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jekas [21]
3 years ago
10

When air transfers energy to a cooler object what happens to the air temperature?

Physics
2 answers:
tensa zangetsu [6.8K]3 years ago
4 0

The answer is: It decreases.

Katarina [22]3 years ago
3 0

Answer: The temperature of air would decrease.

Explanation:

The energy gets transferred from hot object to cold object until the two objects are in thermal equilibrium. When air transfers energy to a cooler object, it means the air it self is cooling down till the air and the object are in thermal equilibrium. Thus, the temperature of air would decrease.

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An AC adapter for a telephone-answering unit uses a transformer to reduce the line voltage of 120 V (rms) to a voltage of 9.0 V.
Illusion [34]

Answer:

1. The number of turns on the secondary output is 18

2. The root mean square power delivered to the transformer is 48 Watts.

Explanation:

A transformer is an electronic device that can be used for increasing or decreasing the value of a given voltage. It consists of primary coils and secondary coil of a definte number of turns. When voltage flows in the primary coil, it induces voltage in the secondary coil. The two types are: step-up and step down transformers.

1. For a given transformer,

          \frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }

where V_{s} is the value of the induced voltage in the secondary coil, V_{p} is he voltage in the primary coil, and N_{s} is the number of turns of the secondary coil, N_{p} is the number of turns in the primary coil.

From the question,

V_{s}  = 9.0 V, V_{p} = 120 V, N_{p} = 240, N_{s} = ?

So that,

           N_{s} = \frac{V_{s}*N_{p}  }{V_{p} }

                = \frac{9 * 240}{120}

               = 18

The number of turns on the secondary output is 18.

2. Power_{rms} = I_{rms} × V_{rms}

                    = 0.4 × 120

                   = 48 W

The rms power delivered to the transformer is 48 Watts.

8 0
3 years ago
Why is the energy stored in the wire per unit volume=1/2*strain*stress​
Sav [38]

Answer:

have tried it and the units end up being Nm^-2 (the unit of stress) however the unit for joules is Nm.

Source https://www.physicsforums.com/threads/why-does-1-2-stress-strain-equal-the-energy-stored-per-unit-volume-in-a-wire.565495/

6 0
3 years ago
(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas
aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = \frac{k*q}{r}

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

⇒k*q1* (1.63m - x) = -k*q2*x:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

3 0
3 years ago
Read 2 more answers
In an emergency, a driver brings a car to a full stop in 5.0s. The car is traveling at a rate of 38m/s when the breaking begins.
Lana71 [14]

(a) By definition of average acceleration,

<em>a</em> = ∆<em>v</em> / ∆<em>t</em> = (0 - 38 m/s) / (5.0 s)

<em>a</em> = -7.6 m/s²

(b) Recall that

<em>v</em>² - <em>u</em>² = 2 <em>a</em> ∆<em>x</em>

where <em>u</em> and <em>v</em> are initial and final velocities, respectively; <em>a</em> is acceleration; and ∆<em>x</em> is the change in position. So

0² - (38 m/s)² = 2 (-7.6 m/s²) ∆<em>x</em>

∆<em>x</em> = 95 m

6 0
4 years ago
a force is applied to an object at rest with a mass of 100 kg. the same force is applied to an object at rest with a mass of 1 k
Dafna1 [17]
The object with the mass ok 1kg will move more quickly because it is lighter than the 100kg object
3 0
4 years ago
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