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Taya2010 [7]
3 years ago
8

A resistor rated at 250 k Ω is connected across two D cell batteries (each 1.50 V) in series, with a total voltage of 3.00 V. Th

e manufacturer advertises that their resistors are within 5% of the rated value. What are the possible minimum current and maximum current through the resistor?
Physics
1 answer:
Andrej [43]3 years ago
6 0

Given Information:  

Resistance = R = 250 Ω

Voltage = V = 3 V

Tolerance =   ±5%

Required Information:  

Maximum current = Imax = ?

Minimum current = Imin = ?

Answer:  

Imax = 12.6 uA

Imin = 11.4 uA

Explanation:  

As we know from Ohm's law

I = V/R

The current will be maximum when R is minimum and the current will be minimum when R is maximum

Rmax = 250,000 + 0.05*250,000

Rmax = 250,000 + 12500

Rmax = 262500 Ω

Rmin = 250,000 - 0.05*250,000

Rmin = 250,000 - 12500

Rmin = 237500 Ω

Imax = V/Rmin

Imax = 3/237500

Imax = 12.6 uA

Imin = V/Rmax

Imin = 3/262500

Imin = 11.4 uA

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o-na [289]

Explanation:

When treating the electron as a wave, the concept of electrons orbiting the nucleus allows for the distinct energy levels.The diameter of electron orbits matches the orbit radii which also discrete the energy levels.

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A 2000 kg car is driving at 5 m/s on wet asphalt, but then makes a turn on some ice and loses control. The driver applies brakes
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10,000kgm/s

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The Bohr model of the hydrogen atom pictures the electron as a tiny particle moving in a circular orbit about a stationary proto
igor_vitrenko [27]

a) The angular velocity of the electron is 4.12\cdot 10^{16} rad/s

b) The number of revolutions per second is 6.54\cdot 10^{15} rev/s

c) The centripetal acceleration of the electron is 8.98\cdot 10^{22} m/s^2

Explanation:

a)

This is a problem of uniform circular motion: in fact, the electron orbits around the proton in a uniform circular motion.

The angular velocity of an object in uniform circular motion is given by

\omega = \frac{v}{r}

where

v is the linear speed

r is the radius of the trajectory

For the electron orbiting around the proton, we have

v=2.18 \cdot 10^6 m/s

r=5.29\cdot 10^{-11} m

Therefore, the angular velocity is

\omega=\frac{2.18\cdot 10^6}{5.29\cdot 10^{-11}}=4.12\cdot 10^{16} rad/s

b)

The period of revolution of the electron is given by

T=\frac{2\pi}{\omega}

where

\omega = 4.12\cdot 10^{16}rad/s is the angular velocity

Substituting,

T=\frac{2\pi}{4.12\cdot 10^{16}}=1.53\cdot 10^{-16}s

The period is the time the electron takes to make one complete orbit around the proton; therefore, the number of revolutions of the electrons in one second is:

f=\frac{1}{T}=\frac{1}{1.53\cdot 10^{-16}}=6.54\cdot 10^{15} rev/s

c)

The centripetal acceleration of an object in circular motion is

a=\frac{v^2}{r}

where

v is the linear speed

r is the radius of the circle

For the electron, we have

v=2.18 \cdot 10^6 m/s

r=5.29\cdot 10^{-11} m

Therefore, the centripetal acceleration is

a=\frac{(2.18\cdot 10^6)^2}{5.29\cdot 10^{-11}}=8.98\cdot 10^{22} m/s^2

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

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