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34kurt
2 years ago
11

In a thunderstorm, the air must be ionized by a high voltage before a conducting path for a lightning bolt can be created. an el

ectric field of about 1.0 × 106 v/m is required to ionize dry air. what would the break- down voltage in air be if a thundercloud were 1.60 km above ground? assume that the electric field between the cloud and the ground is uniform.
Physics
1 answer:
nydimaria [60]2 years ago
4 0
We know that the electric field is equal to 1 E 6 V/m

The distance between the thundercloud and the ground is 1.6km = 1600m

Electric field = Voltage/distance

This means that the breakdown voltage must be equal to

V = 1 E 6 V/m * 1600m = 1.6 E 9 V = 1.6 GV
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Our two intrepid relacar drivers are named Pam and Ned. We use these names to make it easy to remember: measurements made by Pam
zhannawk [14.2K]

The velocity of Ned as measured by Pam is the interpretation of v.

Answer: Option D

<u>Explanation:</u>

According to question, we know that this is an issue depending on the logical and translation of the factors. From the measured information taken what is gathered by the two people is communicated and we have given as:

The Ned reference framework : (x, t)  

The Pam reference framework :  \left(x^{\prime}, t^{\prime}\right)

From the reference framework, we realize that ν is the speed of Pam (the other reference framework) as estimation by Ned.  

At that point, v^{\prime} is the speed of Ned (from the other arrangement of the reference) as estimation by Pam.

3 0
2 years ago
10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of
Rina8888 [55]

The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

W=Fd cos \theta

where :

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

\theta=22^{\circ} is the angle between the direction of the force and the displacement

Substituting, we find the work done

W=(87)(44)(cos 22^{\circ})=3549 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
2 years ago
Why are we unable to work long without food
nataly862011 [7]

Answer:

Without food your not able to produce energy thats why you cannot work so long

7 0
2 years ago
Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is
ad-work [718]

Answer:

T_{out}=27.0000077 ºC

Explanation:

First, let's write the energy balance over the duct:

H_{out}=H_{in}+Q

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

m*Cp(T_{out}-T_{in})=Q

So, let's isolate T_{out}:

T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}

The Cp of the air at 27ºC is 1007\frac{J}{kgK} (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are T_{out} and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

P=2*H+2*A=2*0.004m+2*0.016m=0.04m

Surface area:

A=P*L=0.04m*1m=0.04m^2

Then, the heat Q is:

600\frac{W}{m^2} *0.04m^2=24W

Finally, find the exit temperature:

T_{out}=T_{in}+\frac{Q}{m*Cp}\\T_{out}=27+\frac{24W}{3104\frac{kg}{s} *1007\frac{J}{kgK} }\\T_{out}=27.0000077

T_{out}=27.0000077 ºC

The temperature change so little because:

  • The mass flow is so big compared to the heat flux.
  • The transfer area is so little, a bigger length would be required.
3 0
2 years ago
How is a seismograph similar to an X-ray?
aleksklad [387]

Answer: Seismograph is an instrument that is used to measure the vibration of the earthquake. It is based on seismic waves. X ray is an electromagnetic energy wave that is used for CAT ( computerized axial tomography) scan.

Hence, both seismic wave and X ray are energy waves.

The velocity of seismic waves is different in different media. Similarly, X ray loses its amplitude depending upon the dense layer of the tissue.

5 0
2 years ago
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