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NNADVOKAT [17]
3 years ago
11

A typical cell phone charger is rated to transfer a maximum of 1.0 Coulomb of charge per second. Calculate the maximum number of

electrons that can be transferred by this charger in 1.0 hour
Physics
1 answer:
Alexandra [31]3 years ago
3 0

Answer:

The maximum no. of electrons- 2.25\times 10^{22}

Solution:

As per the question:

Maximum rate of transfer of charge, I = 1.0 C/s

Time, t = 1.0 h = 3600 s

Rate of transfer of charge is current, I

Also,

I = \frac{Q}{t}

Q = ne

where

n = no. of electrons

Q = charge in coulomb

I = current

Thus

Q = It

Thus the charge flow in 1. 0 h:

Q = 1.0\times 3600 = 3600\ C

Maximum number of electrons, n is given by:

n = \frac{Q}{e}

where

e = charge on an electron = 1.6\times 10^{- 19}\ C

Thus

n = \frac{3600}{1.6\times 10^{- 19}} = 2.25\times 10^{22}

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lyudmila [28]
Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
W= \int\limits^{0.540m}_{0} {F} \, dx =  \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =
=16000x+10000  \frac{x^2}{2} - 26000  \frac{x^3}{3}
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
W=16000(0.540m)+10000  \frac{(0.540m)^2}{2} - 26000  \frac{(0.540m)^3}{3}  =
=8733 J=8.73 kJ

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
W=16000(0.95m)+10000  \frac{(0.95m)^2}{2} - 26000  \frac{(0.95m)^3}{3}  =
=12280 J=12.28 kJ
5 0
3 years ago
A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s . Assume in this problem that air drag
Readme [11.4K]

Answer:

(a)0.0675  J

(b)0.0675 J

(c)0.0675 J

(d)0.0675 J

(e)-0.0675 J

(f)0.459 m

Explanation:

15g = 0.015 kg

(a) Kinetic energy as it leaves the hand

E_k = \frac{mv^2}{2} = \frac{0.015*3^2}{2} = 0.0675  J

(b) By the law of energy conservation, the work done by gravitational energy as it rises to its peak is the same as the kinetic energy as the ball leave the hand, which is 0.0675 J

(c) The change in potential energy would also be the same as 0.0675J in accordance with conservation law of energy.

(d) The gravitational energy at peak point would also be the same as 0.0675J

(e) In this case as the reference point is reversed, we would have to negate the original potential energy. So the potential energy as the ball leaves hand is -0.0675J

(f) Since at the maximum height the ball has potential energy of 0.0675J. This means:

mgh = 0.0675

0.015*9.81h = 0.0675

h = 0.459 m

The ball would reach a maximum height of 0.459 m

4 0
3 years ago
What happens if the air pressure in your throat and outside body is less than the air pressure in your middle ear when you swall
andrezito [222]

Answer:

When the air pressure in the throat and outside the body is less than the air pressure in the middle ear, barotrauma occurs.

Explanation:

Ear barotrauma is a medical condition that describes discomfort in the ear which is caused by pressure differences in the inner and outer ear drum.

Usually, the air pressure in the middle ear is the same as the air pressure in the throat and outside the body.

When we swallow, the eustachian tube opens up and air flows out of and into the middle ear, this balances the pressure. But if the eustachian tube is blocked, the air pressure in the throat and outer body become different from the air pressure in the middle ear.

8 0
3 years ago
A 2500-lb vehicle has a drag coefficient of 0.35 and a frontal area of 20 ft2. What is the minimum tractive effort required for
emmasim [6.3K]

Answer:

i put this in the calculator and my answer is 600. hope this helps

Explanation:

4 0
4 years ago
Before the start of a trip, air in a tire is at 320 kPa gage pressure and 27°C temperature. At the end of the trip the tire pres
tia_tia [17]

Answer:

The temperature of air in the tire is 55.57 ºC

Explanation:

Please look at the solution in the attached Word file

Download docx
7 0
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