Answer:
A flat, horizontal line
Explanation:
A flat, horizontal line indicates a phase change.
The temperature does not increase because the added heat goes into converting one phase into another.
A is wrong. A downward-sloping line indicates that the temperature is decreasing with time.
C is wrong. An upward-sloping line indicates that the temperature is increasing with time.
Answer:
frequency of the sound = f = 1,030.3 Hz
phase difference = Φ = 229.09°
Explanation:
Step 1: Given data:
Xini = 0.540m
Xfin = 0.870m
v = 340m/s
Step 2: frequency of the sound (f)
f = v / λ
λ = Xfin - Xini = 0.870 - 0.540 = 0.33
f = 340 / 0.33
f = 1,030.3 Hz
Step 3: phase difference
phase difference = Φ
Φ = (2π/λ)*(Xini - λ) = (2π/0.33)* (0.540-0.33) = 19.04*0.21 = 3.9984
Φ = 3.9984 rad * (360°/2π rad)
Φ = 229.09°
Hope this helps!
Answer
given,
Time period= T = 1.5 s
If it's moving through equilibrium point at t₀= 0 with v = 1.0 m/s
v_max=1.00 m/s
we know,
v_ max=A ω
v = A sin (ωt)
-0.50= -1.00 sin (ωt)
sin (ωt) = 0.5
![\omega t = sin^{-1}(0.5)](https://tex.z-dn.net/?f=%5Comega%20t%20%3D%20sin%5E%7B-1%7D%280.5%29)
![\dfrac{2\pi}{T}\times t =0.524](https://tex.z-dn.net/?f=%5Cdfrac%7B2%5Cpi%7D%7BT%7D%5Ctimes%20t%20%3D0.524)
![\dfrac{2\pi}{1.5}\times t =0.524](https://tex.z-dn.net/?f=%5Cdfrac%7B2%5Cpi%7D%7B1.5%7D%5Ctimes%20t%20%3D0.524)
t = 0.125 s
we have time period T=1.5 it is the time to complete one oscillation
means from eq to right,then left,then eq,then left,then from right to eq
time taken for left = t/4 = 0.125/4 = 0.375 s
smallest value of time
=0.375 + 0.125
= 0.50 sec
Answer:
Explanation:
It means that you only need apply 1/4th of the actual force required to operate the lever as you have an mechanical advantage which permits you to do 4 times the work with the same amount of effort.
Explanation:
According to the energy conservation,
![F_{centripetal} = F_{electric}](https://tex.z-dn.net/?f=F_%7Bcentripetal%7D%20%3D%20F_%7Belectric%7D)
![\frac{mv^{2}}{r} = \frac{kq^{2}}{d^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7Bmv%5E%7B2%7D%7D%7Br%7D%20%3D%20%5Cfrac%7Bkq%5E%7B2%7D%7D%7Bd%5E%7B2%7D%7D)
![v^{2} = \frac{kq^{2}r}{d^{2}m}](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%20%5Cfrac%7Bkq%5E%7B2%7Dr%7D%7Bd%5E%7B2%7Dm%7D)
= ![\frac{9 \times 10^{9} N.m^{2}/C^{2} \times 1.6 \times 10^{-19} C \times 0.75 \times 10^{-9} m}{(1.50 \times 10^{-9}m)^{2} \times 9.11 \times 10^{-31} kg}](https://tex.z-dn.net/?f=%5Cfrac%7B9%20%5Ctimes%2010%5E%7B9%7D%20N.m%5E%7B2%7D%2FC%5E%7B2%7D%20%5Ctimes%201.6%20%5Ctimes%2010%5E%7B-19%7D%20C%20%5Ctimes%200.75%20%5Ctimes%2010%5E%7B-9%7D%20m%7D%7B%281.50%20%5Ctimes%2010%5E%7B-9%7Dm%29%5E%7B2%7D%20%5Ctimes%209.11%20%5Ctimes%2010%5E%7B-31%7D%20kg%7D)
= ![8.430 \times 10^{10} m^{2}/s^{2}](https://tex.z-dn.net/?f=8.430%20%5Ctimes%2010%5E%7B10%7D%20m%5E%7B2%7D%2Fs%5E%7B2%7D)
v = ![\sqrt{8.430 \times 10^{10} m^{2}/s^{2}}](https://tex.z-dn.net/?f=%5Csqrt%7B8.430%20%5Ctimes%2010%5E%7B10%7D%20m%5E%7B2%7D%2Fs%5E%7B2%7D%7D)
= ![2.903 \times 10^{5} m/s](https://tex.z-dn.net/?f=2.903%20%5Ctimes%2010%5E%7B5%7D%20m%2Fs)
Formula for distance from the orbit is as follows.
S = ![2 \pi r](https://tex.z-dn.net/?f=2%20%5Cpi%20r)
= ![2 \times 3.14 \times 0.75 \times 10^{-9} m](https://tex.z-dn.net/?f=2%20%5Ctimes%203.14%20%5Ctimes%200.75%20%5Ctimes%2010%5E%7B-9%7D%20m)
= ![4.71 \times 10^{-9} m](https://tex.z-dn.net/?f=4.71%20%5Ctimes%2010%5E%7B-9%7D%20m)
Now, relation between time and distance is as follows.
T = ![\frac{S}{v}](https://tex.z-dn.net/?f=%5Cfrac%7BS%7D%7Bv%7D)
![\frac{1}{f} = \frac{S}{v}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%20%5Cfrac%7BS%7D%7Bv%7D)
or, f =
=
= ![6.164 \times 10^{13} Hz](https://tex.z-dn.net/?f=6.164%20%5Ctimes%2010%5E%7B13%7D%20Hz)
Thus, we can conclude that the orbital frequency for an electron and a positron that is 1.50 apart is
.