The protons from the glass rod are transferred to the silk cloth.
The glass is the material creating the charge not the cloth. And the charge from the glass is positive while the charge from the silk is negative.
Answer:
![\theta = 33.5 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2033.5%20degree)
Explanation:
As we know that net displacement is
![d = 3.25 km](https://tex.z-dn.net/?f=d%20%3D%203.25%20km)
altitude is given as
![h = 2.15 km](https://tex.z-dn.net/?f=h%20%3D%202.15%20km)
so its horizontal displacement is given as
![x^2 + h^2 = d^2](https://tex.z-dn.net/?f=x%5E2%20%2B%20h%5E2%20%3D%20d%5E2)
![x^2 + 2.15^2 = 3.25^2](https://tex.z-dn.net/?f=x%5E2%20%2B%202.15%5E2%20%3D%203.25%5E2)
![x = 2.44 km](https://tex.z-dn.net/?f=x%20%3D%202.44%20km)
now we have
![v_x = 280 cos\theta](https://tex.z-dn.net/?f=v_x%20%3D%20280%20cos%5Ctheta)
![v_y = 280 sin\theta](https://tex.z-dn.net/?f=v_y%20%3D%20280%20sin%5Ctheta)
now in x direction we have
![2.44 \times 10^3 = (280 cos\theta) t](https://tex.z-dn.net/?f=2.44%20%5Ctimes%2010%5E3%20%3D%20%28280%20cos%5Ctheta%29%20t)
in y direction we have
![2.15 \times 10^3 = (280 sin\theta) t + 4.9 t^2](https://tex.z-dn.net/?f=2.15%20%5Ctimes%2010%5E3%20%3D%20%28280%20sin%5Ctheta%29%20t%20%2B%204.9%20t%5E2)
from above equations we have
![2.15 \times 10^3 = 2.44 \times 10^3 tan\theta + 4.9 (\frac{2.44 \times 10^3}{280 cos\theta})^2](https://tex.z-dn.net/?f=2.15%20%5Ctimes%2010%5E3%20%3D%202.44%20%5Ctimes%2010%5E3%20tan%5Ctheta%20%2B%204.9%20%28%5Cfrac%7B2.44%20%5Ctimes%2010%5E3%7D%7B280%20cos%5Ctheta%7D%29%5E2)
by solving above equation
![\theta = 33.5 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2033.5%20degree)
Volume % = mole %
Molar mass of oxygen = 32kg/kmol
Molar mass of nitrogen = 28 kg/kmol
Density of nitrogen = Density of
oxygen x molar mass of oxygen x molar ratio of air x molar mass of nitrogen
Density of nitrogen = (1.43kgm^-3)(1 kmol O2/32 kg O2)(80 kmol N2/ 20
kmol O2)(28 kg N2/1 kmol O2) = 5.01 kg/m3
Answer:
B) Its horizontal velocity component is the same as it was just as it was launched
Explanation:
From dynamic we know that free falling objects had two components of velocity
The only acceleration that takes place in this case is gravity in the y-component of velocity
![v_{vx}=v_{ox}+at\\v_{y}=v_{oy}+gt](https://tex.z-dn.net/?f=v_%7Bvx%7D%3Dv_%7Box%7D%2Bat%5C%5Cv_%7By%7D%3Dv_%7Boy%7D%2Bgt)
Since a=0 and
that means that the velocity at the y-component changes with time but the horizontal component remain the same
This means that the horizontal velocity is not 0 but is <u><em>the same as it was just as it was launched</em></u>
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