Magnetic flux can be calculated by the product of the magnetic field and the area that is perpendicular to the field that it penetrates. It has units of Weber or Tesla-m^2. For the first question, when there is no current in the coil, the flux would be:
ΦB = BA
A = πr^2
A = π(.1 m)^2
A = π/100 m^2
ΦB = 2.60x10^-3 T (π/100 m^2 ) ΦB = 8.17x10^-5 T-m^2 or Wb (This is only for one loop of the coil)
The inductance on the coil given the current flows in a certain direction can be calculated by the product of the total number of turns in the coil and the flux of one loop over the current passing through. We do as follows:
L = N (ΦB ) / I
L = 30 (8.17x10^-5 T-m^2) / 3.80 = 6.44x10^-4 mH
<span>To find the wavelength of a neutron can be calculated by using the formula
Wavelength=h/m x v
Where h is planck's constant
m=mass of neutron
v= velocity of the particle
By substituting the given values
Wavelength= 6.63 × 10–34 j s / 1.675 × 10–27 kg x 2 m/s^-1
Wavelength of a neutron=1.979 x 10^-7 m</span>
Answer:
T = 686.7N
Explanation:
For this exercise we will use Newton's second law in this case there is no acceleration,
∑ F = ma
T -W = 0
The gymnast's weight is
W = mg
We clear and calculate the tension
T = mg
T = 70 9.81
T = 686.7N
The first problem cannot be solve because you did give the distance or length of the rope, because work = distance x force. i can only solve the the second problem. since the bucket is moving up then force due to gravity is going down, then the net force is:
Fnet = F1 - Fg
where Fg = mg
g is the accelaration due to gravity ( 9.81 m/s^2)
Fnet = 57.5 N - (3.9 kg)(9.81) N
Fnet = 19.24 N
Answer:
a) parallel to the ground True
c) parallel to the ground towards man True
Explanation:
To examine the possibilities, we propose the solution of the problem.
Let's use Newton's second law
F = m a
The force is exerted by the arm and the centripetal acceleration of the golf club, which in this case varies with height.
In our case, the stick is horizontal in the middle of the swing, for this point the centripetal acceleration is directed to the center of the circle or is parallel to the arm that is also parallel to the ground;
Ask the acceleration vector
a) parallel to the ground True
b) down. False
c) parallel to the ground towards True men
d) False feet
e) the head. False
