Answer:
see explanation below
Explanation:
Given that,
500°C
= 25°C
d = 0.2m
L = 10mm = 0.01m
U₀ = 2m/s
Calculate average temperature
262.5 + 273
= 535.5K
From properties of air table A-4 corresponding to 
= 535.5K 
k = 43.9 × 10⁻³W/m.k
v = 47.57 × 10⁻⁶ m²/s
A)
Number for the first strips is equal to
Calculating heat transfer coefficient from the first strip
The rate of convection heat transfer from the first strip is
The rate of convection heat transfer from the fifth trip is equal to
Calculating 
The rate of convection heat transfer from the tenth strip is
Calculating
Calculating the rate of convection heat transfer from the tenth strip
The rate of convection heat transfer from 25th strip is equal to
Calculating 
Calculating 
Calculating the rate of convection heat transfer from the tenth strip
Answer:
A: 4 times as much
B: 200 N/m
C: 5000 N
D: 84,8 J
Explanation:
A.
In the first question, we have to caculate the constant of the spring with this equation:
Getting the k:
![k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Bm%2Ag%7D%7Bx%7D%20%3D%5Cfrac%7B0%2C2%5Bkg%5D%2A9%2C81%5B%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20%5D%7D%7B0%2C05%5Bm%5D%7D%20%3D39%2C24%5B%5Cfrac%7BN%7D%7Bm%7D%5D)
Then we can calculate how much the spring stretch whith the another mass of 0,2kg:
![x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\](https://tex.z-dn.net/?f=x%3D%5Cfrac%7Bm%2Ag%7D%7Bk%7D%20%3D%5Cfrac%7B0%2C4%5Bkg%5D%2A9%2C81%5B%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20%5D%7D%7B39%2C24%5B%5Cfrac%7BN%7D%7Bm%7D%5D%7D%20%3D0%2C1%5Bm%5D%5C%5C)
The energy of a spring:
For the first case:
![E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%7D%20%2A39%2C24%5B%5Cfrac%7BN%7D%7Bm%7D%5D%2A%280%2C05%5Bm%5D%29%5E%7B2%7D%20%3D0%2C049%20%5BJ%5D)
For the second case:
![E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%7D%20%2A39%2C24%5B%5Cfrac%7BN%7D%7Bm%7D%5D%2A%280%2C1%5Bm%5D%29%5E%7B2%7D%20%3D0%2C0196%20%5BJ%5D)
If you take the relation E2/E1 = 4.
B.
We have the next facts:
x=0,005 m
E = 0,0025 J
Using the energy equation for a spring:
⇒![k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]](https://tex.z-dn.net/?f=k%3D%5Cfrac%7BE%2A2%7D%7Bx%5E%7B2%7D%20%7D%20%3D%5Cfrac%7B0%2C0025%5BJ%5D%2A2%7D%7B%280%2C005%5Bm%5D%29%5E%7B2%7D%20%7D%20%3D200%5B%5Cfrac%7BN%7D%7Bm%7D%20%5D)
C.
The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.
![E=W*h=500[N]*10[m]=5000[J]](https://tex.z-dn.net/?f=E%3DW%2Ah%3D500%5BN%5D%2A10%5Bm%5D%3D5000%5BJ%5D)
Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.
D.
The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.
The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:
W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J
Use stronger magnets
increase current
push magnets closer to coil
adding more sets of coils
Answer:
Explanation:
One charge is situated at x = 1.95 m . Second charge is situated at y = 1.00 m
These two charges are situated outside sphere as it has radius of .365 m with center at origin. So charge inside sphere = zero.
Applying Gauss's theorem
Flux through spherical surface = charge inside sphere / ε₀
= 0 / ε₀
= 0 Ans .
Answer:
Given that
V= 0.06 m³
Cv= 2.5 R= 5/2 R
T₁=500 K
P₁=1 bar
Heat addition = 15000 J
We know that heat addition at constant volume process ( rigid vessel ) given as
Q = n Cv ΔT
We know that
P V = n R T
n=PV/RT
n= (100 x 0.06)(500 x 8.314)
n=1.443 mol
So
Q = n Cv ΔT
15000 = 1.433 x 2.5 x 8.314 ( T₂-500)
T₂=1000.12 K
We know that at constant volume process
P₂/P₁=T₂/T₁
P₂/1 = 1000.21/500
P₂= 2 bar
Entropy change given as
Cp-Cv= R
Cp=7/2 R
Now by putting the values
a)ΔS= 20.79 J/K
b)
If the process is adiabatic it means that heat transfer is zero.
So
ΔS= 20.79 J/K
We know that
Process is adiabatic
