Answer:
3.82 Hertz
Explanation:
![y = 2.5 Sin\left (3x-24t+\frac{\pi }{2} \right )](https://tex.z-dn.net/?f=y%20%3D%202.5%20Sin%5Cleft%20%283x-24t%2B%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%20%5Cright%20%29)
This is the equation of a wave which varies sinusoidally.
The standard equation of a wave is given by
![y = A Sin\left ( kx-\omega t+\phi \right )](https://tex.z-dn.net/?f=y%20%3D%20A%20Sin%5Cleft%20%28%20kx-%5Comega%20t%2B%5Cphi%20%20%5Cright%20%29)
where, A be the amplitude of the wave, k be the wave number, x be the displacement of wave, ω be the angular frequency and t be the time taken, and Ф be the phase angle.
now compare the given equation by the standard equation, we get
k = 3
ω = 24
Ф = π / 2
So, the angular frequency = 24
The relation between the angular frequency and the frequency is given by
ω = 2 x π x f
24 = 2 x 3.14 x f
f = 3.82 Hertz
It depends on the objects chemical composition.
Answer:
(B) cash inflows are moved earlier in time.
Explanation:
The payback period stated time-frame during which the initial amount of investment should be recovered. It is expressed in the year form
The formula to compute the payback period is shown below:
Payback period = Initial investment ÷ Net cash flow
where,
The net cash flow = annual net operating income + depreciation expenses
The payback period of the project decreases when the accumulated starting year cash flows increases that results the movement of the cash inflows earlier in time
Answer:
The height of the image will be "1.16 mm".
Explanation:
The given values are:
Object distance, u = 25 cm
Focal distance, f = 1.8 cm
On applying the lens formula, we get
⇒ ![\frac{1}{v} -\frac{1}{u} =\frac{1}{f}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bv%7D%20-%5Cfrac%7B1%7D%7Bu%7D%20%3D%5Cfrac%7B1%7D%7Bf%7D)
On putting estimate values, we get
⇒ ![\frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bv%7D%20-%5Cfrac%7B1%7D%7B%28-25%29%7D%20%3D%5Cfrac%7B1%7D%7B1.8%7D)
⇒ ![\frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bv%7D%20%3D%5Cfrac%7B1%7D%7B1.8%7D%20-%5Cfrac%7B1%7D%7B25%7D)
⇒ ![v=1.94 \ cm](https://tex.z-dn.net/?f=v%3D1.94%20%5C%20cm)
As a result, the image would be established mostly on right side and would be true even though v is positive.
By magnification,
and
⇒ ![\frac{v}{u} =\frac{h_{1}}{h_{0}}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%7D%7Bu%7D%20%3D%5Cfrac%7Bh_%7B1%7D%7D%7Bh_%7B0%7D%7D)
⇒ ![\frac{1.94}{25}=\frac{{h_{1}}}{15}](https://tex.z-dn.net/?f=%5Cfrac%7B1.94%7D%7B25%7D%3D%5Cfrac%7B%7Bh_%7B1%7D%7D%7D%7B15%7D)
⇒ ![{h_{1}}=1.16 \ mm](https://tex.z-dn.net/?f=%7Bh_%7B1%7D%7D%3D1.16%20%5C%20mm)
Answer:
Person B has four times the power output of person A.