Answer:
Here is the complete question:
https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632
a) Current for long straight wire 
b) Current at the center of the circular coil 
c) Current near the center of a solenoid 
Explanation:
⇒ Magnetic Field due to long straight wire is given by (B),where

Plugging the values,
Conversion
,and 

⇒Magnetic Field at the center due to circular coil (at center) is given by,
So 
⇒Magnetic field due to the long solenoid,
Then
So the value of current are
,
and
respectively.
Answer:
t=20s
Explanation:
To solve this problem we must apply the first law of thermodynamics, which indicates that the energy that enters a system is the same that must come out, resulting in the following equation
For this problem we will assume that the water is in a liquid state, since it is a domestic refrigerator
q=m.cp.(T2-T1)
q=heat
m=mass of water =600g=0.6Kg
cp=
specific heat of water=4186J/kgK
T2=temperature in state 2=20°C
T1=temperature in state 1=0°C
solving:
q=(0.6)(4186)(20-0)=50232J
A refrigerator is a device that allows heat to be removed to an enclosure (Qin), by means of the input of an electrical energy (W) and the heat output (Qout), the coefficient of performance COP, allows to know the ratio between the heat removed ( Qin) and the added electrical power (W), the equation for the COP is

To solve this exercise we must know the value of the heat removed to the water (Qin)
solving for Qin
Qin=(COP)(Win)
Qin=(5)(500W)=2500W
finally we remember that the definition of power is the ratio of work over time
w=work
p=power=500w

The energy transferred to the spring is given by:

where
k is the spring constant
x is the elongation of the spring with respect its initial length
Let's convert the data into the SI units:


so now we can use these data inside the equation ,to find the energy transferred to the spring:
The answer is true
Hope it helped
Answer:
(177.94 ± 3.81) cm^2
Explanation:
l + Δl = 21.7 ± 0.2 cm
b + Δb = 8.2 ± 0.1 cm
Area, A = l x b = 21.7 x 8.2 = 177.94 cm^2
Now use error propagation



So, the area with the error limits is written as
A + ΔA = (177.94 ± 3.81) cm^2