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cricket20 [7]
3 years ago
8

A cell phone charger operates at 5 volts and can deliver a maximum current of 1.0 Amp. If your cell phone has an energy storage

capacity of 38.9~38.9 kiloJoules, what is the shortest possible time (in hours) it would take to fully charge your empty phone battery using this charger.
Physics
1 answer:
madam [21]3 years ago
7 0

Answer:

<em>it</em><em> </em><em>takes</em><em> </em><em>2</em><em>.</em><em>1</em><em>6</em><em> </em><em>hrs</em><em> </em><em>to</em><em> </em><em>fully</em><em> </em><em>charge</em><em> </em><em>the</em><em> </em><em>phone</em>

Explanation:

V = 5v

I = 1A

E = 38.9 Kj = 38900 j

t = ?

E = IVt

t = E/IV

t = 38900 ÷ (1 × 5)

t = 38900 ÷ 5

t = 7780 seconds

t = 7780/3600............( converting time in seconds to hours)

t = 2.16 hrs

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The current in a hair dryer measures 15 amps. The resistance of the hair dryer is 14 ohms. What is the voltage?
ZanzabumX [31]
We have that there is a formula connecting these three. V=I*R where V is voltage, I is the current and R is the resistance. Substituting, we get that V=210 Volta, which is the unit of measurement for voltage. You can think of the relationship in the following way : The energy of the field is equal to the flow of the field times the resistance that it meets.
6 0
3 years ago
By calculating its wavelength (in nm), show that the second line in the Lyman series is UV radiation.
Rashid [163]

Answer:

 λ = 102.78  nm

This radiation is in the UV range,

Explanation:

Bohr's atomic model for the hydrogen atom states that the energy is

           E = - 13.606 / n²

where 13.606 eV   is the ground state energy and n is an integer

an atom transition is the jump of an electron from an initial state to a final state of lesser emergy

            ΔE = 13.606 (1 / n_{f}^{2} - 1 / n_{i}^{2})

the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon

            DE = 13.606 (1/1 - 1/3²)

            DE = 12.094 eV

let's reduce the energy to the SI system

            DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J

let's find the wavelength is this energy, let's use Planck's equation to find the frequency

            E = h f

             f = E / h

            f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴

            f = 2.9186 10¹⁵ Hz

now we can look up the wavelength

           c = λ f

           λ = c / f

           λ = 3 10⁸ / 2.9186 10¹⁵

           λ = 1.0278  10⁻⁷ m

let's reduce to nm

            λ = 102.78  nm

This radiation is in the UV range, which occurs for wavelengths less than 400 nm.

5 0
3 years ago
What is the total resistance in this circuit?
saul85 [17]

Answer:B

Explanation:

6 0
3 years ago
On a calm day with no wind, you can run a 1500-m race at a velocity of 4.0 m/s. If you run the same race on a day when you have
lesantik [10]

Answer:

The time taken to finish the race is 750 s.

Explanation:

The velocity of the person on the day of wind is slowed down by 2.0 m/s. So the person's velocity on the day of wind is 4-2=2 m/s.

The relation between time, speed and distance is t=v/d

Given d=1500 m and calculated v= 2 m/s.

t=1500/2

t=750 s.

Learn more about distance formula.

brainly.com/question/11954435

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3 0
2 years ago
Read 2 more answers
What is the magnitude of the electric force between a proton and an electron when they are at a distance of 4.09 angstrom from e
Zinaida [17]

Answer:

F=1.38*10^{-9}N

Explanation:

According to Coulomb's law, the magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

F=\frac{kq_1q_2}{d^2}

Here k is the Coulomb constant. In this case, we have q_1=-e, q_2=e and d=4.09*10^-10m. Replacing the values:

F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-1.6*10^{-19}C)(1.6*10^{-19}C)}{(4.09*10^{-10})^2}\\F=-1.38*10^{-9}N

The negative sign indicates that it is an attractive force. So, the magnitude of the electric force is:

F=1.38*10^{-9}N

5 0
3 years ago
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