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lana [24]
2 years ago
7

Consider a circuit driven by a 110~V110 V ideal battery and dissipating a total of 1200~W1200 W of power. What is the total curr

ent supplied by the battery?
Physics
1 answer:
Leto [7]2 years ago
6 0

Answer : The total current supplied by the battery is, 10.9 A

Explanation :

The relation between power, voltage and current is:

P=I\times V

where,

P = power of circuit = 1200 W

I = current = ?

V = voltage = 110 V

Now put all the given values in the above formula, we get the total current supplied by the battery.

1200W=I\times 110V

I=10.9A

Thus, the total current supplied by the battery is, 10.9 A

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Calculate the current through a 10.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V
Kipish [7]

Answer:

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

Explanation:

Given:

Length = l = 10 meter

Radius = r = 0.321\ mm =0.321\times 10^{-3}\ meter

Resistivity=\rho=1.00\times 10^{-6}\ ohm\ meter

V = 12 Volt

To Find:

Current, I =?

Solution:

Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as

R=\dfrac{\rho\times l}{A}

Where,

R = Resistance

l = length

A = Area of cross section = πr²

\rho=Resistivity=1.00\times 10^{-6}\ ohm\ meter

Substituting the values we get

R=\dfrac{1\times 10^{-6}\times 10}{3.14\times (0.321\times 10^{-3})^{2}}

R=\dfrac{1\times 10^{-5}}{3.23\times 10^{-7}}

R=\dfrac{1\times 10^{2}}{3.23}

R=30.95\ ohm

Now by Ohm's Law,

V= I\times R

Substituting the values we get

I=\dfrac{V}{R}=\dfrac{12}{30.95}=0.3876\ Ampere

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

4 0
2 years ago
Charge q1 is distance r from a positive point charge q. charge q2=q1/3 is distance 2r from q. what is the ratio u1/u2 of their p
makvit [3.9K]
We need the power law for the change in potential energy (due to the Coulomb force) in bringing a charge q from infinity to distance r from charge Q. We are only interested in the ratio U₁/U₂, so I'm not going to bother with constants (like the permittivity of space). 

<span>The potential energy of charge q is proportional to </span>
<span>∫[s=r to ∞] qQs⁻²ds = -qQs⁻¹|[s=r to ∞] = qQr⁻¹, </span>

<span>so if r₂ = 3r₁ and q₂ = q₁/4, then </span>
<span>U₁/U₂ = q₁Qr₂/(r₁q₂Q) = (q₁/q₂)(r₂/r₁) </span>
<span>= 4•3 = 12.</span>
5 0
2 years ago
Determine the change in thermal energy of 100 g of copper (M = 63,5, Debye 348K) if it is cooled from
Setler [38]

Answer:

given,

mass of copper = 100 g

latent heat of liquid (He) = 2700 J/l

a) change in energy

Q = m Cp (T₂ - T₁)

Q = 0.1 × 376.812 × (300 - 4)

Q = 11153.63 J

He required

Q = m L

11153.63 = m × 2700

m = 4.13 kg

b) Q = m Cp (T₂ - T₁)

Q = 0.1 × 376.812 × (78 - 4)

Q = 2788.41 J

He required

Q = m L

2788.41 = m × 2700

m = 1.033 kg

c) Q = m Cp (T₂ - T₁)

Q = 0.1 × 376.812 × (20 - 4)

Q = 602.90 J

He required

Q = m L

602.9 = m × 2700

m =0.23 kg

8 0
2 years ago
A space vehicle deploys its re–entry parachute when it's traveling at a vertical velocity of –150 meters/second (negative becaus
dexar [7]

Answer:

a=5m/s^2

Explanation:

Aceleration is a change on the velocity of the object in a given time.

For this case: the initial velocity is

v_{1}=150m/s

and the final velocity is :

v_{2}=0 m/s

so, the change in velocity is:

\Delta v =v_{2}-v_{1}=0m/s - (-150m/s) =  150 m/s

and the change in time , according to the problem:

\Delta t=30s

So, the aceleration is:

a=\frac{\Delta v}{\Delta t} = \frac{150m/s}{30s} = 5m/s^2

6 0
2 years ago
Which of the following statements accurately describes the atmospheric patterns that influence local weather?
timurjin [86]

Answer: A

Explanation:

Well the high and lows effect the humidity the more humidity the more hot it is so the high brings higher temperatures.

4 0
1 year ago
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