The effiency of a machine is
(output work or energy) / (input work or energy) .
For the system described in the question, that's
(123 J) / (150 J) = 0.82 = 82% .
Density = Mass / Volume
A) Density = 888/800 = 1.11g/mL > 1.0g/mL
B) Density = 7.8 / 8.7 = 0.897g/mL < 1.0g/mL
C) density = 725 / 715 = 1.01 g/mL > 1.0g/mL
D)density = 1.3/1.1= 1.18 g/mL > than 1.0 g/mL
E) Density = 18/19 = 0.95g/mL < 1.0g/mL
F) Density = 1.25/1.78 = 0.7 g/mL < 1.0g/mL
SUMMARY:
A C D have densities greater than 1.0g/mL
-- First, it can never swing to a higher elevation off the floor
than where it was when she let it go. The higher it is off the
floor, the more potential energy it has, and that's all the energy
she gave it when she lifted it to the height of her chin.
-- Second, it can't even return to THAT height, because during
its swing out and back, it's losing energy by plowing through air.
So each swing is slightly narrower, and ends slightly lower, than
the one before it.
The lifetime of a star is determined by its mass. The larger it’s mass, the shorter it’s lifetime
Answer:
11.4 m
Explanation:
We are given that
Total distance traveled by cab,d=214 m
Maximum speed of cab=v=315 m/min=![\frac{315}{60}=5.25m/s](https://tex.z-dn.net/?f=%5Cfrac%7B315%7D%7B60%7D%3D5.25m%2Fs)
1 min =60 s
Acceleration of cab=![1.21 m/s^2](https://tex.z-dn.net/?f=1.21%20m%2Fs%5E2)
We have to find the distance traveled by the cab while accelerating to full speed from rest.
Initial speed o f cab=u=0
![v^2-u^2=2ax](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2ax)
Substitute the values in the formula
![(5.25)^2-0=2(1.21)x](https://tex.z-dn.net/?f=%285.25%29%5E2-0%3D2%281.21%29x)
![27.5625=2.42x](https://tex.z-dn.net/?f=27.5625%3D2.42x)
![x=\frac{27.5625}{2.42}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B27.5625%7D%7B2.42%7D)
![x=11.4 m](https://tex.z-dn.net/?f=x%3D11.4%20m)
Hence, the cab travel 11.4 m from rest to full speed.