Answer:
C.subscript
Explanation:
The number 3 is a subscript on the hydrogen atom in ammonia NH₃.
Subscript is a lower notation on a reference atom. It states the number of hydrogen atom covalently bonding with the single N atom.
The superscript is located up. For example P², the 2 is a superscript.
The coefficient is the number before a compound or atom e.g 2C represents two atoms of Carbon.
The numerator is the upper value in a fraction. The denominator is the one beneath.
Work done by each is the same for they reach the same height. The one who climbs in 30s uses more power because work is done in a shorter time.
12.8 rad
Explanation:
The angular displacement
through which the wheel turned can be determined from the equation below:
(1)
where
![\omega_0 = 0](https://tex.z-dn.net/?f=%5Comega_0%20%3D%200)
![\omega = 34.7\:\text{rad/s}](https://tex.z-dn.net/?f=%5Comega%20%3D%2034.7%5C%3A%5Ctext%7Brad%2Fs%7D)
![\alpha = 47.0\:\text{rad/s}^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%2047.0%5C%3A%5Ctext%7Brad%2Fs%7D%5E2)
Using these values, we can solve for
from Eqn(1) as follows:
![2\alpha\theta = \omega^2 - \omega_0^2](https://tex.z-dn.net/?f=2%5Calpha%5Ctheta%20%3D%20%5Comega%5E2%20-%20%5Comega_0%5E2)
or
![\theta = \dfrac{\omega^2 - \omega_0^2}{2\alpha}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Cdfrac%7B%5Comega%5E2%20-%20%5Comega_0%5E2%7D%7B2%5Calpha%7D)
![\:\:\:\:= \dfrac{(34.7\:\text{rad/s})^2 - 0}{2(47.0\:\text{rad/s}^2)}](https://tex.z-dn.net/?f=%5C%3A%5C%3A%5C%3A%5C%3A%3D%20%5Cdfrac%7B%2834.7%5C%3A%5Ctext%7Brad%2Fs%7D%29%5E2%20-%200%7D%7B2%2847.0%5C%3A%5Ctext%7Brad%2Fs%7D%5E2%29%7D)
![\:\:\:\:= 12.8\:\text{rad}](https://tex.z-dn.net/?f=%5C%3A%5C%3A%5C%3A%5C%3A%3D%2012.8%5C%3A%5Ctext%7Brad%7D)
0.013888888888889
I believe this is the answer
Answer:
The acceleration of the crate is
.
Explanation:
Given that,
Force, F = 750 N
Mass of the crate, m = 250 kg
The coefficient of friction is 0.12.
We need to find the acceleration of the crate. The net force acting on the crate is given by :
![F=ma\\\\F-f=ma](https://tex.z-dn.net/?f=F%3Dma%5C%5C%5C%5CF-f%3Dma)
f is frictional force, ![f=\mu N=\mu mg](https://tex.z-dn.net/?f=f%3D%5Cmu%20N%3D%5Cmu%20mg)
![F-\mu mg=ma\\\\a=\dfrac{F-\mu mg}{m}\\\\a=\dfrac{750-0.12\times 250\times 9.8}{250}\\\\a=1.82\ m/s^2](https://tex.z-dn.net/?f=F-%5Cmu%20mg%3Dma%5C%5C%5C%5Ca%3D%5Cdfrac%7BF-%5Cmu%20mg%7D%7Bm%7D%5C%5C%5C%5Ca%3D%5Cdfrac%7B750-0.12%5Ctimes%20250%5Ctimes%209.8%7D%7B250%7D%5C%5C%5C%5Ca%3D1.82%5C%20m%2Fs%5E2)
So, the acceleration of the crate is
. Hence, this is the required solution.