An indicator of average kinetic energy is temperature. Temperature is directly proportional to Kinetic energy of the molecules of an element.
Explanation:
Given that,
Mass = 0.254 kg
Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]
Force = 0.5 N
y = 0.628
We need to calculate the A and d
Using formula of A and d
.....(I)
....(II)
Put the value of
in equation (I) and (II)
![A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}](https://tex.z-dn.net/?f=A%3D%5Cdfrac%7B%5Cdfrac%7B0.5%7D%7B0.254%7D%7D%7B%5Csqrt%7B%2810.0%5E2-0.628%29%5E2%2B0.628%5E2%5Ctimes0.628%5E2%7D%7D)
![A=0.0198](https://tex.z-dn.net/?f=A%3D0.0198)
From equation (II)
![tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}](https://tex.z-dn.net/?f=tan%20d%3D%5Cdfrac%7B0.628%5Ctimes0.628%7D%7B%28%2810.0%5E2-0.628%29%5E2%29%7D)
![d=0.0023](https://tex.z-dn.net/?f=d%3D0.0023)
Put the value of
in equation (I) and (II)
![A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}](https://tex.z-dn.net/?f=A%3D%5Cdfrac%7B%5Cdfrac%7B0.5%7D%7B0.254%7D%7D%7B%5Csqrt%7B%2810.0%5E2-3.14%29%5E2%2B0.628%5E2%5Ctimes3.14%5E2%7D%7D)
![A=0.0203](https://tex.z-dn.net/?f=A%3D0.0203)
From equation (II)
![tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}](https://tex.z-dn.net/?f=tan%20d%3D%5Cdfrac%7B0.628%5Ctimes3.14%7D%7B%28%2810.0%5E2-3.14%29%5E2%29%7D)
![d=0.0120](https://tex.z-dn.net/?f=d%3D0.0120)
Put the value of
in equation (I) and (II)
![A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}](https://tex.z-dn.net/?f=A%3D%5Cdfrac%7B%5Cdfrac%7B0.5%7D%7B0.254%7D%7D%7B%5Csqrt%7B%2810.0%5E2-6.28%29%5E2%2B0.628%5E2%5Ctimes6.28%5E2%7D%7D)
![A=0.0209](https://tex.z-dn.net/?f=A%3D0.0209)
From equation (II)
![tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}](https://tex.z-dn.net/?f=tan%20d%3D%5Cdfrac%7B0.628%5Ctimes6.28%7D%7B%28%2810.0%5E2-6.28%29%5E2%29%7D)
![d=0.0257](https://tex.z-dn.net/?f=d%3D0.0257)
Put the value of
in equation (I) and (II)
![A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}](https://tex.z-dn.net/?f=A%3D%5Cdfrac%7B%5Cdfrac%7B0.5%7D%7B0.254%7D%7D%7B%5Csqrt%7B%2810.0%5E2-9.42%29%5E2%2B0.628%5E2%5Ctimes9.42%5E2%7D%7D)
![A=0.0217](https://tex.z-dn.net/?f=A%3D0.0217)
From equation (II)
![tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}](https://tex.z-dn.net/?f=tan%20d%3D%5Cdfrac%7B0.628%5Ctimes9.42%7D%7B%28%2810.0%5E2-9.42%29%5E2%29%7D)
![d=0.0413](https://tex.z-dn.net/?f=d%3D0.0413)
Hence, This is the required solution.
Answer:
612000 C
Explanation:
Current, I, is given as the rate of flow of charge, that is:
I = Δq / Δt
where q = electric charge
t = time taken
This implies that:
Δq = I * Δt
The battery rating is 170 Ampere-hours, therefore:
Δq = 170 * 1 hour
But 1 hour = 3600 seconds;
=> Δq = 170 * 3600 = 612000 C
The total charge that the battery can provide is 612000 C.