To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:
Where,
B= Magnetic Field
l = length
= Vacuum permeability
= Vacuum permittivity
Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to
Recall that the speed of light is equivalent to
Then replacing,
Our values are given as
Replacing we have,
Therefore the magnetic field around this circular area is
Don't text while driving
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don't get distracted
You are given the mass of a sphere that is 26 kg sphere and it is released from rest when θ = 0°. You are also given the force of the spring that is F = 100 N. You are asked to find the tension of the spring. Imagine that the sphere is connected to a spring. The spring exerts a tension and the spring exerts gravitational pull. This will follow the second law of newton.
T - F = ma
T = ma + F
T = 26kg (9.81m/s²) + 100 N
T = 355.06 N
Answer:
90 degree hope it help :))
Answer:
31677.2 lb
Explanation:
mass of hammer (m) = 3.7 lb
initial velocity (u) = 5.8 ft/s
final velocity (v) = 0
time (t) = 0.00068 s
acceleration due to gravity (g) 32 ft/s^{2}
force = m x ( a + g )
where
- m is the mass = 3.7 lb
- g is the acceleration due to gravity = 32 ft/s^{2}
- a is the acceleration of the hammer
from v = u + at
a = (v-u)/ t
a = (0-5.8)/0.00068 = -8529.4 ( the negative sign showa the its decelerating)
we can substitute all required values into force= m x (a+g)
force = 3.7 x (8529.4 + 32) = 31677.2 lb
