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Temka [501]
3 years ago
11

What is the applied voltage of a circuit that has 10 amps of current and 12 ohms of resistance?

Physics
1 answer:
Anettt [7]3 years ago
7 0
Voltage (V)= Current (I) * Resistance (R)
V=IR=10*12=120volt

*This is a common answer of mine*
Hope this helps!
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In what type of job might being a difficult person not be much of a liability ?
zhannawk [14.2K]
A possible answer would be a stocker. a person who sorts thing or delivery person who just drops things off. there are little to no jobs with no human interaction. some good options for jobs where you work by yourself are: 
<span>Embalmer.Accountant.Travel Photographer.Tree Planter.Freelance Writer.Truck Driver.Data Scientist.<span>Taxidermist. I HOPE THIS HELPS!!!</span></span>
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3 years ago
What is the required force to bring a 1550 kg vehicle moving at 32 m/s to a stop in a distance of 45 m?
Vilka [71]
The answer is 9 i think.
8 0
3 years ago
Two identical charges q placed 2.0 mapart exert forces of magnitude 4.0 N on each other What is the value of the charge q? a) q
katen-ka-za [31]

Answer:

c) 4.2*10^{-5}C

Explanation:

Coulomb's law says that the force exerted between two charges is inversely proportional to the square of distance between them, and is given by the expression:

F=\frac{kq_{1}q_{2}}{r^{2}}

where k is a proportionality constant with the value k=9*10^{9}\frac{Nm^{2}}{C^{2}}

In this case q_{1}=q_{2}=q, so we have:

F=\frac{kq^{2}}{r^{2}}

Solving the equation for q, we have:

kq^{2}=Fr^{2}

q^{2}=\frac{Fr^{2}}{k}

q=\sqrt{\frac{Fr^{2}}{k}}

Replacing the given values:

q=\sqrt{\frac{4.0N*(2.0m)^{2}}{9*10^{-9}\frac{Nm^{2}}{C^{2}}}}

q=4.2*10^{-5}C

3 0
3 years ago
In any electromagnetic wave,
puteri [66]

A). Both the energy and the wave travel in the same direction.

If they didn't, they'd wind up in different cities almost instantly.

6 0
3 years ago
A ship leaves the island of Guam and sails a distance 255 km at an angle 49.0 o north of west. Part A: In which direction must i
kiruha [24]

Answer:

Explanation:

We shall represent displacement in vector form .Consider east as x axes and north as Y axes west as - ve x axes and south as - ve Y axes . 255 km can be represented by the following vector

D₁ = - 255 cos 49 i  + 255 sin49 j

= - 167.29 i + 192.45 j

Let D₂ be the further displacement which lands him 125 km east . So the resultant displacement is

D = 125 i

So

D₁ + D₂ = D

- 167.29 i + 192.45 j + D₂ = 125 i

D₂ = 125 i + 167.29 i - 192.45 j

= 292.29 i - 192.45 j

Angle of D₂ with x axes θ

tan θ = -192.45 / 292.29

= - 0.658

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Magnitude of D₂

D₂² = ( 192.45)² + ( 292.29)²

D₂ = 350 km approx

Tan

7 0
3 years ago
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