Answer:
Substance 2 is a still figure and substance 1 isn't
Explanation:
Answer:
The value of heat transfer during the process Q = - 29.49 KJ
Explanation:
Given data
= 50 
= 344.7 k pa
= 0.113 
F = 366.4 K
= 477.6 K
Poly tropic index n = 1.2
gas constant for oxygen = 0.26 
From ideal gas equation
= m R 
Put all the values in above equation we get
⇒ 344.7 × 0.113 = m × 0.26 × 366.4
⇒ m = 0.408 kg
Heat transfer in poly tropic process is given by
Q = ![\frac{\gamma - n}{( \gamma - 1)( n - 1)} [ {m R (T_{1} - T_{2} ) ]](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cgamma%20-%20n%7D%7B%28%20%5Cgamma%20-%201%29%28%20n%20-%201%29%7D%20%5B%20%7Bm%20R%20%28T_%7B1%7D%20-%20T_%7B2%7D%20%20%29%20%5D)
Put all the values in above formula we get
⇒ Q = ![\frac{1.4 - 1.2}{( 1.4 - 1)( 1.2 - 1)} [ {m R (T_{1} - T_{2} ) ]](https://tex.z-dn.net/?f=%5Cfrac%7B1.4%20-%201.2%7D%7B%28%201.4%20-%201%29%28%201.2%20-%201%29%7D%20%5B%20%7Bm%20R%20%28T_%7B1%7D%20-%20T_%7B2%7D%20%20%29%20%5D)
⇒ Q = 2.5 × 0.408 × 0.26 × ( 366.4 - 477.6 )
⇒ Q = - 29.49 KJ
This is the value of heat transfer during the process & negative sign shows that heat is lost during the process.
Answer:
A) I = Io 0.578, B) he light that leaves the polarized is completely polarized, being perpendicular to the axis of the second filter
Explanation:
A) Light passing through a polarizer must comply with the / bad law
I = Io cos2 tea
Where is at the angle of the polarizer and incident light
I = Io cos2 45
I = Io 0.578
Therefore the beam intensity is 0.578 of the incident intensity
.B) the light that leaves the polarized is completely polarized, being perpendicular to the axis of the second filter
Answer:
These planets rotate around the sun in a circular path. Likewise in a heliocentric model it is believed that the sun is at the center of the universe and the planet earth along with all other planet move around it. Thus in both geocentric model and heliocentric model bodies in space move in circular orbits.
Answer:
a) W = 10995.6 J
b) W = - 9996 J
c) Kf = 999.6 J
d) v = 5.77 m/s
Explanation:
Given
m = 60 Kg
h = 17 m
a = g/10
g = 9.8 m/s²
a) We can apply Newton's 2nd Law as follows
∑Fy = m*a ⇒ T - m*g = m*a ⇒ T = (g + a)*m
where T is the force exerted by the cable
⇒ T = (g + (g/10))*m = (11/10)*g*m = (11/10)*(9.8 m/s²)*(60 Kg)
⇒ T = 646.8 N
then we use the equation
W = F*d = T*h = (646.8 N)*(17 m)
W = 10995.6 J
b) We use the formula
W = m*g*h ⇒ W = (60 Kg)(9.8 m/s²)(-17 m)
⇒ W = - 9996 J
c) We have to obtain Wnet as follows
Wnet = W₁ + W₂ = 10995.6 J - 9996 J
⇒ Wnet = 999.6 J
then we apply the equation
Wnet = ΔK = Kf - Ki = Kf - 0 = Kf
⇒ Kf = 999.6 J
d) Knowing that
K = 0.5*m*v² ⇒ v = √(2*Kf / m)
⇒ v = √(2*999.6 J / 60 Kg)
⇒ v = 5.77 m/s
