Answer:
Option C = internal energy stays the same.
Explanation:
The internal energy will remain the same or unchanged because this question has to do with a concept in physics or classical chemistry (in thermodynamics) known as Free expansion.
So, the internal energy will be equals to the multiplication of the change in temperature, the heat capacity (keeping volume constant) and the number of moles. And in free expansion the internal energy is ZERO/UNCHANGED.
Where, the internal energy, ∆U = 0 =quantity of heat, q - work,w.
The amount of heat,q = Work,w.
In the concept of free expansion the only thing that changes is the volume.
Answer:
The correct answer will be "400.4 N". The further explanation is given below.
Explanation:
The given values are:
Mass of truck,
m = 600 kg
g = 9.8 m/s²
On equating torques at the point O,
⇒
So that,
On putting the values, we get
⇒ ![T\times Cos(15^{\circ})\times 5.3=600\times 9.8\times Sin(5^{\circ})\times 4](https://tex.z-dn.net/?f=T%5Ctimes%20Cos%2815%5E%7B%5Ccirc%7D%29%5Ctimes%205.3%3D600%5Ctimes%209.8%5Ctimes%20Sin%285%5E%7B%5Ccirc%7D%29%5Ctimes%204)
⇒ ![T=400.4 \ N](https://tex.z-dn.net/?f=T%3D400.4%20%5C%20N)
Answer:
for i in range(0,10):
if SimonPattern[i] == UserPattern[i]:
score = score + 1
i = i + 1
else:
break
if i == 9:
score = score + 1
print("Total Score: {}".format(score))
Explanation:
This for loop was made using Python. Full code attached.
- For loop requires a range of numbers to define the end points. For this Simon Says game, we are talking about 10 characters, so that must be the range for the for loop: from 0 to 10.
- Conditional if tests if Simon pattern matches User's one characheter by one and add point for each match.
- Break statement is ready to escape the for loop at first mismatch.
- As we are starting from index 0, if the users matched all the characters correctly, then we need to add 1, otherwise the maximun score would be 9 and it should be 10.
Answer:
eccentrcity of orbit is 0.22
Explanation:
GIVEN DATA:
Initial velocity of satellite = 8333.3 m/s
distance from the sun is 600 km
radius of earth is 6378 km
as satellite is acting parallel to the earth therefore![\theta angle = 0](https://tex.z-dn.net/?f=%20%5Ctheta%20angle%20%3D%200)
and radial component of given velocity is zero
we have![h = r_o v_r_o = 6378+600 =6.97*10^6 m](https://tex.z-dn.net/?f=%20h%20%3D%20r_o%20v_r_o%20%3D%206378%2B600%20%3D6.97%2A10%5E6%20m)
h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s
we know that
![\frac{1}{r} =\frac{GM}{h^2} \times ( i + \epsilon cos\theta)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Br%7D%20%3D%5Cfrac%7BGM%7D%7Bh%5E2%7D%20%5Ctimes%20%28%20i%20%2B%20%5Cepsilon%20cos%5Ctheta%29)
![GM = gr^2 = 9.81*(6.37*10^6)^2 = 398*10^{12} m^3/s](https://tex.z-dn.net/?f=GM%20%3D%20gr%5E2%20%3D%209.81%2A%286.37%2A10%5E6%29%5E2%20%3D%20398%2A10%5E%7B12%7D%20m%5E3%2Fs)
so
![\frac{1}{6.97*10^6} =\frac{398*10^{12}}{(58.08*10^9)^2} \times ( i + \epsilon cos0)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B6.97%2A10%5E6%7D%20%3D%5Cfrac%7B398%2A10%5E%7B12%7D%7D%7B%2858.08%2A10%5E9%29%5E2%7D%20%5Ctimes%20%28%20i%20%2B%20%5Cepsilon%20cos0%29)
solvingt for ![\epsilon)](https://tex.z-dn.net/?f=%20%5Cepsilon%29)
![\epsilon = 0.22)](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%200.22%29)
therefore eccentrcity of orbit is 0.22
Answer:
17.799°
Explanation:
When the bullet hits the block at that time the momentum is conserved
So, initial momentum = final momentum
![P_i=P_f](https://tex.z-dn.net/?f=P_i%3DP_f)
So ![28\times 10^{-3}\times 210=(3.6+0.028)v_f](https://tex.z-dn.net/?f=28%5Ctimes%2010%5E%7B-3%7D%5Ctimes%20210%3D%283.6%2B0.028%29v_f)
![v_f=1.6207\ m/sec](https://tex.z-dn.net/?f=v_f%3D1.6207%5C%20m%2Fsec)
Now energy is also conserved
So ![\frac{1}{2}\times (3.6+0.028)\times 1.6207^2=(3.6+0.028)\times 9.81\times 2.8(1-cos\Theta )](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%283.6%2B0.028%29%5Ctimes%201.6207%5E2%3D%283.6%2B0.028%29%5Ctimes%209.81%5Ctimes%202.8%281-cos%5CTheta%20%29)
![cos\Theta =0.8521So\ \Theta =17.799^{\circ}](https://tex.z-dn.net/?f=cos%5CTheta%20%3D0.8521So%5C%20%5CTheta%20%3D17.799%5E%7B%5Ccirc%7D)