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Lesechka [4]
3 years ago
7

What is the current in the 10.0 , resistor?

Physics
2 answers:
In-s [12.5K]3 years ago
5 0

Answer:

B. 12.0A

Explanation:

The circuit shown in the image is a parallel circuit which mean that the voltage of the source is equal across each resistance in the circuit because they are directly connected to the terminals of the voltage source.

The voltage across the 10 Ohms resistor is 120V

So, let's proceed to calculate the current

By using Ohm's Law

I=\frac{V}{R} \\I= \frac{120.0V}{10.0Ohm} \\I= 12.0A

12A is the amount of current flowing through the 10 ohms resistor.

olga nikolaevna [1]3 years ago
4 0

Current = (voltage) / (resistance)

The ends of the 10-ohm resistor are connected directly to both terminals of the battery.  

So the voltage across the resistor is the voltage of the battery ... 120 V.  None of that other stuff in the circuit has any effect on it.

Current = (120 V) / (10 ohms)

<em>Current = 12 A .  (B)</em>

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An electron in an atom has an uncertainty of 0.2 nm. If it is doubled to 0.4 nm by what factor does the uncertainty in momentum
fenix001 [56]

Answer:

The uncertainty in momentum changes by a factor of 1/2.

Explanation:

By Heisenberg's uncertainty principle, ΔpΔx ≥ h/2π where Δp = uncertainty in momentum and Δx = uncertainty in position = 0.2 nm. The uncertainty in momentum is thus Δp ≥ h/2πΔx. If the uncertainty in position is doubled, that is Δx₁ = 2Δx = 0.4 nm, the uncertainty in momentum Δp₁ now becomes Δp₁ ≥ h/2πΔx₁ = h/2π(2Δx) = (h/2πΔx)/2 = Δp/2.

So, the uncertainty in momentum changes by a factor of 1/2.

4 0
3 years ago
A 250 kg flatcar 25 m long is moving with a speed of 3.0 m/s along horizontal frictionless rails. A 61 kg worker starts walking
Anna11 [10]

Answer:

x=31.09m

Explanation:

p1=p2

The momentum of flatcar and the momentum of the worker so

The velocity of the worker is:

m_{f}*v_{f}=m_{w}*v_{w}\\\\v_{f}=\frac{m_{f}*v_{f}}{m_{w}}\\v_{f}=\frac{61kg*3.0\frac{m}{s}}{250kg}\\v_{f}=0.732\frac{m}{s}

The total motion has a total velocity and is

Vt=v_{w}+v_{f}\\Vt=0.732\frac{m}{s}+3.0\frac{m}{s}\\Vt=3.732\frac{m}{s}

The time the worker take walking is

t=\frac{x}{v_{w}}\\t=\frac{25m}{3\frac{m}{s}}=8.33s

Now the total time and the total velocity determinate the motion of tha flatcar how far has moved

x=t*Vt\\x=8.33s*3.732\frac{m}{s} \\x=31.09m

5 0
3 years ago
What are the 3 tools used to collect weather data
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Anemometer, Psychrometer, <span>Barometer</span>
5 0
3 years ago
How much work is done on 10.0 c of charge to move it through a potential diffrence of 9.0 v in 10.0s
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We need more evidence to be provided
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3 years ago
Which of the following provides evidence that there must be at least two types of electrical charge, but that there is only one
polet [3.4K]

Answer:

Option D (On the...............dominate) would be the right approach.

Explanation:

The Gravitational constant (G) will be:

= 6.67\times 10^{-11}

The Coulomb's law constant (K) will be:

= 9\times 10^9

  • Throughout particular, these have been determined that among 2 substances with almost the similar form of charge, the combination of electromagnetic as well as the force does seem to be usually the following:

⇒ \frac{f_e}{f_g}\sim 10^{42}

  • By that same argument, the electrostatic force including its planet's atmosphere would have strongly influenced the effect, as well as maybe the planet's atmosphere, would have crashed, or perhaps the earth would have shifted at a much longer exposure from one another and.
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The latter three choices aren't connected to either the situation mentioned in the clarification segment elsewhere here.

5 0
3 years ago
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