Answer:
3.68 m/s
Explanation:
Full answer in the attached picture
The voltage across an inductor ' L ' is
V = L · dI/dt .
I(t) = I(max) sin(ωt)
dI/dt = I(max) ω cos(ωt)
V = L · ω · I(max) cos(ωt)
L = 1.34 x 10⁻² H
ω = 2π · 60 = 377 /sec
I(max) = 4.80 A
V = L · ω · I(max) cos(ωt)
V = (1.34 x 10⁻² H) · (377 / sec) · (4.8 A) · cos(377 t)
<em>V = 24.25 cos(377 t)</em>
V is an AC voltage with peak value of 24.25 volts and frequency = 60 Hz.
Answer:
A. It is zero.
Explanation:
D Later in the day, more power is developed in lifting each box. 12 A manometer is used to indicate the pressure in a steel vessel, as shown in the diagram. What value does the liquid manometer give for the pressure in the vessel? It is zero
no it can't do this why because I think that it is water and it can not go any where.
Answer:
E = 31.329 N/C.
Explanation:
The differential electric field 
at the center of curvature of the arc is
<em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>
where 
is the radius of curvature.
Now
,
where 
is the charge per unit length, and it has the value
Thus, the electric field at the center of the curvature of the arc is:
Now, we find 
and 
. To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:
and this is
radians.
Therefore,
evaluating the integral, and putting in the numerical values we get:
