The question is incomplete. The complete question is :
A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magnetic field. A field of 1.2 T is directed along the positive z-direction, which is upward. (a)If the loop is removed from the field region in a time interval of 2.8 ms, find the average emf that will be induced in the wire loop during the extraction process.
Solution :
Let us consider a 
which has a 
of r = 
cm.
It is oriented horizontally along the xy-plane and is located in the region of an 
, such that it points in the positive z direction and having a magnitude of B = 1.2 T.
Now if the loop 
in a time interval of Δt = 2.8 ms. Initially the magnetic field and the area points is in the same direction, so that the angle between them is Ф = 0°, thus the initial and the final fluxes are :
and 
Area A = 
The induced emf equals to the change in the flux, and is divided by the time that it takes to go from the initial flux, Δt and multiplied by the number of turns N = 1, i.e. ,
= 30.27 V
Therefore, the emf generated is 30.27 V.
Answer:
Explanation: so how many minutes are in an hour 60 right, and the bus travels 10km in 7 minutes right so use math the bus travels 14km in 10 minutes so the bus travels 98km in an hour
Explanation:
Draw a free body diagram for each disc.
Disc A has three forces acting on it: 86.5 N up, T₁ down, and Wa down.
∑F = ma
86.5 N − T₁ − Wa = 0
Wa = 86.5 N − T₁
ma × 9.8 m/s² = 86.5 N − 55.6 N
ma = 3.2 kg
Disc B has three forces acting on it: T₁ up, T₂ down, and Wb down.
∑F = ma
T₁ − T₂ − Wb = 0
Wb = T₁ − T₂
mb × 9.8 m/s² = 55.6 N − 36.5 N
mb = 1.9 kg
Disc C has three forces acting on it: T₂ up, T₃ down, and Wc down.
∑F = ma
T₂ − T₃ − Wc = 0
Wc = T₂ − T₃
mc × 9.8 m/s² = 36.5 N − 9.6 N
mc = 2.7 kg
Disc D has two forces acting on it: T₃ up and Wd down.
∑F = ma
T₃ − Wd = 0
Wd = T₃
md × 9.8 m/s² = 9.6 N
md = 0.98 kg
The part that moves are called anti-nodes. The stationary pars are nodes
