Answer:
t1 = t2 + 3.02 V = 41.5
V t1 - 1/2 g t1^2 = V t2 - 1/2 g t2^2
Both stones reach the same height after the specified times
V (t1 - t2) = g/2 (t1^2 - t2^2) = g/2 (t1 - t2) (t1 + t2)
2 V / g = t1 + t2 = 2t1 + 3.02
t1 = V / g - 1.51 = 41.5 / 9.8 -1.51 = 2.72 s
t2 = t1 + 3.02 = 5.74 sec
Check:
41.5 * 2.72 - 4.9 * 2.72^2 = 76.6 m
41.5 * 5.74 - 4.9 * 5.74^2 = 76.8 m
Speed of second stone = 41.5 - 9.8 * 2.72 = 14.8 m/s
Speed = (distance) / (time)
Speed = (
Velocity = speed, and its direction
The velocity of the plane is 10.2 miles per second East.
(about 48 times the speed of sound)
Answer: V = 15 m/s
Explanation:
As stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. The observed frequency the car will be experiencing will be addition of the two frequency. That is,
F = 2.1 × 10^10 + 1030 = 2.100000103×10^10Hz
Using doppler effect formula
F = C/ ( C - V) × f
Where
F = observed frequency
f = source frequency
C = speed of light = 3×10^8
V = speed of the car
Substitute all the parameters into the formula
2.100000103×10^10 = 3×10^8/(3×10^8 -V) × 2.1×10^10
2.100000103×10^10/2.1×10^10 = 3×108/(3×10^8 - V)
1.000000049 = 3×10^8/(3×10^8 - V)
Cross multiply
300000014.7 - 1.000000049V = 3×10^8
Collect the like terms
1.000000049V = 14.71429
Make V the subject of formula
V = 14.71429/1.000000049
V = 14.7 m/s
The speed of the car is 15 m/s approximately.
<span> Light energy is verified by many scientists to be made of particles called photons. The amount of energy in each photon is related to its wavelength using the Planck-Einstein equation. </span><span>Nuclear energy the binding energy of atomic nuclei which holds the subatomic particles within the nucleus.</span>
Answer:
the spring constant k = ![5.409*10^4 \ N/m](https://tex.z-dn.net/?f=5.409%2A10%5E4%20%5C%20N%2Fm)
the value for the damping constant ![\\ \\b = 1.518 *10^3 \ kg/s](https://tex.z-dn.net/?f=%5C%5C%20%5C%5Cb%20%3D%201.518%20%2A10%5E3%20%5C%20kg%2Fs)
Explanation:
From Hooke's Law
![F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m](https://tex.z-dn.net/?f=F%20%3D%20kx%5C%5C%5C%5Ck%20%3D%5Cfrac%7BF%7D%7Bx%7D%5C%5C%5C%5Cwhere%20%5C%20F%20%3D%20mg%5C%5C%5C%5Ck%20%3D%20%5Cfrac%7Bmg%7D%7Bx%7D%5C%5C%5C%5Cgiven%20%5C%20that%3A%5C%5C%5C%5Cmass%20%5C%20of%20%5C%20each%20%5C%20wheel%20%3D%20425%20%5C%20kg%5C%5C%5C%5Cx%20%3D%207.7cm%20%3D%200.077%20m%5C%5C%5C%5Cg%20%3D%209.8%20%5C%20m%2Fs%5E2%5C%5C%5C%5CThen%3B%5C%5C%5C%5Ck%20%3D%20%5Cfrac%7B425%20%5C%20kg%20%2A%209.8%20%5C%20m%2Fs%5E2%7D%7B0.077%20%5C%20m%7D%5C%5C%5C%5Ck%20%3D%205.409%2A10%5E4%20%5C%20N%2Fm)
Thus; the spring constant k = ![5.409*10^4 \ N/m](https://tex.z-dn.net/?f=5.409%2A10%5E4%20%5C%20N%2Fm)
The amplitude is decreasing 37% during one period of the motion
![e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}](https://tex.z-dn.net/?f=e%5E%7B%5Cfrac%7B-bT%7D%7B2m%7D%7D%3D%20%5Cfrac%7B37%7D%7B100%7D%5C%5C%5C%5Ce%5E%7B%5Cfrac%7B-bT%7D%7B2m%7D%7D%3D%200.37%5C%5C%5C%5C%5Cfrac%7B-bT%7D%7B2m%7D%20%3D%20In%280.37%29%5C%5C%5C%5C%5Cfrac%7B-bT%7D%7B2m%7D%20%3D%20-0.9943%5C%5C%5C%5Cb%20%3D%20%5Cfrac%7B2m%280.9943%29%7D%7BT%7D%5C%5C%5C%5Cb%20%3D%20%5Cfrac%7B2m%280.9943%29%7D%7B%5Cfrac%7B2%20%5Cpi%7D%7B%5Comega%7D%7D%5C%5C%5C%5Cb%20%3D%20%5Cfrac%7Bm%280.9943%29%20%5C%20%28%20%5Comega%29%20%29%7D%7B%20%5Cpi%7D)
![b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) } }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s](https://tex.z-dn.net/?f=b%20%3D%20%5Cfrac%7Bm%280.9943%29%28%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%29%7D%7D%7B%5Cpi%7D%5C%5C%5C%5Cb%20%3D%20%5Cfrac%7B425%2A%280.9943%29%28%5Csqrt%7B%5Cfrac%7B5.409%2A10%5E4%7D%7B425%7D%29%20%7D%20%20%20%20%7D%7B3.14%7D%5C%5C%5C%5Cb%20%3D%201518.24%20%5C%20kg%2Fs%5C%5C%5C%5Cb%20%3D%201.518%20%2A10%5E3%20%5C%20kg%2Fs)
Therefore; the value for the damping constant ![\\ \\b = 1.518 *10^3 \ kg/s](https://tex.z-dn.net/?f=%5C%5C%20%5C%5Cb%20%3D%201.518%20%2A10%5E3%20%5C%20kg%2Fs)