Answer: The magnitude of the current in the second wire 2.67A
Explanation:
Here is the complete question:
Two straight parallel wires are separated by 7.0 cm. There is a 2.0-A current flowing in the first wire. If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is the magnitude of the current in the second wire?
Explanation: Please see the attachments below
Answer:
268.2 m/s (1dp)
Explanation:
600 miles = 965,606m (1mile = 1609.34m)
965606/60 = 16,0943.43 ..... (metres traveled every minute)
16,0943.43/60 = 268.2238 .....
let the length of the beam be "L"
from the diagram
AD = length of beam = L
AC = CD = AD/2 = L/2
BC = AC - AB = (L/2) - 1.10
BD = AD - AB = L - 1.10
m = mass of beam = 20 kg
m₁ = mass of child on left end = 30 kg
m₂ = mass of child on right end = 40 kg
using equilibrium of torque about B
(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)
30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)
L = 1.98 m
Answer:a) 34.5 N; b) 24.5 N; c) 10 N; d) 1J
Explanation: In order to solve this problem we have to used the second Newton law given by:
∑F= m*a
F-f=m*a where f is the friction force (uk*Normal), from this we have
F= m*a+f=5 Kg*2 m/s^2+0.5*5Kg*9.8 m/s^2= 34.5 N
then f=uk*N=0.5*5Kg*9.8 m/s^2= 24.5N
the net Force = (34.5-24.5)N= 10 N
Finally the work done by the net force is equal to kinetic energy change so
W=∫Force net*dr= 10 N* 0.1 m= 1J
Answer:
Answer for A
Explanation:
F1=GmM/r1^2
If r2 becomes r2=5r
F2=GmM/(25r^2)
Multiply with 25 gives to maintain the same force
I.e.,25F2=F1
F2=G(25m)M/25r^2=F1
By the factor 25 would change to increase to same.
