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fgiga [73]
3 years ago
11

Find V V=25m. _____=?? 3.5s.

Physics
1 answer:
matrenka [14]3 years ago
4 0

Answer:

?

Explanation:

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When the sphere makes a complete revolution around its circular path, does it spend:_______
topjm [15]

Answer:

Explanation:

the same amount of time in both halves of the circle

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2 years ago
What is the international standard for measurement it is a modified version of the metric system
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SI system i think it is right

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2 years ago
Group of physics students are asked to demonstrate Ohm’s Law in parallel circuits.  Using their knowledge of Ohm’s Law and the r
astra-53 [7]

The student’s suggestion who provides enough evidence to be able to determine the value of each resistor is student D.

<h3>What is current?</h3>

The current is the stream of charges which flow inside the conductors when connected across the end of voltage.

For the given set of parallel resistors, we need to find the resistance of each resistor.

From the Ohm's law, V =IR

R = V/I

Resistance value depends upon the voltage difference across the resistor and the current flowing through that resistance.

Thus, the student D gives enough evidence to find resistance of the circuit is

Learn more about current.

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8 0
1 year ago
A certain copper wire has a resistance of 10.5 ω. at what fraction of the length l must the wire be cut so that the resistance o
Delvig [45]
The answer for that questions will be 759. Stars itir
6 0
3 years ago
A circular loop of wire with a diameter of 0.626 m is rotated in a uniform electric field to a position where the electric flux
Mila [183]

Answer:

C) 2.44 × 106 N/C

Explanation:

The electric flux through a circular loop of wire is given by

\Phi = EA cos \theta

where

E is the electric field

A is the cross-sectional area

\theta is the angle between the direction of the electric field and the normal to A

The flux is maximum when \theta=0^{\circ}, so we are in this situation and therefore cos \theta =1, so we can write

\Phi = EA

Here we have:

\Phi = 7.50\cdot 10^5 N/m^2 C is the flux

d = 0.626 m is the diameter of the coil, so the radius is

r = 0.313 m

and so the area is

A=\pi r^2 = \pi (0.313 m)^2=0.308 m^2

And so, we can find the magnitude of the electric field:

E=\frac{\Phi}{A}=\frac{7.50\cdot 10^5 Nm^2/C}{0.308 m^2}=2.44\cdot 10^6 N/C

3 0
3 years ago
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