Answer:
capacitance of the capacitor = 0.18 μ F
Explanation:
Area of the plate A = 0.063 m x 5.4 m = 0.3402 m²
distance between the plate d = 3.5 × 10–5 m
dielectric value for Teflon K = 2.1
capacitance of capacitor = ?
Formula for capacitance of parallel plate is as follows ,
(Where
)
putting the values in the equation,
C =
=0.18 μ F
Answer:
Final Temperature of the steam tank = 456.4°C
Explanation:
Assuming it to be a uniform flow process, kinetic and potential energy to be zero, and work done and heat input to be zero also. We can conclude that,
Enthalpy of the steam in pipe = Internal Energy of the steam in tank
Using the Property tables and Charts - Steam tables,
At Pressure= 1 MPa and Temperature= 300°C,
Enthalpy = 3051.2 kJ/kg
At Pressure= 1 MPa and Internal Energy= 3051.2 kJ/kg,
Temperature = 456.4°C.
Answer:

Explanation:
The magnitude of the gravitational force between two objects is given by the equation:

where
G is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between the objects
The gravitational force is always attractive.
In this problem, we have:
is the mass of the Earth
is the mass of the Moon
is the separation between the Earth and the Moon
Therefore, the gravitational force between them is

in this since your volume remains at a constant you'll need to use Gay-Lussacs law, p1/t1=p2/t2.
your temp should be converted in kelvin
variables:
p1=3.0×10^6 n/m^2
t1= 270k
just add 273 to your celcius
p2= ? your solving for this
t2= 315k
then you set up the equation
(3.0×10^6)/270= (x)(315)
you then cross multiply
(3.0×10^6)315=270x
distribute the 315 to the pressure.
9.45×10^8=270x then you divide 270 o both sides to get
answer
3.5×10^6 n/m^2
Answer:
F = 0.78[N]
Explanation:
The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.
<u>For F₁</u>
<u />
<u />
<u>For F₂</u>
![F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N]](https://tex.z-dn.net/?f=F_%7Bx%7D%3D2%2Acos%2860%29%5C%5CF_%7Bx%7D%3D1%5BN%5D%5C%5CF_%7By%7D%3D-2%2Asin%2860%29%5C%5CF_%7By%7D%3D-1.73%5BN%5D)
<u>For F₃</u>
<u />
<u />
Now we can sum each one of the forces in the given axes:
![F_{x}=1-0.866=0.134[N]\\F_{y}=2-1.73+0.5\\F_{y}=0.77[N]](https://tex.z-dn.net/?f=F_%7Bx%7D%3D1-0.866%3D0.134%5BN%5D%5C%5CF_%7By%7D%3D2-1.73%2B0.5%5C%5CF_%7By%7D%3D0.77%5BN%5D)
Now using the Pythagorean theorem we can find the total force.
![F=\sqrt{(0.134)^{2} +(0.77)^{2}}\\F= 0.78[N]](https://tex.z-dn.net/?f=F%3D%5Csqrt%7B%280.134%29%5E%7B2%7D%20%2B%280.77%29%5E%7B2%7D%7D%5C%5CF%3D%200.78%5BN%5D)