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2 years ago

The charge of an electron is 1.6x10^'' C. How many electrons does it take to make 1 C of charge?

Physics

1 answer:

olasank [31]2 years ago

6 0

Answer:

If the charge on an electron be 1.6X10^-19 C, find the approximate number of electrons in 1 C. ANSWER: - As it is given that 1.6 X 10-19charge is of 1 electron, So 1 C charge is of = 1/1.6X10-19electron Now number of electrons = 1019/1.6 = (100 X 1018) / 16

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I need help with this please

kondaur [170]

Answer:

1. The respiratory system functions when our involuntary nervous system sends impulses to the muscles in the diaphragm; thereby, causing the lungs to expand and contract.

2. The respiratory system oxygenates the blood which is vital for bodily function as oxygenated blood is carried from your lungs to the left side of your heart, to be circulated throughout the body. Furthermore deoxygenated blood is carried back to the right side of your heart to get oxygenated once more.

3. The other body systems that are crucial for the lungs to function are the nervous system and the muscular system.

4. without the raspatory system the body wouldn't receive any oxygen and the brain would slowly die. therefore, without the brain the heart would stop functioning and atrophy etc.

Explanation:

6 0

3 years ago

One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F

Alja [10]

Answer:

$\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}$

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

$\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}$

$f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}$

b) Wave speed

(i) Calculate the wavelength

In a fundamental vibration, the length of the string is half the wavelength.

$\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}$

(b) Calculate the speed s

$\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}$

$\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}$

$\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}$

4 0

3 years ago

When the mass of the bottle is 0.125 kg, the average maximum height of the beanbag is m. When the mass of the bottle is 0.250 kg

Jet001 [13]

Answer:

when the mass of the bottle is 0.125 kg, the average height of the beanbag is 0.35 m.

when the mass of the bottle is 0.250 kg, the average maximum height of the beanbag is 0.91m.

when the mass of the bottle is 0.375 kg, the average maximum height of the beanbag is 1.26m.

when the mass of the bottle is 0.500 kg, the average maximum height of the beanbag is 1.57m.

Explanation:

5 0

3 years ago

A man lifts a 25.9 kg bucket from a well and does 5.92 kJ of work.The acceleration of gravity is 9.8 m/s^2 . How deep is the wel

Jet001 [13]

Answer:

The well is 23.3 m

Explanation:

As the bucket is lifted out of the well, energy in the man is being transferred to the bucket as gravitational potential energy.

Work done against gravity = mass * height * acceleration due to to gravity

W = mgh

5 920 J = 25.9 kg * h * 9.8 m/s²

h = 23.3 m

4 0

3 years ago

Agri-Chem's contract with Enerco specified a maximum of 90,000 cu.ft. X 103 per day for its complexes. However, curtailments are

zvonat [6]

Answer:

Caustic soda would be least affected by a gas curtailment

Explanation:

Let,

X1= ammonia

X2= ammonium phosphate

X3= ammonium nitrate

X4 = Urea

X5= hydro-fluoric acid

X6= Chlorine

X7= Caustic soda

X8= Vinyl chloride monomer.

Agri chem’s current natural gas usage

= (1200 × 8 + 540 × 10 + 490 × 12 + …)

= 85,680,000 cu. ft. per day

When, the curtailment is 20%, availability is

= 0.8 X 85,680

= 68,554,000 cu. ft. per day

Therefore, the gas constant

= 8X1 + 10X2 + 12X3 + 12X4 + 7X5 + 18X6 + 20X7 + 14X8 ≤68,544

When, the curtailment is 40% availability is

= 0.6 X 85,680

= 51,408,000 cu. ft. per day

Constant = 8X1 + 10X2 + 12X3 + 12X4 + 7X5 + 18X6 + 20X7 + 14X8≤ 51,408

7 0

3 years ago