A transformer supplies 60 watts of power to a device that is rated at 20 volts. The primary coil is connected to a 120-volt ac s
1 answer:
Answer:
The current in the primary coil would be .
Explanation:
Given the power supplied by a transformer is 60 watts.
And the voltage in the primary coil is 120 Volts.
We need to find the current supply in the primary coil.
We will use the formula
Where,
is the power in Watts.
is the voltage in Volts.
is the current supply in Ampere.
So, the current in the primary coil would be .
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Answer:
the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.
Explanation:
We can answer this exercise using Gauss's law
Ф = ∫ e . dA = / ε₀
field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell. the flow must be zero since the charge of the sphere is equal induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field
From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.
Answer:
<em>1.</em> <em>39068.07 N</em>
<em>2. 19534.036 N</em>
Explanation:
depth of water h = 10 m
atmospheric pressure Patm = 101325 Pa
density of water p = 1000 kg/m^3
acceleration due to gravity g = 9.81 m/s^2
pressure due to depth of water = pgh
P = 1000 x 9.81 x 10 = 98100 Pa
total pressure on the bottom of the tank is Patm + p = 101325 + 98100 = 199425 Pa
Left plug has diameter = 50 cm = 0.5 m
radius = 0.5/2 = 0.25 m
height = 1 cm = 0.01 m
<em>height below tank surface = 10 - 0.01 = 9.99</em>
pressure at this depth = 1000 x 9.81 x 9.99 = 98001.9 Pa
<em>total pressure = Patm + P = 101325 + 98001.9 = 199326.9 Pa</em>
surface area of plug = π = 3.142 x
= 0.196
<em>force required to lift left plug = pressure x area</em>
F = 199326.9 x 0.196 = <em>39068.07 N</em>
<em>The right side is a hemisphere with the same diameter, therefore surface area is half of the left plug</em>
A = 0.196/2 = 0.098
force F required to lift right plug = 199326.9 x 0.098 =<em> 19534.036 N</em>