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Hatshy [7]
3 years ago
9

Who invented the telephone?

Physics
2 answers:
krok68 [10]3 years ago
5 0
<h2>Answer:</h2><h3><em><u>Alexander Graham Bell</u></em></h3><h2>Explanation:</h2>

Alexander Graham Bell is often credited as the inventor of the telephone since he was awarded the first successful patent.

tia_tia [17]3 years ago
3 0
Alexander graham bell
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1The density of an of an object with a volume of 60.0 and mass of 400.0g is______
Kobotan [32]
D is the answer
To get the density you divide mass by volume
So the equation is 400/60=d
4 0
3 years ago
You have two photos of a person walking. One shows the person at the corner of Third and Main streets, the other shows the perso
SCORPION-xisa [38]
  <span>average speed 
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8 0
3 years ago
A V = 108-V source is connected in series with an R = 1.1-kΩ resistor and an L = 34-H inductor and the current is allowed to rea
soldi70 [24.7K]

Answer:

Explanation:

Given an RL circuit

A voltage source of.

V = 108V

A resistor of resistance

R = 1.1-kΩ = 1100 Ω

And inductor of inductance

L = 34 H

After he inductance has been fully charged, the switch is open and it connected to the resistor in their own circuit, so as to discharge the inductor

A. Time the inductor current will reduce to 12% of it's initial current

Let the initial charge current be Io

Then, final current is

I = 12% of Io

I = 0.12Io

I / Io = 0.12

The current in an inductor RL circuit is given as

I = Io ( 1—exp(-t/τ)

Where τ is time constant and it is given as

τ = L/R = 34/1100 = 0.03091A

So,

I = Io ( 1—exp(-t/τ))

I / Io = ( 1—exp(-t/τ))

Where I/Io = 0.12

0.12 = 1—exp(-t/τ)

0.12 — 1 = —exp(-t/τ)

-0.88 = -exp(-t/0.03091)

0.88 = exp(-t/0.03091)

Take In of both sides

In(0.88) = In(exp(-t/0.03091)

-0.12783 = -t/0.030901

t = -0.12783 × 0.030901

t = 3.95 × 10^-3 seconds

t = 3.95 ms

B. Energy stored in inductor is given as

U = ½Li²

So, the current at this time t = 3.95ms

I = Io ( 1—exp(-t/τ))

Where Io = V/R

Io = 108/1100 = 0.0982 A

Now,

I = Io ( 1—exp(-t/τ))

I = 0.0982(1 — exp(-3.95 × 10^-3 / 0.030901))

I = 0.0982(1—exp(-0.12783)

I = 0.0982 × 0.12

I = 0.01178

I = 11.78mA

Therefore,

U = ½Li²

U = ½ × 34 × 0.01178²

U = 2.36 × 10^-3 J

U = 2.36 mJ

8 0
3 years ago
Suppose there was a star with a parallax angle of 1 arcsecond. How far away would it be? Select all that apply.
lesya692 [45]

Answer:

option E

Explanation:

given,                          

Parallax angle(d) = 1 arcsecond

using Parallax formula                  

      d = \dfrac{1}{p}

 p is the parsecs angle which is measured in 1 arcsecond

 d is the distance in parsec

now,                                            

      P = \dfrac{1}{d}

      P = \dfrac{1}{1}

      P = 1 \ parsec

we know,                                  

    1 parsec = 3.26 light year

hence, the answer will be option E

6 0
4 years ago
Starting at t = 0 a net external force in the +x-direction is applied to an object that has mass 5.00 kg. A graph of the force a
Reptile [31]

Answer:

  15√2 N

Explanation:

The acceleration is given by ...

  a = F/m = 5t/5 = t . . . . meters/second^2

The velocity is the integral of acceleration:

  v = ∫a·dt = (1/2)t^2

This will be 9 m/s when ...

  9 = (1/2)t^2

  t = √18 . . . . seconds

And the force at that time is ...

  F = 5(√18) = 15√2 . . . . newtons

6 0
3 years ago
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