Answer:
The value is 
Explanation:
From the question we are told that
The initial pressure is
The initial temperature is ![T_1 = 50 \ F = (50 - 32) * [\frac{5}{9} ] + 273 = 283 \ K](https://tex.z-dn.net/?f=T_1%20%3D%20%2050%20%5C%20F%20%3D%20%2850%20-%2032%29%20%2A%20%5B%5Cfrac%7B5%7D%7B9%7D%20%5D%20%2B%20273%20%3D%20283%20%20%5C%20%20K)
The final temperature is ![T_2 = 320 \ F = (320 - 32) * [\frac{5}{9} ] + 273 =433 \ K](https://tex.z-dn.net/?f=T_2%20%3D%20%20320%20%5C%20F%20%3D%20%28320%20-%2032%29%20%2A%20%5B%5Cfrac%7B5%7D%7B9%7D%20%5D%20%2B%20273%20%3D433%20%20%5C%20%20K)
Generally the equation for adiabatic process is mathematically represented as

=> 
Generally for a monoatomic gas 
So
![14 * 283^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} } =P_2 * 433^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} }](https://tex.z-dn.net/?f=14%20%2A%20283%5E%7B%5Cfrac%7B%5Cfrac%7B5%7D%7B3%7D%20%7D%7B1-%20%5B%5Cfrac%7B5%7D%7B3%7D%20%5D%7D%20%7D%20%3DP_2%20%2A%20433%5E%7B%5Cfrac%7B%5Cfrac%7B5%7D%7B3%7D%20%7D%7B1-%20%5B%5Cfrac%7B5%7D%7B3%7D%20%5D%7D%20%7D)
=> 
=> 
Answer:
1.843 x 10^-5 C
Explanation:
<u><em>Givens:
</em></u>
It is given that the air starts ionizing when the electric field in the air exceeds a magnitude of 3 x 10^6 N/C, which means that the max electric field can stand without forming a spark is 3 x 10^6 N/C.
Also it is given that the radius of the disk is 50 cm, it is required to find out the max amount of charge that the disk can hold without forming spark, which means the charge that would produce the max magnitude of the electric field that air can stand without forming spark, and since we know that the electric field in between 2 disk "Capacitor" is given by the following equation
E = (Q/A)/∈o (1)
Where,
Q: total charge on the disk.
A: the area of the disk.
<u><em>Calculations: </em></u>
We want to find the quantity of charge on the disk that would produce an electric field of 3 x 10^6 N/C, knowing the radius of the disk we can find the cross-section of the disk, thus substituting in equation (1) we find the maximum quantity of charge the disk can hold
Q = EA∈o
= (3 x 10^6) x (π*0.50) x (8.85 x 10^-12)
= 1.843 x 10^-5 C
note:
calculations maybe wrong but method is correct
Answer:
Explanation:
The force exerted in a magnetic field is given as
F = q (v × B)
Where
F is the force entered
q is the charge
v is the velocity
B is the magnetic field
Given that,
The magnetic field is
B = 2•i + 4•j. T
The velocity of the electron is
v = 2•i + 6•j + 8•k. m/s
Also, the charge of an electron is
q = -1.602 × 10^-19 C.
Then note that,
V×B is the cross product of the speed and the magnetic field
Then,
F = q (V×B)
F = -1.602 × 10^-19( 2•i + 4•j +8•k × 2•i + 4•j)
Note
i×i=j×j×k×k=0
i×j=k. j×i=-k
j×k=i. k×j=-i
k×i=j. i×k=-j
F = -1.602 × 10^-19[(2•i + 4•j +8•k) × (2•i + 4•j)]
F = -1.602 × 10^-19 [2×2•(i×i) + 2×4•(i×j) + 4×2•(j×i) + 4×4•(j×j) + 8×2•(k×i) + 8×4•(k×j)]
F = -1.602 × 10^-19[4•0 + 8•k + 8•-k + 16•0 + 16•j + 32•-i]
F = -1.602 × 10^-19(0 + 8•k - 8•k + 0 + 16•j - 32•i)
F = -1.602 × 10^-19(16•j - 32•i)
F = -1.602 × 10^-19 × ( -32•i + 16•j)
F = 5.126 × 10^-18 •i - 2.563 × 10^-18 •j
Then, the x component of the force is
Fx = 5.126 × 10^-18 N
Also, the y component of the force is
Fy = -2.563 × 10^-18 N
Answer:
The speed of the electron is
.
Explanation:
Given that,
The magnitude of electric field, 
The magnitude of magnetic field, B = 0.516 T
Both the magnetic and electric fields are acting on the moving electron. Then, the magnitude of electric field and magnetic field is balanced such that :

or

So, the speed of the electron is
. Hence, this is the required solution.
Answer:
Explanation:
We shall apply law of conservation of momentum in space to know the velocity of combination after the impact
m₁v₁ = m₂v₂
.1 x 4 = ( 1 + .1 ) v₂
v₂ = .3636 m /s
1 )
Kinetic energy of the combination
= 1/2 x 1.1 x ( .3636)²
= 7.3 x 10⁻² J
2 )
Initial kinetic energy of the system
= 1/2 x 0.1 x 4²
= 0.8 J
Final kinetic energy of the system = 7.3 x 10⁻²
Loss of energy = .8 - .073
= .727 J
This energy was converted into internal energy of the system .
3 )
increase in entropy = dQ / T
Here dQ = .727 J
T = 300 ( Constant )
dQ / T = 2.42 X 10⁻³ J/K