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mylen [45]
2 years ago
5

A standard AA battery provides 1.5 V. How many pennies and nickels would you need to include in a voltaic pile to produce the sa

me voltage as a standard AA battery?
Physics
1 answer:
GrogVix [38]2 years ago
8 0

Answer: In a battery, voltage determines how strongly electrons are pushed through a circuit, much like pressure determines how strongly water is pushed through a hose. Most AAA, AA, C, and D batteries are around 1.5 volts. Imagine the batteries shown in the diagram are rated at 1.5 volts and 500 milliamp-hours.

Explanation: Today "AA" is frequently used as a size designation, irrespective of the battery's electrochemical system. The main numbers used for the most common NiMH and NiCad battery

You might be interested in
The velocity of sound in air saturated with water vapour at 30°C
Luba_88 [7]

Explanation:

The velocity of sound depends on the density of the medium.  So we need to find the density of air at each set of conditions.  The density of air is:

ρ = (Pd / (Rd T)) + (Pv / (Rv T))

where Pd and Pv are the partial pressures of dry air and water vapor,

Rd and Rv are the specific gas constants of dry air and water vapor,

and T is the absolute temperature.

At the first condition:

Pv = 31.7 mmHg = 4226.3 Pa

Pd = 650 mmHg - 31.7 mmHg = 618.3 mmHg = 82433 Pa

Rv = 461.52 J/kg/K

Rd = 287.00 J/kg/K

T = 30°C = 303.15°C

ρ = (82433 / 287.00 / 303.15) + (4226.3 / 461.52 / 303.15)

ρ = 0.94746 + 0.03021

ρ = 0.97767 kg/m³

At the second condition:

Pv = 0 Pa

Pd = 650 mmHg = 86660 Pa

Rv = 461.52 J/kg/K

Rd = 287.00 J/kg/K

T = 0°C = 273.15°C

ρ = (86660 / 287.00 / 273.15) + (0 / 461.52 / 273.15)

ρ = 1.1054 + 0

ρ = 1.1054 kg/m³

The square of the velocity of sound is proportional to the ratio between pressure and density:

v² = k P / ρ

Since the atmospheric pressure is constant, we can say it's proportional to just the density:

v² = k / ρ

Using the first condition to find the coefficient:

(340)² = k / 0.97767

k = 113018.652

Now finding the velocity of sound at the second condition:

v² = 113018.652 / 1.1054

v = 319.75

6 0
3 years ago
A 4.40-m-long, 500 kg steel uniform beam extends horizontally from the point where it has been bolted to the framework of a new
icang [17]

Answer:

Torque=13798.4 N.m

Explanation:

Given data

Mass of beam m₁=500 kg

Mass of the person m₂=70 kg

length of steel r₁=4.40m

center of gravity of the beam is at r₂=r₁/2 =4.40/2 = 2.20m

To find

Torque

Solution

Torque due to beam own weight

T_{1}=m_{1}gr_{1}\\ T_{1}=500*2.2*9.8\\T_{1}=10780N.m

Torque due to person

T_{2}=m_{2}r_{2}g\\T_{2}=70*(4.40)*(9.8)\\T_{2}=3018.4 N.m

Now for total torque

T_{total}=T_{1}+T_{2}\\T_{total}=10780+3018.4\\T_{total}=13798.4N.m

4 0
3 years ago
The chart shows the temperatures of four different substances.
garik1379 [7]

Answer:

C) 20

Explanation:

Happy Halloween LOL

4 0
3 years ago
Read 2 more answers
an electron and a 0.0320-kg bullet each have a velocity of magnitude 510 m/s, accurate to within 0.0100%. within what lower limi
MArishka [77]

for this we apply, Heisenberg's uncertainty principle.

it states that physical variables like position and momentum, can never simultaneously know both variables at the same moment.

the formula is,

Δp * Δx = h/4π

m(e).Δv * Δx = h/4π

by rearranging,

Δx = h / 4π * m(e).Δv

Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 5.10*10^-2

Δx = 6.63*10^-34 / 583.9 X 10 ⁻³¹

Δx = 0.011 X 10⁻³

for the bullet

Δx = (6.63*10^-34) / 4 * 3.142 * 0.032*10^-31 * 5.10*10^-2

Δx = 6.63*10^-34 /2.05

Δx =3.23 X 10⁻³² m

therefore, we can say that the lower limits are   0.011 X 10⁻³  m for the electron and  3.23 X 10⁻³² m   for the bullet

To know more about bullet problem,

brainly.com/question/21150302

#SPJ4

6 0
1 year ago
A 615.00 kg race car is uniformly traveling around a circular race track. It takes the race car 20.00 seconds to do one lap arou
koban [17]

Answer:

The value is  f  =  0.05 \ Hz

Explanation:

From the question we are told that

     The mass of the car is  m  =  615 \  kg

      The period of the circular motion is  T  =  20 \  s

      The radius  is r =  80 \  m/s

Generally the frequency of the circular motion is  

       f  =   \frac{1}{T }

=>    f  =   \frac{1}{ 20  }

=>    f  =  0.05 \ Hz

3 0
3 years ago
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