Answer:
The volume of radius is
× π × radius³ Proved
Explanation:
Given as :
We know that volume of sphere is v =
× π × radius³
Or, v =
× π × r³
Let prove the volume of sphere
So, From the figure of sphere
At the height of z , there is shaded disk with radius x
Let Find the area of triangle with side x , z , r
<u>From Pythagorean theorem</u>
x² + z² = r²
Or, x² = r² - z²
Or, x = ![\sqrt{r^{2}-z^{2} }](https://tex.z-dn.net/?f=%5Csqrt%7Br%5E%7B2%7D-z%5E%7B2%7D%20%20%7D)
Now, Area of shaded disk = Area = π × x²
Where x is the radius of disk
Or, Area of shaded disk = π × (
) ²
∴ Area of shaded disk = π × (r² - z²)
Again
<u>If we calculate the area of all horizontal disk, we can get the volume of sphere</u>
So, we simply integrate the area of all disk from - r to + r
i.e volume = ![\int_{-r}^{r} \Pi(r^{2}-z^{2} )dz](https://tex.z-dn.net/?f=%5Cint_%7B-r%7D%5E%7Br%7D%20%5CPi%28r%5E%7B2%7D-z%5E%7B2%7D%20%29dz)
Or, v =
- ![\int_{-r}^{r} \Pi z^{2}dz](https://tex.z-dn.net/?f=%5Cint_%7B-r%7D%5E%7Br%7D%20%5CPi%20z%5E%7B2%7Ddz)
Or, v = π r² (r + r) - π ![\frac{r^{3} -(-r)^{3})}{3}](https://tex.z-dn.net/?f=%5Cfrac%7Br%5E%7B3%7D%20-%28-r%29%5E%7B3%7D%29%7D%7B3%7D)
Or, v = π r² (r + r) - π ![\frac{2r^{3}}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B2r%5E%7B3%7D%7D%7B3%7D)
Or, v = 2πr³ - π ![\frac{2r^{3}}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B2r%5E%7B3%7D%7D%7B3%7D)
Or, v = 2πr³ (
)
Or, v = 2πr³ × ![\frac{2}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B3%7D)
∴ v =
× π × r³
Hence, The volume of radius is
× π × radius³ Proved . Answer
Answer:
241 kPa
Explanation:
The ideal gas law states that:
![pV=nRT](https://tex.z-dn.net/?f=pV%3DnRT)
where
p is the gas pressure
V is its volume
n is the number of moles
R is the gas constant
T is the absolute temperature of the gas
We can rewrite the equation as
![\frac{pV}{T}=nR](https://tex.z-dn.net/?f=%5Cfrac%7BpV%7D%7BT%7D%3DnR)
For a fixed amount of gas, n is constant, so we can write
![\frac{pV}{T}=const.](https://tex.z-dn.net/?f=%5Cfrac%7BpV%7D%7BT%7D%3Dconst.)
Therefore, for a gas which undergoes a transformation we have
![\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7Bp_1%20V_1%7D%7BT_1%7D%3D%5Cfrac%7Bp_2%20V_2%7D%7BT_2%7D)
where the labels 1 and 2 refer to the initial and final conditions of the gas.
For the sample of gas in this problem we have
![p_1 = 98 kPa=9.8\cdot 10^4 Pa\\V_1 = 750 mL=0.75 L=7.5\cdot 10^{-4}m^3\\T_1 = 30^{\circ}C+273=303 K\\p_2 =?\\V_2 = 250 mL=0.25 L=2.5\cdot 10^{-4} m^3\\T_2 = -25^{\circ}C+273=248 K](https://tex.z-dn.net/?f=p_1%20%3D%2098%20kPa%3D9.8%5Ccdot%2010%5E4%20Pa%5C%5CV_1%20%3D%20750%20mL%3D0.75%20L%3D7.5%5Ccdot%2010%5E%7B-4%7Dm%5E3%5C%5CT_1%20%3D%2030%5E%7B%5Ccirc%7DC%2B273%3D303%20K%5C%5Cp_2%20%3D%3F%5C%5CV_2%20%3D%20250%20mL%3D0.25%20L%3D2.5%5Ccdot%2010%5E%7B-4%7D%20m%5E3%5C%5CT_2%20%3D%20-25%5E%7B%5Ccirc%7DC%2B273%3D248%20K)
So we can solve the formula for
, the final pressure:
![p_2 = \frac{p_1 V_1 T_2}{T_1 V_2}=\frac{(9.8\cdot 10^4 Pa)(7.5\cdot 10^{-4} m^3)(248 K)}{(303 K)(2.5\cdot 10^{-4} m^3)}=2.41\cdot 10^5 Pa = 241 kPa](https://tex.z-dn.net/?f=p_2%20%3D%20%5Cfrac%7Bp_1%20V_1%20T_2%7D%7BT_1%20V_2%7D%3D%5Cfrac%7B%289.8%5Ccdot%2010%5E4%20Pa%29%287.5%5Ccdot%2010%5E%7B-4%7D%20m%5E3%29%28248%20K%29%7D%7B%28303%20K%29%282.5%5Ccdot%2010%5E%7B-4%7D%20m%5E3%29%7D%3D2.41%5Ccdot%2010%5E5%20Pa%20%3D%20241%20kPa)
Answer:
C. To wear shoes with hard soles
Explanation:
Answer:
B . 68.7 N
Explanation:
Given in the question that;
Force = 22 N
Mass = 7.0 kg
Velocity = 4.0 m/s
Gravitational force is the weight of wooden block , calculated as mass times acceleration due to gravity.
F= m* a
F= 7 * 9.81
F= 68.67 N
F= 68.7 N