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
Higher frequency,higher energy,shorter wavelength
Answer:
u need to make sure that comparison is = to shapes and then find the shapes sizes and add them
Answer:22.36 m/s
Explanation:
Given
the diameter of loop d=50 m
the radius of loop r=25 m
At the top position, we can write,
weight and Normal reaction combination will provide the centripetal force i.e.
![R=W\quad \quad [\text{apparent weight =Actual weight}]](https://tex.z-dn.net/?f=R%3DW%5Cquad%20%5Cquad%20%5B%5Ctext%7Bapparent%20weight%20%3DActual%20weight%7D%5D)
Frictional force = vertical force x coeff of friction = 590N x 0.045 = 26.55N
Mass of boy and sled = 590N / g = 590N / 9.8m/s^2 = 60.20 kg
Deceleration due to friction = 26.55N / 60.20kg = 0.44 m/s^2.
For constant acceleration we have:
v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and ti is time. In this case v = 0, u = 5.9 m/s, a = -0.44 m/s^2. So we have
0 = 5.9 - 0.44t
which gives t = 5.9 / 0.44 = 13.409 s.
Distance traveled in this time d = ut + 0.5at^2 = 5.9 x 13.409 - 0.5 x 0.44 x 13.409^2 = 39.56 m
<span>Answer: </span>40 meters
