = 8.36 (c)
So C = 7032 J ÷ 8.36
= 841 J/kg.K