Answer:
Assuming the typical saw used to saw boards in home construction and renovation is a circular saw, the diameter of the blade can be estimated as 7-10 inches (17.8-25.4 cm). The linear speed in m/s of a point on the rim of the circular saw blade is then approximately: (π × 25.4 cm) × (2600 rev/min) / (60 s/min) = 28.0 m/s.
Answer:
![L_A = 9 L_B](https://tex.z-dn.net/?f=L_A%20%3D%209%20L_B)
Explanation:
The formula that relates the luminosity of a star (L) to its radius (R) and the temperature (T) is
![L=\frac{R^2}{T^4}](https://tex.z-dn.net/?f=L%3D%5Cfrac%7BR%5E2%7D%7BT%5E4%7D)
For star B, we can write:
![L_B=\frac{R_B^2}{T_B^4}](https://tex.z-dn.net/?f=L_B%3D%5Cfrac%7BR_B%5E2%7D%7BT_B%5E4%7D)
For star A, we have
![T_A = 2 T_B\\R_A = 12 R_B](https://tex.z-dn.net/?f=T_A%20%3D%202%20T_B%5C%5CR_A%20%3D%2012%20R_B)
So the luminosity of star A is
![L_A=\frac{R_A^2}{T_A^4}=\frac{(12 R_B)^2}{(2 T_B)^4}=\frac{144 R_B^2}{16 T_B^4}=9\frac{R_B^2}{T_B^4}=9 L_B](https://tex.z-dn.net/?f=L_A%3D%5Cfrac%7BR_A%5E2%7D%7BT_A%5E4%7D%3D%5Cfrac%7B%2812%20R_B%29%5E2%7D%7B%282%20T_B%29%5E4%7D%3D%5Cfrac%7B144%20R_B%5E2%7D%7B16%20T_B%5E4%7D%3D9%5Cfrac%7BR_B%5E2%7D%7BT_B%5E4%7D%3D9%20L_B)
Answer:
Explanation:
Given
mass of bullet ![m=130 gm](https://tex.z-dn.net/?f=m%3D130%20gm)
Length of barrel ![L=0.56 m](https://tex.z-dn.net/?f=L%3D0.56%20m%20)
Force ![F=11000+13000x-28000x^2](https://tex.z-dn.net/?f=F%3D11000%2B13000x-28000x%5E2)
Work done ![W=\int_{x_0}^{x_f}Fdx](https://tex.z-dn.net/?f=W%3D%5Cint_%7Bx_0%7D%5E%7Bx_f%7DFdx)
![W_1=\int_{0}^{0.56}\left ( 11000+13000x-28000x^2\right )dx](https://tex.z-dn.net/?f=W_1%3D%5Cint_%7B0%7D%5E%7B0.56%7D%5Cleft%20%28%2011000%2B13000x-28000x%5E2%5Cright%20%29dx)
![W_1=\left [ 11000x+6500x^2-\frac{28000}{3}x^3\right ]_0^{0.56}](https://tex.z-dn.net/?f=W_1%3D%5Cleft%20%5B%2011000x%2B6500x%5E2-%5Cfrac%7B28000%7D%7B3%7Dx%5E3%5Cright%20%5D_0%5E%7B0.56%7D)
![W_1=6160+2038.4-1639.082=6599.318 J\approx 6.599 kJ](https://tex.z-dn.net/?f=W_1%3D6160%2B2038.4-1639.082%3D6599.318%20J%5Capprox%206.599%20kJ)
(b) When barrel is 1 m long
![W_2=\int_{0}^{1}\left ( 11000+13000x-28000x^2\right )dx](https://tex.z-dn.net/?f=W_2%3D%5Cint_%7B0%7D%5E%7B1%7D%5Cleft%20%28%2011000%2B13000x-28000x%5E2%5Cright%20%29dx)
![W_2=\left [ 11000x+6500x^2-\frac{28000}{3}x^3\right ]_0^{1}](https://tex.z-dn.net/?f=W_2%3D%5Cleft%20%5B%2011000x%2B6500x%5E2-%5Cfrac%7B28000%7D%7B3%7Dx%5E3%5Cright%20%5D_0%5E%7B1%7D)
![W_2=11000+6500-\frac{28000}{3}](https://tex.z-dn.net/?f=W_2%3D11000%2B6500-%5Cfrac%7B28000%7D%7B3%7D)
![W_2=8166.66 kJ](https://tex.z-dn.net/?f=W_2%3D8166.66%20kJ)
![\frac{W_2}{W_1}=\frac{8166.66}{6599.318}=1.237 kJ](https://tex.z-dn.net/?f=%5Cfrac%7BW_2%7D%7BW_1%7D%3D%5Cfrac%7B8166.66%7D%7B6599.318%7D%3D1.237%20kJ)
Answer:
I believe he shouted eureka after he stepped into a bath and noticed that the water rose. He then realized that the volume of water displaced must be equal to the volume of the part of his body he had submerged
Explanation: