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algol [13]
2 years ago
7

A current of 0.4 A flows through a wire. How many electrons flow through a cross section of

Physics
1 answer:
Free_Kalibri [48]2 years ago
4 0

9 × 10²¹ electrons flow through a cross section of the wire in one hour.

<h3>What is the relation between current and charge?</h3>
  • Mathematically, current = charge / time
  • In S.I. unit, Charge is written in Coulomb and time in second.

<h3>What is the amount of charge flown through a wire for one hour if it carries 0.4 A current?</h3>
  • Charge= current × time
  • Current= 0.4 A, time = 1 hour= 3600 s
  • Charge= 0.4× 3600

= 1440 C

<h3>How many numbers of electrons present in 1440C of charge?</h3>
  • One electron= 1.6 × 10^(-19) C
  • So, 1440 C = 1440/1.6 × 10^(-19)

= 9 × 10²¹ electrons

Thus, we can conclude that the 9 × 10²¹ electrons flow through a cross section of the wire in one hour.

Learn more about current here:

brainly.com/question/25922783

#SPJ1

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A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck
Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing}) where M_{truck} is the mass of the truck, V_{truck} is velocity of the truck, V_{common} is the common velocity of moving and standing truck after collision and M_{standing} is the mass of the standing truck

Making V_{truck} the subject we obtain

V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}

Substituting M_{truck} as 25000 Kg, V_{common} as 22.3 m/s, M_{standing} as 2000 Kg we obtain

V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0
3 years ago
Two stones resembling diamonds are suspected of being fakes. To determine if the stones might be real, the mass and volume of ea
dimaraw [331]

Answer:

stone A is diamond.

Explanation:

given,

Volume of the two stone =  0.15 cm³

Mass of stone A = 0.52 g

Mass of stone B = 0.42 g

Density of the diamond =  3.5 g/cm³

So, to find which stone is gold we have to calculate the density of both the stone.

We know,

density\density = \dfrac{mass}{volume}

density of stone A

\rho_A = \dfrac{0.52}{0.15}

\rho_A = 3.467\ g/cm^3

density of stone B.

\rho_B = \dfrac{0.42}{0.15}

\rho_B = 2.8\ g/cm^3

Hence, the density of the stone A is the equal to Diamond then stone A is diamond.

6 0
3 years ago
Name three categories that are used to classify the elements in the periodic table?
nikitadnepr [17]

Answer:

metals,nonmetals, and inert gases

Explanation:

6 0
3 years ago
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I need help with number 2 and 3
Naddik [55]
1. Gas
2. Hot
Are the answers.

8 0
3 years ago
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A 72 kg skydiver can be modeled as a rectangular "box" with dimensions 21 cm × 41 cm × 170 cm . if he falls feet first, his drag
LuckyWell [14K]

Formula for terminal velocity is:


Vt = √(2mg/ρACd) 
<span>Vt = terminal velocity = ?
<span>m = mass of the falling object = 72 kg
<span>g = gravitational acceleration = 9.81 m/s^2
<span>Cd = drag coefficient = 0.80
<span>ρ = density of the fluid/gas = 1.2 kg/m^3</span>
<span>A = projected area of the object (feet first) = 0.21 m * 0.41 m = 0.0861 m^2

Therefore:</span></span></span></span></span>

Vt =  √(2 * 72 * 9.81 / 1.2 * 0.0861 * 0.80) 

<span>Vt = 130.73 m/s</span>

4 0
3 years ago
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