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WINSTONCH [101]

2 years ago

6

The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is:

I = ( 0.600 A)e^(-t/6hr).
What is the total number of electrons transported from the positive electrode to the negative electrode by the charge escalator from the time the battery is first used until it is completely dead (expressed as a number of electrons)?

1 answer:

ludmilkaskok [199]2 years ago

5 0

Answer: 8.1 x 10^24

Explanation:

I(t) = (0.6 A) e^(-t/6 hr)

I'll leave out units for neatness: I(t) = 0.6e^(-t/6)

If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).

For neatness let k = 1/(6x3600) = 4.63x10^-5, then:

I(t) = 0.6e^(-kt)

Providing t is in seconds, total charge Q in coulombs is

Q= ∫ I(t).dt evaluated from t=0 to t=∞.

Q = ∫(0.6e^(-kt)

= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.

= -(0.6/k)[e^-∞ - e^-0]

= -0.6/k[0 - 1]

= 0.6/k

= 0.6/(4.63x10^-5)

= 12958 C

Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.

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jenyasd209 [6]

To solve this problem we will apply the first law of thermodynamics which details the relationship of energy conservation and the states that the system's energy has. Energy can be transformed but cannot be created or destroyed.

Accordingly, the rate of work done in one cycle and the heat transferred can be expressed under the function,

$\dot{U} = \dot{Q}-\dot{W}$

Substitute 1W for $\dot{Q}$ and 1.5 W for $\dot{W}$

$\dot{U} = 1-1(1.5)$

$\dot{U} = 2.5W$

Now calculcate the rate of specific internal energy increase,

$\dot{u} = \frac{\dot{U}}{m}$

$\dot{u} = \frac{2.5}{1.5}$

$\do{u} = 1.6667W/kg$

The rate of specific internal energy increase is 1.6667W/kg

8 0

2 years ago

While playing basketball in PE class, Logan lost his balance after making a lay-up and colliding with the padded wall behind the

olga2289 [7]

Answer:

a.) F = 3515 N

b.) F = 140600 N

Explanation: given that the

Mass M = 74kg

Initial velocity U = 7.6 m/s

Time t = 0.16 s

Force F = change in momentum ÷ time

F = (74×7.6)/0.16

F = 3515 N

b.) If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds

Change in momentum = 74×7.6 + 74×7.6

Change in momentum = 562.4 + 562.4 = 1124.8 kgm/s

F = 1124.8/0.0080 = 140600 N

6 0

2 years ago

Where would the weight of an explorer be greater?The summit of Chimborazo, in Ecuador, which is at a distance of about 6,384 km6

FromTheMoon [43]

Answer:

The weight would be greater in the Mariana because the radius of the earth is lower.

Explanation:

We will make a comparison through constants and equations to see which one is more viable.

Using the force of gravity

$F = \frac{Gm}{R^2}$

F = Gravity force

G= Gravitational constant

m= mass

R is the radius of the earth, $R_E = 6384$ km and $R_M =6370$ km

Density $(\rho)$ is equal in both places,

-Gravity Force at Ecuador, $F_E=\frac{Gm}{R_E^2}$

-Gravity Force at bottom of Mariana trench, $F_M=\frac{Gm}{R_M^2}$

Making the relation,

$\frac{F_E}{F_M}= \frac{\frac{Gm}{R_E^2}}{\frac{Gm}{R_M^2}}$

$\frac{F_E}{F_M}= (\frac{R_M}{R_E})^2$

For the Ecuador,

$F_E= F_M * \frac{R_M^2}{R_E^2}$

If we take $F_M * R_M^2$ as a constant X then

$F_E= \frac{X}{R_E^2}$

So the gravity force of the place is inversely proportion of the radius

Same for the Mariana trench,

$FM= \frac{X}{R_M^2}$

Then, the weight would be greater in the Mariana because the radius of the earth is lower.

3 0

2 years ago

Two astronauts, A and B, both with mass of 60Kg, are moving along a straight line in the same direction in a weightless spaceshi

sveta [45]

Answer:

$V=14m/s$

Explanation:

From the Question we are told that

Mass of A and B is 60kg

Speed of A=2m/s

Speed of B=1m/s

Mass of bag =5kg

Generally the momentum of the astronaut A and bag is mathematically given as

$M_A=(60+5)*2$

$M_A=130kgm/s$

Generally to avoid collision the speed of astronaut be should be less than or equal to that of astronaut A with the bag

Therefore for minimum requirement speed of astronaut A should be given by astronaut B's speed which is equal to 1

Therefore

$130=(60*1)=(5*v)$

$V=14m/s$

7 0

1 year ago

A farm hand does 972 J of work pulling an empty hay wagon along level ground with a force of 310 N [23° below the horizontal]. T

irina1246 [14]

Answer:

Option d)

Solution:

As per the question:

Work done by farm hand, $W_{FH} = 972J$

Force exerted, F' = 310 N

Angle, $\theta = 23^{\circ}$

Now,

The component of force acting horizontally is F'cos $\theta$

Also, we know that the work done is the dot or scalar product of force and the displacement in the direction of the force acting on an object.

Thus

$W_{FH} = \vec{F'}.\vec{d}$

$972 = 310\times dcos23^{\circ}$

d = 3.406 m = 3.4 m

3 0

2 years ago