Power is the rate work is usually done in.
By adding up all the individual forces of the object
Answer:
1st – Place the film canister on the <u>scale</u>.
2nd – Slide the large <u>weight </u>to the right until the arm drops below the line and then move it back one notch.
3rd – Repeat this process with the <u>top</u> weight. When the arm moves below the line, back it up one groove.
4th – Slide the <u>small </u>weight on the front beam until the <u>lines</u> match up.
5th – Add the amounts on each beam to find the total <u>mass </u>to the nearest tenth of a gram.
Explanation:
The triple beam balance is an instrument that is used in measuring the mass of substances to a very high degree of precision. The reading error is given by ±0.05 grams. The triple beam balance as the name implies has three beams that measure substances of different mass levels.
The beams are categorized as small, medium, and large. There is a balance on which the substance to be weighed is placed directly upon. To use this measuring device, the procedures mentioned above are followed.
Answer:
<em>The distance the car traveled is 21.45 m</em>
Explanation:
<u>Motion With Constant Acceleration
</u>
It occurs when an object changes its velocity at the same rate thus the acceleration is constant.
The relation between the initial and final speeds is:
![v_f=v_o+at\qquad\qquad [1]](https://tex.z-dn.net/?f=v_f%3Dv_o%2Bat%5Cqquad%5Cqquad%20%5B1%5D)
Where:
a = acceleration
vo = initial speed
vf = final speed
t = time
The distance traveled by the object is given by:
![\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3Dv_o.t%2B%5Cfrac%7Ba.t%5E2%7D%7B2%7D%5Cqquad%5Cqquad%20%5B2%5D)
Solving [1] for a:
Substituting the given data vo=0, vf=6.6 m/s, t=6.5 s:
The distance is now calculated with [2]:
x = 21.45 m
The distance the car traveled is 21.45 m
Answer:
Equation for SHM can be written
V = w A cos w t where w is the angular frequency and the velocity is a maximum at t = 0
V1 = w1 A cos w1 t
V2 = w2 A cos w2 t
V2 / V1 = w2 / w1 since cos X t = 1 if t = zero
V2 / V1 = 2 pi f2 / (2 pi f1) = f2 / f1 = T1 / T2
If the velocity is twice as large the period will be 1/2 long
