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uranmaximum [27]
4 years ago
13

How long does it take electrons to get from the car battery to the starting motor? Assume the current is 143 A and the electrons

travel through copper wire with cross sectional area 40.2 mm2 and length 84.5 cm . The mass density of copper is 8960 kg/m3 and the molar mass is 63.5 g/mol . Avogadro’s number is 6.022 × 1023 . Assume that each copper atom contributes one conduction electron.
Physics
1 answer:
Ostrovityanka [42]4 years ago
5 0

Answer

the concentration of electron/ volume (n) = 8.5 × 10²⁸ /m³

the drift velocity of electron

v = \dfrac{I}{neA}

v = \dfrac{143}{8.5\times 10^{28}\times 1.6\times 10^{-19}\times 40.2 \times 10^{-6}}

v = 2.615 × 10⁻⁴ m/s

time require is

t=\dfrac{L}{v}

t=\dfrac{0.845}{2.615\times 10^{-4}}

t = 3231.35 s

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What is the range of wavelengths that our eyes can see on the spectrum(this is called visible light spectrum)
Nookie1986 [14]

Answer:

380 to 700 nanometers

Explanation:

The visible light spectrum is the segment of the electromagnetic spectrum that the human eye can view. More simply, this range of wavelengths is called visible light. Typically, the human eye can detect wavelengths from 380 to 700 nanometers.

6 0
3 years ago
In a thunderstorm, the thunder occurs at the same time as lightning. Explain how does a person hears the thunder 5s after he see
sergey [27]

Answer:

If you count the number of seconds between the flash of lightning and the sound of thunder, and then divide by 5, you'll get the distance in miles to the lightning: 5 seconds = 1 mile, 15 seconds = 3 miles, 0 seconds = very close. Keep in mind that you should be in a safe place while counting.

Explanation:

4 0
3 years ago
Object's velocity change
marshall27 [118]

the definition of acceleration in kinematics, allows to find that the correct answer is:

     2.  speed up  ( acceleration)

The kinematics study the movement of the body, the acceleration is defined as the change of the speed by the time in the interval

             a = Δv / t

Where the bold letters indicate vectors, a is the acceleration, v the velocity and t the time

Analyzing this expression we see that for there to be a change in velocity, there must be an acceleration of the body.

Let's analyze the different claims

1. True. If it is stopped and you start to move there is an acceleration, therefore there is a change in speed, but after you are moving the acceleration becomes zero and there is no change in speed

2. True. Whenever there is acceleration there is a change in speed

3. False. Moving slowly does not change the acceleration, therefore there is no change in speed

4. True. If you are moving and you stop at this moment there is an acceleration, therefore there is a change in speed, but after being stopped the acceleration is zero and there is no change in speed

5. False. If you change the direction at the instant of change there is an acceleration but after you go in the opposite direction there is not acceleration therefore there is no change in speed.

In conclusion using the definition of acceleration in kinematics, we can find the answer the condition for a change in velocity, the correct statement is:

     2.   speed up  ( acceleration)

Learn more here:  brainly.com/question/5063616

3 0
2 years ago
You walk with a velocity of 2 m/s north. You see a man approaching you, and from your frame of
solong [7]

Answer:

The velocity of the man from the frame of  reference of a stationary observer is, V₂ = 5 m/s

Explanation:

Given,

Your velocity, V₁ = 2 m/

The velocity of the person, V₂ =?

The velocity of the person relative to you, V₂₁ = 3 m/s

According to the relative velocity of two

                                V₂₁ = V₂ -V₁

∴                               V₂ =  V₂₁ + V₁

On substitution

                                 V₂ = 3 + 2

                                      = 5 m/s

Hence, the velocity of the man from the frame of reference of a stationary observe is, V₂ = 5 m/s

8 0
3 years ago
Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass
Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

\Delta W = \Delta PE + \Delta KE

Where,

PE = Potential Energy

KE = Kinetic Energy

\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

P = \frac{\Delta W}{\Delta t}

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

The rate of mass flow is,

\frac{\Delta m}{\Delta t} = \rho_w Av

Where,

\rho_w = Density of water

A = Area of the hose \rightarrow A=\pi r^2

The given radius is 0.83cm or 0.83 * 10^{-2}m, so the Area would be

A = \pi (0.83*10^{-2})^2

A = 0.0002164m^2

We have then that,

\frac{\Delta m}{\Delta t} = \rho_w Av

\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)

\frac{\Delta m}{\Delta t} = 1.16856kg/s

Final the power of the pump would be,

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)

P = 57.1192W

Therefore the power of the pump is 57.11W

6 0
3 years ago
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