I think D. None of the above
I think that it
Explanation:
iy finds so she said !gWelcome to Gboard clipboard, any text that you copy will be saved here.Welcome to Gboard clipboard, any text that you copy will be saved here.Welcome to Gboard clipboard, any text that you copy will be saved here.Tap on a clip to paste it in the text box.Tap on a clip to paste it in the text box.Touch and hold a clip to pin it. Unpinned clips will be deleted after 1 hour.Use the edit icon to pin, add or delete clips.Touch and hold a clip to pin it. Unpinned clips will be deleted after 1 hour.Use the edit icon to pin, add or delete clips.
The temperature of the gas is 41.3 °C.
Answer:
The temperature of the gas is 41.3 °C.
Explanation:
So on combining the Boyle's and Charles law, we get the ideal law of gas that is PV=nRT. Here P is the pressure, V is the volume, n is the number of moles, R is gas constant and T is the temperature. The SI unit of pressure is atm. So we need to convert 1 Pa to 1 atm, that is 1 Pa = 9.86923×
atm. Thus, 171000 Pa = 1.6876 atm.
We know that the gas constant R = 0.0821 atmLMol–¹K-¹. Then the volume of the gas is given as 50 L and moles are given as 3.27 moles.
Then substituting all the values in ideal gas equation ,we get
1.6876×50=3.27×0.0821×T
Temperature = 
So the temperature is obtained to be 314.3 K. As 0°C = 273 K,
Then 314.3 K = 314.3-273 °C=41.3 °C.
Thus, the temperature is 41.3 °C.
Apple hits the surface with speed 16.2 m/s
The angle made by the apple velocity with normal to the incline surface is given as 20 degree
now the component of velocity which is parallel to the surface and perpendicular to the surface is given as


so here we have


<em>so its velocity along the incline plane will be 5.5 m/s</em>
Answer:
Explanation:
The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.
So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.
force on it due to rest of the charges will be equal and opposite so
k3q Q / x² =k 8q Q / (L+x)²
8x² = 3 (L+x)²
2√2 x = √3 (L+x)
2√2 x - √3 x = √3 L
x(2√2 - √3 ) = √3 L
x = √3 L / (2√2 - √3 )
Let us consider the balancing force on 3q
force on it due to -Q and -8q will be equal
kQ . 3q / x² = k3q 8q / L²
Q = 8q (x² / L²)
so charge required = - 8q (x² / L²)
and its distance from x on negative x side = √3 L / (2√2 - √3 )