Answer:
159241.048 cm³/s
Explanation:
r = Radius = 3×height = 3h
h = height = 16 cm
Height of the pile increases at a rate = ![\frac{dh}{dt}=22\ cm/s](https://tex.z-dn.net/?f=%5Cfrac%7Bdh%7D%7Bdt%7D%3D22%5C%20cm%2Fs)
![\text{Volume of cone}=\frac{1}{3}\pi r^2h\\\Rightarrow V=\frac{1}{3}\pi (3h)^2h\\\Rightarrow V=3\pi h^3](https://tex.z-dn.net/?f=%5Ctext%7BVolume%20of%20cone%7D%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%20r%5E2h%5C%5C%5CRightarrow%20V%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%20%283h%29%5E2h%5C%5C%5CRightarrow%20V%3D3%5Cpi%20h%5E3)
Differentiating with respect to time
![\frac{dv}{dt}=9\pi h^2\frac{dh}{dt}\\\Rightarrow \frac{dv}{dt}=9\pi 16^2\times 22\\\Rightarrow \frac{dv}{dt}=159241.048\ cm^3/s](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdt%7D%3D9%5Cpi%20h%5E2%5Cfrac%7Bdh%7D%7Bdt%7D%5C%5C%5CRightarrow%20%5Cfrac%7Bdv%7D%7Bdt%7D%3D9%5Cpi%2016%5E2%5Ctimes%2022%5C%5C%5CRightarrow%20%5Cfrac%7Bdv%7D%7Bdt%7D%3D159241.048%5C%20cm%5E3%2Fs)
∴ Rate is the sand leaving the bin at that instant is 159241.048 cm³/s
Answer:
Option (c)
Explanation:
Both the bullets have same acceleration because they both falls under the influence of acceleration due to gravity.
The bullet which is fired from the gun has some initial velocity but the bullet which is dropped has zero initial velocity.
the acceleration is acting on both the bullets which is equal to the acceleration due to gravity and they both in motion in the influence of gravity.
Answer: ![0.05\ mm](https://tex.z-dn.net/?f=0.05%5C%20mm)
Explanation:
Given
Cross-sectional area of wire ![A_1=4\ mm^2](https://tex.z-dn.net/?f=A_1%3D4%5C%20mm%5E2)
Extension of wire ![\delta l=0.1\ mm](https://tex.z-dn.net/?f=%5Cdelta%20l%3D0.1%5C%20mm)
Extension in a wire is given by
![\Rightarrow \delta l=\dfrac{FL}{AE}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cdelta%20l%3D%5Cdfrac%7BFL%7D%7BAE%7D)
where, ![E=\text{Youngs modulus}](https://tex.z-dn.net/?f=E%3D%5Ctext%7BYoungs%20modulus%7D)
![\Rightarrow \delta_1=\dfrac{FL}{A_1E}\quad \ldots(i)](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cdelta_1%3D%5Cdfrac%7BFL%7D%7BA_1E%7D%5Cquad%20%5Cldots%28i%29)
for same force, length and material
![\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cdelta_2%3D%5Cdfrac%7BFL%7D%7BA_2E%7D%5Cquad%20%5Cldots%28ii%29)
Divide (i) and (ii)
![\Rightarrow \dfrac{0.1}{\delta_2}=\dfrac{A_2}{A_1}\\\\\Rightarrow \delta_2=0.1\times \dfrac{4}{8}\\\\\Rightarrow \delta_2=0.05\ mm](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cdfrac%7B0.1%7D%7B%5Cdelta_2%7D%3D%5Cdfrac%7BA_2%7D%7BA_1%7D%5C%5C%5C%5C%5CRightarrow%20%5Cdelta_2%3D0.1%5Ctimes%20%5Cdfrac%7B4%7D%7B8%7D%5C%5C%5C%5C%5CRightarrow%20%5Cdelta_2%3D0.05%5C%20mm)
Answer:
doubled
Explanation:
F=ma1----------(1)
2F = ma2-------(2)
Divide 2nd equation by 1st one
we get a1×2=a2