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Montano1993 [528]
2 years ago
13

1. Describe the methods by which an electric potential develops in primary cells and dry cells.

Physics
1 answer:
Andreyy892 years ago
6 0

Answer:

In primary cells, an electric potential develops through chemical action between the plates within the cell. Positively charged ions of zinc enter the acid and free electrons released from zinc atoms collect on the zinc plate, which results in a negative charge. At the same time, positively charged ions of hydrogen from the acid remove free electrons from the copper plate, which becomes positively charged. Through a conducting material connecting the plates, free electrons move from the zinc plate to the copper plate as long as the chemical reaction lasts.

Dry cells also develop electric potential via chemical actions within the cell. Free electrons removed from the carbon rod collect on a zinc can. The rod exhibits a positive charge and the can becomes negatively charged; this allows for an electric potential to develop between these two items. Through a conducting material connecting the can to the rod, free electrons move from the can to the rod as long as the conducting path exists.

Electric generators develop an electric potential via magnetic induction. Moving a conducting rod through a magnetic field that exists between the poles of a horseshoe magnet causes an electric potential to be set up in the rod. Free electrons move through this rod from one end to the other for as long as movement of the rod is maintained. The direction of this movement depends on whether the rod is moved across the lines of force in the magnetic field in either the opposite direction or the same direction. Generators usually consist of multiple conductors mounted on a cylinder that rotates in a magnetic field.

Thermocouples utilize heat to develop an electric potential. Two strips of different metals are connected at one end to form a junction and the other ends are kept apart. A heat source is applied to the junction; this causes each metal strip’s temperature to rise at the junction. The free ends aren’t as hot and electric charges are produced at these free ends. Because the strips consist of different materials, there's a difference of potential between these free ends; when connected by a conducting wire, the electrons can move through the pathway. The voltage that's produced will become greater as the difference in temperature between the free ends and the junction increases.

a. Increase

b. Decrease

c. Decrease

Since 1 Btu = 0.293 Wh, dividing the given amount of Wh by 0.293 will convert this amount into Btu. Therefore, 0.8 ÷ 0.293 = 2.73 Btu

365 days × 10 hours × 40 W = 146,000 Wh or 146 kWh

Explanation:

Penn Foster

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8 0
2 years ago
If the angle of elevation of the cannon is decreased from 35 to 30 degrees, the vertical component of the ball's initial velocit
stealth61 [152]

Answer:

Determine the initial horizontal and vertical velocity for a dart launched with a ... an angle of 30 degrees. Vi=4m/s Vix Vicoso Mas cos 30 = = 30°. Viy=visine ... What is the resultant direction (angle) of the cannon ball (relative to a fixed frame)? ... Which of the following launch angles would result in the greatest time of flight iſ the.

Explanation:

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4 0
2 years ago
A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helic
andreev551 [17]

Answer:

a) 447.21m

b) -62.99 m/s

c)94.17 m/s

Explanation:

This situation we can divide in 2 parts:

⇒ Vertical : y =-200 m

y =1/2 at²

-200 = 1/2 *(-9.81)*t²

t= 6.388766 s

⇒Horizontal: Vx = Δx/Δt

Δx = 70 * 6.388766 = 447.21 m

b) ⇒ Horizontal

Vx = Δx/Δt ⇒ 70 = 400 /Δt

Δt= 5.7142857 s

⇒ Vertical:

y = v0t + 1/2 at²

-200 = v(5.7142857) + 1/2 *(-9.81) * 5.7142857²

v0= -7 m/s  ⇒ it's negative because it goes down.

v= v0 +at

v= -7 + (-9.81) * 5.7142857

v= -62.99 m/s

c) √(70² + 62.99²) = 94.17 m/s

8 0
2 years ago
A block of mass 2.00 kg is initially at rest at x=0 on a slippery horizontal surface for which there is no friction. Starting at
Allisa [31]

Answer:

   x = 1,185 m ,     t = 4/3 s ,  F = - 4 N

Explanation:

For this exercise we use Newton's second law

         F = m a = m dv /dt

        β - α t = m dv / dt

        dv = (β – α t) dt

     

We integrate

        v = β t - ½ α t²

We evaluate between the lower limits v = v₀ for t = 0 and the upper limit v = v for t = t

       v-v₀ = β t - ½ α t²

the farthest point of the body is when v = v₀ = 0

  0 = β t - ½ α t²

  t = 2 β / α

  t = 2 4/6

  t = 4/3 s

Let's find the distance at this time

   v = dx / dt

   dx / dt = v₀ + β t - ½ α t2

   dx = (v₀ + β t - ½ α t2) dt

We integrate

   x = v₀ t + ½ β t - ½ 1/3 α t³

   x = v₀ 4/3 + ½ 4 (4/3)² - 1/6 6 (4/3)³

The body comes out of rest

    x = 3.5556 - 2.37

    x = 1,185 m

The value of force is

    F = β - α t

    F = 4 - 6 4/3

   F = - 4 N

8 0
2 years ago
What is the resistance ofa wire made of a material with resistivity of 3.2 x 10^-8 Ω.m if its length is 2.5 m and its diameter i
Katarina [22]

R = 0.407Ω.

The resistance  R of a particular conductor is related to the resistivity ρ of the material by  the equation R = ρL/A, where ρ is the material resistivity, L is the length of the material and A is the cross-sectional area of ​​the material.

To calculate the resistance R of a wire made of a material with resistivity of 3.2x10⁻⁸Ω.m, the length of the wire is 2.5m and its diameter is 0.50mm.

We have to use the equation R = ρL/A but first we have to calculate the cross-sectional area of the wire which is a circle. So, the area of a circle is given by A = πr², with r = d/2. The cross-sectional area of the wire is A = πd²/4.  Then:

R =[(3.2x10⁻⁸Ω.m)(2.5m)]/[π(0.5x10⁻³m)²/4]

R = 8x10⁻⁸Ω.m²/1.96x10⁻⁷m²

R = 0.407Ω

5 0
2 years ago
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