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2 years ago

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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.

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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The value of specific heat for copper is 390 J/kg⋅C∘, for aluminun is 900 J/kg⋅C∘, and for water is 4186 J/kg⋅C∘.

abruzzese [7]

Answer:

The equilibrium temperature is

21.97°c

Explanation:

This problem bothers on the heat capacity of materials

Given data

specific heat capacities

copper is Cc =390 J/kg⋅C∘,

aluminun Ca = 900 J/kg⋅C∘,

water Cw = 4186 J/kg⋅C∘.

Mass of substances

Copper Mc = 235g

Aluminum Ma = 135g

Water Mw = 825g

Temperatures

Copper θc = 255°c

Water and aluminum calorimeter θ1= 16°c

Equilibrium temperature θf =?

Applying the principle of conservation of heat energy, heat loss by copper equal heat gained by aluminum calorimeter and water

McCc(θc-θf) =(MaCa+MwCw)(θf-θ1)

Substituting our data into the expression we have

235*390(255-θf)=

(135*900+825*4186)(θf-16)

91650(255-θf)=(3574950)(θf-16)

23.37*10^6-91650*θf=3.57*10^6θf- +57.2*10^6

Collecting like terms and rearranging

23.37*10^6+57.2*10^6=3.57*10^6θf+91650θf

8.2*10^6=3.66*10^6θf

θf=80.5*10^6/3.6*10^6

θf =21.97°c

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1 year ago

Which statement about energy conservation BEST explains why a bouncing basketball will not remain in motion forever?

bearhunter [10]

Answer: d

Explanation:

7 0

1 year ago

During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s

Ray Of Light [21]

Answer:

part a : <em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

part b : <em>The void ratio is 0.77</em>

part c : <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

$\gamma_d$ is the dry unit weight which is to be calculated
γ is the bulk unit weight given as

$\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

w is the moisture content in percentage, given as 12%

Substituting values

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

<em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

Part b

Void Ratio

The void ratio is given as

$e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

e is the void ratio which is to be calculated

$\gamma_d$ is the dry unit weight which is calculated in part a
$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

Substituting values

$e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

<em>The void ratio is 0.77</em>

Part c

Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

e is the void ratio which is calculated in part b

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

$S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261$

<em>Degree of Saturation is 0.43</em>

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}$

Here

$\gamma_{zav}$ is the zero air unit weight which is to be calculated

$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\$

Now as the volume is known, the the overall weight is given as

$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

4 0

1 year ago

A helicopter flies 250 km on a straight path in a direction 60° south of east. The east component of the helicopter’s displaceme

GaryK [48]

Given that,

Distance in south-west direction = 250 km

Projected angle to east = 60°

East component = ?

since,

cos ∅ = base/hypotenuse

base= hyp * cos ∅

East component = 250 * cos 60°

East component = 125 km

8 0

2 years ago

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A gas is compressed from 600 cm3 to 200cm3 at a constant pressure of 400 kpa. at the same time, 100 j of heat energy is transfer

Mekhanik [1.2K]

The initial volume of the gas is
$V_i = 600 cm^3$
while its final volume is
$V_f = 200 cm^3$
so its variation of volume is
$\Delta V = V_f - V-i = 200 cm^3 - 600 cm^3 = -400 cm^3 = -400 \cdot 10^{-6} m^3$

The pressure is constant, and it is
$p=400 kPa = 400 \cdot 10^3 Pa$

Therefore the work done by the gas is
$W=p\Delta V = (400 \cdot 10^3 Pa)(-400 \cdot 10^{-6} m^3)=-160 J$
where the negative sign means the work is done by the surrounding on the gas.

The heat energy given to the gas is
$Q=+100 J$

And the change in internal energy of the gas can be found by using the first law of thermodynamics:
$\Delta U = Q-W = 100 J - (-160 J)=+260 J$
where the positive sign means the internal energy of the gas has increased.

7 0

1 year ago