They don't want to be killed or they are hiding
Part A : 4.33 Volt.
Part B : 4764 Amperes.
Given Data :
Field near a typical pulsed-field machine rises from 0 T to 2.5 T in 200 μs.
Axis of his 2.1-cm-diameter wedding band is parallel to the field.
Part A :
induced emf = rate of change of magnetic flux
= d (B A )/dt
= A dB/dt
= { π (2.1 x 10^-2 / 2)^2) (2.5) } / (200 x 10^-6)
E = 4.33 Volt
Part B :
R = rho L / A
R = { (6.2 x 10^-8) ( π x 0.021) } / (4.5 x 10^-6)
R = 9.10 x 10^-4 Ohm
We know, (I = V/ R)
I = 4.33 Volts / 9.10 x 10^-4 Ohm
I = 4764 Amperes
What exactly is the function of TMS?
Transcranial magnetic stimulation (TMS) is a noninvasive procedure that uses magnetic fields to stimulate nerve cells in the brain to alleviate depression symptoms. TMS is usually used when other depression treatments have failed.
Find more on TMS at : brainly.com/question/13851741
#SPJ4
➤ Although your question isn't complete, I may have assumed that you were referring to this specific question.
The complete question is :
A TMS (transcranial magnetic stimulation) device creates very rapidly changing magnetic fields. The field near a typical pulsed-field machine rises from 0 T to 2.5 T in 200 μs. Suppose a technician holds his hand near the device so that the axis of his 2.1-cm-diameter wedding band is parallel to the field.
Part A: What emf is induced in the ring as the field changes? Express your answer to two significant figure E= ___ unit_
Part B: If the band is made of a gold alloy with resistivity 6.2×10−8Ω⋅m and has a cross-section area 4.5 mm2 , what is the induced current? Express your answer to two significant figures and include the appropriate units. I= ___ units__
<h2>
Answer:</h2>
(a) 10N
<h2>
Explanation:</h2>
The sketch of the two cases has been attached to this response.
<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>
In this case, a frictional force
is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;
∑F = ma -------------------(i)
Where;
∑F = effective force acting on the object (box)
m = mass of the object
a = acceleration of the object
∑F = F - 
m = 50kg
a = 0 [At constant velocity, acceleration is zero]
<em>Substitute these values into equation (i) as follows;</em>
F -
= m x a
F -
= 50 x 0
F -
= 0
F =
-------------------(ii)
<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>
In this case, the same frictional force
is opposing the movement of the box.
∑F = 1.5F - 
m = 50kg
a = 0.1m/s²
<em>Substitute these values into equation (i) as follows;</em>
1.5F -
= m x a
1.5F -
= 50 x 0.1
1.5F -
= 5 ---------------------(iii)
<em>Substitute </em>
<em> = F from equation (ii) into equation (iii) as follows;</em>
1.5F - F = 5
0.5F = 5
F = 5 / 0.5
F = 10N
Therefore, the value of F is 10N
<em />
Answer:
16km
Explanation:
First change the minutes into hours then multiply by the distance.
(8÷60)×120=16km
Answer:
<em>Gravity</em><em>.</em><em> </em><em>The</em><em> </em><em>weight-force</em><em> </em><em>or</em><em> </em><em>weight</em><em> </em><em>of</em><em> </em><em>an</em><em> </em><em>object</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>force</em><em> </em><em>because</em><em> </em><em>of</em><em> </em><em>Gravity</em><em>,</em><em> </em><em>which</em><em> </em><em>acts</em><em> </em><em>on</em><em> </em><em>the</em><em> </em><em>object</em><em> </em><em>attracting</em><em> </em><em>it</em><em> </em><em>towards</em><em> </em><em>the</em><em> </em><em>centre</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>earth</em><em>.</em>
<em>Hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>,</em><em> </em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>x</em>