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3 years ago

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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Thing 1 and Thing 2 push with a force of 2000 N to move a rather small 500 kg elephant 30 m across the kitchen floor. How much w

Amanda [17]

Answer:

60000 J

Explanation:

Assuming the force is applied parallel to the displacement of the elephant, the work done to move it across the floor is

$W=Fd$

where

F = 2000 N is the force applied

d = 30 m is the displacement of the elephant

Substituting the numbers into the formula, we find

$W=(2000 N)(30 m)=60,000 J$

3 0

3 years ago

Which is a complete sentence?

8_murik_8 [283]

It takes a noun and a verb to make a complete sentence.
There isn't a single verb in a), b), or c).
"Affords" is the verb (predicate) in d)., the only complete sentence.

6 0

3 years ago

Read 2 more answers

If a car has a momentum of 2.04 x 104 kgm/s and a velocity of 18 m/s, what is its mass?

muminat

Answer:

Mass = 1133.33 kg (Approx.)

Explanation:

Given:

Momentum = 2.04 x 10⁴ kg[m/s]

Velocity = 18 m/s

Find:

Mass

Computation:

Mass = Momentum / Velocity

Mass = [2.04 x 10⁴] / 18

Mass = 1133.33 kg (Approx.)

5 0

3 years ago

Describe grafting, what are the<br> three methods?

Tanya [424]

Answer: cleft grafting, inlay grafting, four-flap grafting, and whip grafting.

Explanation:

I hope I helped , Have a great day or night . Xoxoxo.

6 0

2 years ago

calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame

kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

$\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2$

0 - p × Area for trachea =

$\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}$

$-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)$

⇒ $-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}$

$\frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s$

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

3 0

3 years ago

Read 2 more answers