The magnitude of the magnetic force per unit length on the top wire is
2×10⁻⁵ N/m
<h3>How can we calculate the magnitude of the magnetic force per unit length on the top wire ?</h3>
To calculate the magnitude of the magnetic force per unit length on the top wire, we are using the formula
F=
Here we are given,
= magnetic permeability
= 4×10⁻⁷ H m⁻¹
If= 12 A
d= distance from each wire to point.
=0.12m
Now we put the known values in the above equation, we get
F=
Or, F =
Or, F= 2×10⁻⁵ N/m.
From the above calculation, we can conclude that the magnitude of the magnetic force per unit length on the top wire is 2×10⁻⁵ N/m.
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Density = mass /volume= 1000/100= 10 kg/m'3
Answer:
45.125 m/s
Explanation:
Acceleration is given by
r = 4 m/s
For maximum velocity
Maximum velocity
The maximum velocity of the rocket is 45.125 m/s
Answer:
No it can't .
Explanation:
In accordance to the law of conservation of energy
Answer:
a) 2.87 m/s
b) 3.23 m/s
Explanation:
The avergare velocity can be found dividing the length traveled d by the total time t.
a)
For the first part we easily know the total traveled length which is:
d = 50.2 m + 50.2 m = 100.4 m
The time can be found dividing the distance by the velocity:
t1 = 50.2 m / 2.21 m/s = 22.7149 s
t2 = 50.2 m / 4.11 m/s = 12.2141 s
t = t1 +t2 = 34.9290 s
Therefore, the average velocity is:
v = d/t =2.87 m/s
b)
Here we can easily know the total time:
t = 1 min + 1.16 min = 129.6 s
Now the distance wil be found multiplying each velocity by the time it has travelled:
d1 = 2.21 m/s * 60 s = 132.6 m
d2 = 4.11 m/s *(1.16 * 60 s) = 286.056 m
d = 418.656 m
Therefore, the average velocity is:
v = d/t =3.23 m/s