Answer:
![v = 1.6 \frac{m}{s} *\frac{100cm}{1m}= 160 \frac{cm}{s}](https://tex.z-dn.net/?f=%20v%20%3D%201.6%20%5Cfrac%7Bm%7D%7Bs%7D%20%2A%5Cfrac%7B100cm%7D%7B1m%7D%3D%20160%20%5Cfrac%7Bcm%7D%7Bs%7D)
Explanation:
If we have a periodic wave we need to satisfy the following basic relationship:
![v = \lambda f](https://tex.z-dn.net/?f=%20v%20%3D%20%5Clambda%20f)
From the last formula we see that the velocity is proportional fo the frequency.
For this case we have the following info given by the problem:
![T= 0.2 s, \lambda =32 cm* \frac{1m}{100cm} =0.32 m, A= 3cm*\frac{1m}{100 cm}=0.03 m](https://tex.z-dn.net/?f=%20T%3D%200.2%20s%2C%20%5Clambda%20%3D32%20cm%2A%20%5Cfrac%7B1m%7D%7B100cm%7D%20%3D0.32%20m%2C%20A%3D%203cm%2A%5Cfrac%7B1m%7D%7B100%20cm%7D%3D0.03%20m)
We know that the frequency is the reciprocal of the period so we have this formula:
![f = \frac{1}{T}](https://tex.z-dn.net/?f=%20f%20%3D%20%5Cfrac%7B1%7D%7BT%7D)
And if we replace we got:
![f =\frac{1}{0.2 s}= 5Hz](https://tex.z-dn.net/?f=%20f%20%3D%5Cfrac%7B1%7D%7B0.2%20s%7D%3D%205Hz)
Now since we have the value for the wavelength we can find the velocity like this:
![v = 0.32 m * 5Hz = 1.6 \frac{m}{s}](https://tex.z-dn.net/?f=%20v%20%3D%200.32%20m%20%2A%205Hz%20%3D%201.6%20%5Cfrac%7Bm%7D%7Bs%7D)
And if we convert this into cm/s we got:
![v = 1.6 \frac{m}{s} *\frac{100cm}{1m}= 160 \frac{cm}{s}](https://tex.z-dn.net/?f=%20v%20%3D%201.6%20%5Cfrac%7Bm%7D%7Bs%7D%20%2A%5Cfrac%7B100cm%7D%7B1m%7D%3D%20160%20%5Cfrac%7Bcm%7D%7Bs%7D)
It runs on Hydrogen gas.
Actually, using hydrogen as a fuel is not new. We used to use it on air vehicle like air balloon. But back then, we still cannot figure out how to safely use this because Hydrogen exlodes rather easily
<u>Answer:</u> The weight of the object is 29.4 N
<u>Explanation:</u>
To calculate the weight of the object, we use the equation:
![W=m\times g](https://tex.z-dn.net/?f=W%3Dm%5Ctimes%20g)
where,
m = mass of the object = 3 kg
g = acceleration due to gravity = ![9.8m/s^2](https://tex.z-dn.net/?f=9.8m%2Fs%5E2)
Putting values in above equation, we get:
![W=3kg\times 9.8m/s^2\\\\W=29.4N](https://tex.z-dn.net/?f=W%3D3kg%5Ctimes%209.8m%2Fs%5E2%5C%5C%5C%5CW%3D29.4N)
Hence, the weight of the object is 29.4 N
Answer:
(A) 11 m/s
(B) 1.3 m
Explanation:
Horizontal range, R = 9.6 m
Angle of projection, theta = 28 degree
(A)
Use the formula of horizontal range
R = u^2 Sin 2 theta / g
u^2 = R g / Sin 2 theta
u^2 = 9.6 × 9.8 / Sin ( 2 × 28)
u = 10.65 m/s
u = 11 m/s
(B)
Use the formula for maximum height
H = u^2 Sin ^2 theta / 2g
H =
10.65 × 10.65 × Sin^2 (28) / ( 2 × 9.8)
H = 1.275 m
H = 1 .3 m
Answer:
The sugars produced by photosynthesis can be stored, transported throughout the tree, and converted into energy which is used to power all cellular processes. Respiration occurs when glucose (sugar produced during photosynthesis) combines with oxygen to produce useable cellular energy.
Explanation:
I think this is correct lol.