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Morgarella [4.7K]
3 years ago
9

This is heat or electrical transfer by contact.

Physics
2 answers:
Vedmedyk [2.9K]3 years ago
8 0
Heat is the awser because we are using ate phones
zlopas [31]3 years ago
3 0

Answer:

The correct answer for this problem is conduction.

Explanation:

I just did the test and conduction was the correct answe.

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A lamp draws a current of 0.50 A when it is connected to a 120 V source? What is the resistance of the lamp?
emmainna [20.7K]
Given,
Current (I) = 0.50A
Voltage (V) = 120 volts
Resistance (R) =?
We know that:-
Voltage (V) = Current (I) x Resistance (R)
→Resistance (R) = Voltage (V) / Current (I)
= 120/0.50
= 24Ω
∴ Resistance (R) = 24Ω
8 0
3 years ago
Read 2 more answers
Which best explains parallel forces?
Lyrx [107]
The answer is to this question is B
8 0
3 years ago
How polar bear can survive in the polar region<br>​
ladessa [460]

Answer:

Hey mate.....

Explanation:

This is ur answer...

<em>Polar bears are well adapted for survival in the Arctic. Their adaptations include: a white appearance - as camouflage from prey on the snow and ice. thick layers of fat and fur - for insulation against the cold.</em>

Hope it helps!

Brainliest pls!

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7 0
2 years ago
Find the angle for the third-order maximum for 591 nm wavelength light falling on a diffraction grating having 1460 lines per ce
Marat540 [252]

Answer:

15.32°

Explanation:

We have given the wavelength \lambda =591nm=591\times 10^{-9}m

Diffraction grating is 1460 lines per cm

So  d=\frac{10^{-2}}{1460}=6.71\times 10^{-6}m (as 1 m=100 cm )

For maximum diffraction

dsin\Theta =m\lambda here m is order of diffraction

So 6.71\times 10^{-6}sin\Theta =3\times 591\times 10^{-9}

sin\Theta =0.264

\Theta =15.32^{\circ}

6 0
2 years ago
An engine absorbs 1749 J from a hot reservoir and expels 539 J to a cold reservoir in each cycle. a. What is the engine’s effici
Masja [62]

Answer:

a)η = 69.18 %

b)W= 1210 J

c)P=3967.21 W

Explanation:

Given that

Q₁ = 1749 J

Q₂ = 539  J

From first law of thermodynamics

Q₁   = Q₂ +W

W=Work out put

Q₂=Heat rejected to the cold reservoir

Q₁ =heat absorb by hot  reservoir

W= Q₁- Q₂

W= 1210 J

The efficiency given as

\eta=\dfrac{W}{Q_1}

\eta=\dfrac{1210}{1749}

\eta=0.6918

η = 69.18 %

We know that rate of work done is known as power

P=\dfrac{W}{t}

P=\dfrac{1210}{0.305}\ W

P=3967.21 W

8 0
3 years ago
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