Does this equation satisfy the Laws of Conservation of Matter and Mass?

an airplane is flying through a thundercloud at a height of 2000m (this is very dangerous thing to do because of updrafts, turbu

Vedmedyk [2.9K]

Answer

In this question we have given,

Height of plane, h1=2000m

Height at which charge concentration is 40C, h2=3000m

Height at which charge concentration is -40C, h3=1000m

charge concentaration, q1=40C

charge concentaration, q2=-40C

let the charge concentrations at height h2 and h3 as point charges

Now we will first find the electric feild on plane due to positive charge q1=40

$E1= k*q1/(h1-h2)..............(1)$

Here k=8.98755*10^9N.m^2/C^2

q1=40C

put values of k, q1 , h1 and h2 in equation 1

$[tex]E1=(8.98755*10^9)*(40)/(2000-3000)^2$ \\

$E1=[tex]E= 359502+359502\\E=719004 V/m$ V/m[/tex]

similarly electric feild due to negative charge q2=-40

$[tex]E2=(8.98755*10^9)*(-40)/(2000-1000)^2$ \\

$E2=359502V/m$

Total electric feild E at the aircraft is given as

$E= E1+ E2\\$ ...............(2)

Put values of E1 and E2 in equation2

$\\E=359502+359502\\E= 719004V/m\\$

therefore s Total electric feild E at the aircraft is E= 719004V/m

3 0

2 years ago

If a 50 W light bulb and a 90 W light bulb operate from 120 V, which bulb has a greater current in it?

PIT_PIT [208]

The 50 W bulb would have a greater current than the 90 W bulb

8 0

2 years ago

The magnitude of the electric current is directly proportional to the _____________ of the electric field.

MatroZZZ [7]

Answer;

the potential difference

The magnitude of the electric current is directly proportional to the potential difference of the electric field

Explanation;

An electric current results from the collective movement of free charges under the effect of an electric field. An electric field exists and can be observed in the space around a single charge or a number of charges.

Electric fields cause charges to move. It stands to reason that an electric field applied to some material will cause currents to flow in that material. In other words, the current density is directly proportional to the electric field. The constant of proportionality σ is called the material’s conductivity.

8 0

3 years ago

Read 2 more answers

2.5 g of helium at an initial temperature of 300 K interacts thermally with 9.0 g of oxygen at an initial temperature of 620 K .

muminat

Answer:

Explanation:

2.5 g of He = 2.5 / 4 mole

= .625 moles

9 g of oxygen = 9/32

= .28 mole of oxygen

C_p of He = 3/2 R

C_p of O₂ = 5/2 R

A ) Initial thermal energy of He = 3/2 n R T

= 1.5 x .625 x 8.32 x 300

= 2340 J

Initial thermal energy of O₂ = 5/2 n R T

= 2.5 x .28 x 8.32 x 620

= 3610.88 J

B ) If T be the equilibrium temperature after mixing

gain of heat by helium

= n C_p Δ T

= .625 x 3/2 R x ( T - 300 )

Loss of heat by oxygen

n C_p Δ T

= .28 x 5/2 R x ( 620 - T )

Loss of heat = gain of heat

.625 x 3/2 R x ( T - 300 ) = .28 x 5/2 R x ( 620 - T )

1.875 T- 562.5 = 868- 1.4 T

3.275 T = 1430,5

T = 436.8 K

Thermal energy of He

= 1.5 x .625 x 8.32 x 436.8

= 3407 J

thermal energy of O₂

= 2.5 x .28 x 8.32 x 436.8

= 2543.92 J

C )

Heat energy transferred

= .28 x 5/2 R x ( 620 - T )

= .28 x 5/2 x 8.32 x ( 620 - 436.8 )

1066.95 J

Heat will flow from O₂ to He

Final temperature is 436.8 K

7 0

3 years ago

Which most describes this table?

kati45 [8]

Answer:

In the table there we always eat.

It just based on my own...

3 0

2 years ago