Answer:
The frequency of the standing wave in the second case is higher than that in the first case
Explanation:
The frequency and wavelength of a wave are related.
The moment you sliced the bottle, you've reduced the wavelength of the bottle.
When wavelength decreases, frequency increases and vice versa.
So, When frequency
increases in the second case, more wave crests pass a fixed point each second. That means
the wavelength shortens. So, as frequency increases, wavelength
decreases. The opposite is also true—as frequency decreases,
wavelength increases.
D because it has the lowest wavelength, meaning the wave is the longest. Its crest are more spaced out creating lower energy. The smaller distance between the crests equals the higher the sound. Creating higher energy.
I am 95% sure this is right. Very sorry if it isn't.
Hope it helps. ;3
Answer:
1) a = -1 m/s²
2) v = 12 m/s
Explanation:
Given,
The initial velocity of the object, u = 15 m/s
The final velocity of the object, v = 10 m/s
The time taken by the object to travel is, t = 5 s
Using the first equation of motion
<em>v = u + at</em>
a = (v - u) / t
Substituting the values
a = (10 - 15) / 5
= -1 m/s²
The negative sign indicates the body is decelerating
The acceleration of the object is, a = -1 m/s²
The speed of the object after 2 seconds
From the above equations of motion
v = 15 + (-1) 2
= 12 m/s
Hence, the speed of the object after 2 seconds is, v = 12 m/s
Answer:
5.90 ft/s^2
Explanation:
There are mixed units in this question....convert everything to miles or feet
and hr to s
28 mi / hr = 41.066 ft/s
Displacement = vo t + 1/2 at^2
599 = 41.066 (8.9) + 1/2 a (8.9^2)
solve for a = ~ 5.90 ft/s^2
Answer:
t = 1.098*RC
Explanation:
In order to calculate the time that the capacitor takes to reach 2/3 of its maximum charge, you use the following formula for the charge of the capacitor:
![Q=Q_{max}[1-e^{-\frac{t}{RC}}]](https://tex.z-dn.net/?f=Q%3DQ_%7Bmax%7D%5B1-e%5E%7B-%5Cfrac%7Bt%7D%7BRC%7D%7D%5D)
(1)
Qmax: maximum charge capacity of the capacitor
t: time
R: resistor of the circuit
C: capacitance of the circuit
When the capacitor has 2/3 of its maximum charge, you have that
Q=(2/3)Qmax
You replace the previous expression for Q in the equation (1), and use properties of logarithms to solve for t:
![Q=\frac{2}{3}Q_{max}=Q_{max}[1-e^{-\frac{t}{RC}}]\\\\\frac{2}{3}=1-e^{-\frac{t}{RC}}\\\\e^{-\frac{t}{RC}}=\frac{1}{3}\\\\-\frac{t}{RC}=ln(\frac{1}{3})\\\\t=-RCln(\frac{1}{3})=1.098RC](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B2%7D%7B3%7DQ_%7Bmax%7D%3DQ_%7Bmax%7D%5B1-e%5E%7B-%5Cfrac%7Bt%7D%7BRC%7D%7D%5D%5C%5C%5C%5C%5Cfrac%7B2%7D%7B3%7D%3D1-e%5E%7B-%5Cfrac%7Bt%7D%7BRC%7D%7D%5C%5C%5C%5Ce%5E%7B-%5Cfrac%7Bt%7D%7BRC%7D%7D%3D%5Cfrac%7B1%7D%7B3%7D%5C%5C%5C%5C-%5Cfrac%7Bt%7D%7BRC%7D%3Dln%28%5Cfrac%7B1%7D%7B3%7D%29%5C%5C%5C%5Ct%3D-RCln%28%5Cfrac%7B1%7D%7B3%7D%29%3D1.098RC)
The charge in the capacitor reaches 2/3 of its maximum charge in a time equal to 1.098RC
