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beks73 [17]
2 years ago
10

Suppose the battery in a clock wears out after moving thousand coulombs of charge through the clock at a rate of 0.5 Ma how long

did the clock run on does battery and how many electrons per second slowed?
Physics
1 answer:
Ksivusya [100]2 years ago
8 0

Answer:

Hello your question is poorly written below is the complete question

Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?

answer :

a) 231.48 days

b) n = 3.125 * 10^15

Explanation:

Battery moved 10,000 coulombs

current rate = 0.5 mA

<u>A) Determine how long the clock run on the battery. use the relation below</u>

q = i * t ----- ( 1 )

q = charge , i = current , t = time

10000 = 0.5 * 10^-3 * t

hence  t = 2 * 10^7 secs

hence the time = 231.48  days

<u>B) Determine how many electrons per second flowed </u>

q = n*e ------ ( 2 )

n = number of electrons

e = 1.6 * 10^-19

q = 0.5 * 10^-3 coulomb ( charge flowing per electron )

back to equation 2

n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )

hence : n = 3.125 * 10^15

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monitta

<u>P</u><u>e</u><u>r</u><u>s</u><u>o</u><u>n</u><u>-</u><u>1</u>

  • Initial velocity=u=0m/s
  • Final velocity=v=10m/s
  • Time=10s=t

\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \sf\longmapsto Acceleration=\dfrac{10-0}{10}

\\ \sf\longmapsto Acceleration=\dfrac{10}{10}

\\ \sf\longmapsto Acceleration=1m/s^2

<u>P</u><u>e</u><u>r</u><u>s</u><u>o</u><u>n</u><u>-</u><u>2</u>

  • initial velocity=0m/s=u
  • Final velocity=v=0.25m/s
  • Time=t=2s

\\ \sf\longmapsto Acceleration=\dfrac{0.25-0}{2}

\\ \sf\longmapsto Acceleration=\dfrac{0.25}{2}

\\ \sf\longmapsto Acceleration=0.125m/s^2

Person-1 is accelerating faster.

4 0
3 years ago
Which of the following are single-displacement reactions?
ololo11 [35]

Answer:

A & D

Explanation:

A single-displacement reaction is a chemical reaction whereby one element is substituted for another one in a compound and thereby generating a new element and also a new compound as products.

From the options, only options A & D fits this definition of single-displacement reactions.

For option D: Both left and hand and right hand sides each have one element and one compound. We can see that K is substituted from KBr to join Cl to form KCl and Br2 on the right hand side.

For option A: Both left and hand and right hand sides each have one element and one compound. We can see that OH is substituted from 2H2O to join Mg to form Mg(OH)2 and H2 on the right hand side.

The other options are not correct because they don't involve only and element and a compound on each side of the reaction.

6 0
2 years ago
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An electromagnetic wave of intensity 150 W/m2 is incident normally on a rectangular black card with sides of 25 cm and 30 cm tha
LenKa [72]

Answer:

3.75 × 10⁻⁸ N

Explanation:

Given:

Intensity of the electromagnetic wave, I = 150 W/m²

Sides of the board = 25 cm (= 0.25 m) and 30 cm (= 0.30 m)

therefore,

the area of the rectangular box, A = 0.25 × 0.30 = 0.075 m²

Now,

force exerted on the card by the radiation, F = \frac{IA}{C}

here,

C is the speed of the light = 3 × 10⁸ m/s

on substituting the respective values, we get

F = \frac{150\times0.075}{3\times10^8}

or

F = 3.75 × 10⁻⁸ N

5 0
2 years ago
Which statement corresponds to emission spectra?
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7 0
3 years ago
Suppose that you have a 680 Ω, a 720 Ω and a 1.20 kΩ resistor. (a) What is the maximum resistance you can obtain by combining th
Delvig [45]

Explanation:

As the given data is as follows.

    R_{1} = 680 \ohm ohm\ohm,    R_{2} = 720 \ohm ohm,

   R_{3} = 1.2 k\ohm = 1200 \ohm   (as 1 k ohm = 1000 m)

(a)   We will calculate the maximum resistance by combining the given resistances as follows.

      Max. Resistance = R_{1} + R_{2} + R_{3}

                                  = (680 + 720 + 1200) \ohm ohm

                                  = 2600 ohm

or,                               = 2.6 k\ohm ohm

Therefore, the maximum resistance you can obtain by combining these is 2.6 k\ohm ohm.

(b)   Now, the minimum resistance is calculated as follows.

      Min. Resistance = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}

                                 = \frac{1}{680} + \frac{1}{720} + \frac{1}{1200}

                                 = 3.683 \times 10^{-3} ohm

Hence, we can conclude that minimum resistance you can obtain by combining these is 3.683 \times 10^{-3} ohm.

3 0
2 years ago
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