Answer:
a)-1.014x 
J
b)3.296 x J
Explanation:
For Sphere A:
mass 'Ma'= 47kg
xa= 0
For sphere B:
mass 'Mb'= 110kg
xb=3.4m
a)the gravitational potential energy is given by
= -GMaMb/ d
= - 6.67 x 
x 47 x 110/ 3.4 => -1.014x J
b) at d= 0.8m (3.4-2.6) and =-1.014x J
The sum of potential and kinetic energies must be conserved as the energy is conserved.
+ = 
+ 
As sphere starts from rest and sphere A is fixed at its place, therefore is zero
= +
The final potential energy is
= - GMaMb/d
Solving for ' '
= + GMaMb/d => -1.014x + 6.67 x x 47 x 110/ 0.8
= 3.296 x J
Answer:
Explanation:
The inlet specific volume of air is given by:
The mass flow rates is expressed as:
The energy balance for the system can the be expresses in the rate form as:
![E_{in}-E_{out}=\bigtriangleup \dot E=0\\\\E_{in}=E_{out}\\\\\dot m(h_1+0.5V_1^2)=\dot W_{out}+\dot m(h_2+0.5V_2^2)+Q_{out}\\\\\dot W_{out}=\dot m(h_2-h_1+0.5(V_2^2-V_1^2))=-m({cp(T_2-t_1)+0.5(V_2^2-V_1^2)})\\\\\\\dot W_{out}=-(10.42lbm/s)[(0.25\frac{Btu}{lbm.\textdegree F})(300-900)\textdegree F+0.5((700ft/s)^2-(350ft/s)^2)(\frac{1\frac{Btu}{lbm}}{25037ft^2/s^2})]\\\\\\\\=1486.5\frac{Btu}{s}](https://tex.z-dn.net/?f=E_%7Bin%7D-E_%7Bout%7D%3D%5Cbigtriangleup%20%5Cdot%20E%3D0%5C%5C%5C%5CE_%7Bin%7D%3DE_%7Bout%7D%5C%5C%5C%5C%5Cdot%20m%28h_1%2B0.5V_1%5E2%29%3D%5Cdot%20W_%7Bout%7D%2B%5Cdot%20m%28h_2%2B0.5V_2%5E2%29%2BQ_%7Bout%7D%5C%5C%5C%5C%5Cdot%20W_%7Bout%7D%3D%5Cdot%20m%28h_2-h_1%2B0.5%28V_2%5E2-V_1%5E2%29%29%3D-m%28%7Bcp%28T_2-t_1%29%2B0.5%28V_2%5E2-V_1%5E2%29%7D%29%5C%5C%5C%5C%5C%5C%5Cdot%20W_%7Bout%7D%3D-%2810.42lbm%2Fs%29%5B%280.25%5Cfrac%7BBtu%7D%7Blbm.%5Ctextdegree%20F%7D%29%28300-900%29%5Ctextdegree%20F%2B0.5%28%28700ft%2Fs%29%5E2-%28350ft%2Fs%29%5E2%29%28%5Cfrac%7B1%5Cfrac%7BBtu%7D%7Blbm%7D%7D%7B25037ft%5E2%2Fs%5E2%7D%29%5D%5C%5C%5C%5C%5C%5C%5C%5C%3D1486.5%5Cfrac%7BBtu%7D%7Bs%7D)
Hence, the mass flow rate of the air is 1486.5Btu/s
Explanation:
Given that,
Radius R= 2.00
Charge = 6.88 μC
Inner radius = 4.00 cm
Outer radius = 5.00 cm
Charge = -2.96 μC
We need to calculate the electric field
Using formula of electric field
(a). For, r = 1.00 cm
Here, r<R
So, E = 0
The electric field does not exist inside the sphere.
(b). For, r = 3.00 cm
Here, r >R
The electric field is
Put the value into the formula
The electric field outside the solid conducting sphere and the direction is towards sphere.
(c). For, r = 4.50 cm
Here, r lies between R₁ and R₂.
So, E = 0
The electric field does not exist inside the conducting material
(d). For, r = 7.00 cm
The electric field is
Put the value into the formula
The electric field outside the solid conducting sphere and direction is away of solid sphere.
Hence, This is the required solution.
The Hubble Telescope or Hubble Space Telescope (HST) is designed to take images found within the solar system. It is a low-orbit kind of telescope that helped a lot of scientists in proving some theories in astrophysics and the dimensions of the Universe. It was launched in 1990s and since then has been in space taking up good quality images that would help scientists learn more from space explorations.
