Answer:
(a) A = m/s^3, B = m/s.
(b) dx/dt = m/s.
Explanation:
(a)
![x = At^3 + Bt\\m = As^3 + Bs\\m = (\frac{m}{s^3})s^3 + (\frac{m}{s})s](https://tex.z-dn.net/?f=x%20%3D%20At%5E3%20%2B%20Bt%5C%5Cm%20%3D%20As%5E3%20%2B%20Bs%5C%5Cm%20%3D%20%28%5Cfrac%7Bm%7D%7Bs%5E3%7D%29s%5E3%20%2B%20%28%5Cfrac%7Bm%7D%7Bs%7D%29s)
Therefore, the dimension of A is m/s^3, and of B is m/s in order to satisfy the above equation.
(b) ![\frac{dx}{dt} = 3At^2 + B = 3(\frac{m}{s^3})s^2 + \frac{m}{s} = m/s](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bdt%7D%20%3D%203At%5E2%20%2B%20B%20%3D%203%28%5Cfrac%7Bm%7D%7Bs%5E3%7D%29s%5E2%20%2B%20%5Cfrac%7Bm%7D%7Bs%7D%20%3D%20m%2Fs)
This makes sense, because the position function has a unit of 'm'. The derivative of the position function is velocity, and its unit is m/s.
It's hard to tell exactly what's happening in that 110 cm that you marked over the wave. What is under the ends of the long arrow ? How many complete waves ? I counted 4.5 complete waves ... maybe ?
If there are 4.5 complete waves in 110cm, then the length of 1 wave is (110/4.5)=24.44cm.
Frequency = speed/wavelength
Frequency = 2m/s /0.2444m
Frequency = 8.18 Hz
Answer:
Output voltage is 1.92 volts.
Explanation:
Given that,
Number of turns in primary coil, ![N_p=475](https://tex.z-dn.net/?f=N_p%3D475)
Number of turns in secondary coil, ![N_s=8](https://tex.z-dn.net/?f=N_s%3D8)
Input voltage, ![V_i=114\ V](https://tex.z-dn.net/?f=V_i%3D114%5C%20V)
We need to find the voltage output of a transformer used for rechargeable flashlight batteries. For a transformer, the number of turns and the voltage ratio is given by :
![\dfrac{N_p}{N_s}=\dfrac{V_i}{V_o}\\\\\\V_o=114\cdot\dfrac{8}{475}\\\\V_o=1.92\ V](https://tex.z-dn.net/?f=%5Cdfrac%7BN_p%7D%7BN_s%7D%3D%5Cdfrac%7BV_i%7D%7BV_o%7D%5C%5C%5C%5C%5C%5CV_o%3D114%5Ccdot%5Cdfrac%7B8%7D%7B475%7D%5C%5C%5C%5CV_o%3D1.92%5C%20V)
So, the output voltage is 1.92 volts.
Answer:
Remains same
Explanation:
= Time period of oscillation
= mass
= spring constant
Time period of oscillation is given as
![T = 2\pi \sqrt{\frac{m}{k} }](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%20%7D)
we know that as we move from earth to moon, the value of spring constant "k" and mass "m" remains unchanged because they do not depend on the acceleration due to gravity.
Time period depends on spring constant inversely and directly on the mass.
hence the time period remains the same.