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Dmitry_Shevchenko [17]
2 years ago
7

A 1.5 volt, AAA cell supplies 750 milliamperes of current through a flashlight bulb for 5.0 minutes, while a 1.5 volt, C cell su

pplies 750 milliamperes of current through the same flashlight bulb for 20 minutes. Compared to the total charge transferred by the AAA cell through the bulb, the total charge transferred by the C cell through the bulb is
A) half as great

B) twice as great

C) the same

D) four times as great

Why??
Physics
1 answer:
likoan [24]2 years ago
7 0

Answer:

D) four times as great

Explanation:

The current through a circuit or device is given by the formula:

I = q/t

q = (I)(t)

where,

I = current

q = charge

t = time

<u>FOR AAA CELL</u>:

I = 750 mA = 0.75 A

t = 5 min = (5)(60) sec = 300 sec

q = q₁

Therefore,

q₁ = (0.75 A)(300 sec)

q₁ = 225 Coulomb

<u></u>

<u>FOR C CELL</u>:

I = 750 mA = 0.75 A

t = 20 min = (20)(60) sec = 1200 sec

q = q₂

Therefore,

q₂ = (0.75 A)(1200 sec)

q₂ = 900 Coulomb

Thus, from the values of charges for both cells, it is clear that, Compared to the total charge transferred by the AAA cell through the bulb, the total charge transferred by the C cell through the bulb is:

<u>D) four times as great</u>

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2 years ago
A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg
WINSTONCH [101]

Answer:

=2,012,319.36 \ m/s

Explanation:

-The only relevant force is the electrostatic force

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F = Eq

E is the electric field and q is the magnitude of the charge.

#Since the electric field is the same in both cases, and the charge of the protons and electrons have the same magnitude, you can state that the magnitude of the electric forces acting in both proton and electron are the same.

F_e = F_p\\\\F_e= Force \ on \ electron\\F_p = Force \ on \ proton

-Applying Newton's 2nd Law:

F=ma

F_e=M_ea_e

F_p=M_pa_p

#equate the two forces:

F_e = F_p\\\\M_ea_e=M_pa_p\\\\a_e=\frac{M_pa_p}{M_e}

#The equations for velocity in uniform acceleration:

V_f^2=V_o^2+2ad\\\\V_o^2=0\\\\\therefore V_f^2=2ad

#For the proton:

V_f^2=2a_pd\\\\a_p=\frac{V_f^2}{2d}\\\\a_p=\frac{47000m/s)^2}{2d}

#For the electron:

V_f^2=2{a_e}^2\times 2d\\\\A_e=M_p\times A_p/M_e\\\\V_f^2=M_p\times (47000m/s)^2/2d\times2d/M_e\\\\V_f^2=M_p\times (47000m/s)^2/M_e\\\\V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}

The mass values of the proton and electron are:

M_p=1.67\times 10^{-27} kg\\\\M_e=9.11\times10^{-31}kg

The speed of the ion is therefore calculated as:

V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}\\\\=47000m/s\times\sqrt{\frac{1.67\times10^{-27}}{9.11\times10^{-31}}\\\\=2,012,319.36 \ m/s

Hence, the ion's speed at the negative plate is =2,012,319.36 \ m/s

7 0
2 years ago
If a 1,300 kg car with no people inside is on the edge of a cliff 1,500 m above the ground, what is its potential energy?
Ghella [55]

<u>Given that</u>

mass (m) = 1300 Kg ,

height (h) = 1500 m

Determine the potential energy ?

     P.E = m × g × h

           = 1300 × 9.81 × 1500

           = 19129500  Joules

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3 0
2 years ago
Two billiard balls of equal mass are traveling straight toward each other with the same speed. They meet head-on in an elastic c
Rus_ich [418]

Answer:

0 kg m/s before and after collision

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P = mv - mv = 0 kg m/s

As the collision is elastic. The total momentum after the collision is the same as the total momentum before the collision, which is 0.

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You hang different masses M from the lower end of a vertical spring and measure the period T for each value of M. You use Excel
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Answer:

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T^2=\frac{4\pi^2}{k}M +\frac{4\pi^2}{k}m_{spring}

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k =spring constant

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a) So for the equation we can compare, that is,

y=T^2=0.0569x+0.0010

the hanging mass M is x here, so comparing the equation we know that

\frac{4\pi^2}{k}=0.0569\\k= \frac{4\pi^2}{0.0569}\\k=693.821N/m

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2 years ago
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