Answer:
![\theta = 20.98 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2020.98%20degree)
Explanation:
As we know that the speed of the sound is given as
![v = 332 + 0.6 t](https://tex.z-dn.net/?f=v%20%3D%20332%20%2B%200.6%20t)
now at t = 273 k = 0 degree
![v = 332 m/s](https://tex.z-dn.net/?f=v%20%3D%20332%20m%2Fs)
so we have
![a sin\theta = N\lambda](https://tex.z-dn.net/?f=a%20sin%5Ctheta%20%3D%20N%5Clambda)
![a sin\theta = N(\frac{v_1}{f})](https://tex.z-dn.net/?f=a%20sin%5Ctheta%20%3D%20N%28%5Cfrac%7Bv_1%7D%7Bf%7D%29)
now when temperature is changed to 313 K we have
![t = 313 - 273 = 40 degree](https://tex.z-dn.net/?f=t%20%3D%20313%20-%20273%20%3D%2040%20degree)
now we have
![v = 332 + (0.6)(40)](https://tex.z-dn.net/?f=v%20%3D%20332%20%2B%20%280.6%29%2840%29)
![v_2 = 356 m/s](https://tex.z-dn.net/?f=v_2%20%3D%20356%20m%2Fs)
![a sin\theta' = N(\frac{v_2}{f})](https://tex.z-dn.net/?f=a%20sin%5Ctheta%27%20%3D%20N%28%5Cfrac%7Bv_2%7D%7Bf%7D%29)
now from two equations we have
![\frac{sin19.5}{sin\theta} = \frac{332}{356}](https://tex.z-dn.net/?f=%5Cfrac%7Bsin19.5%7D%7Bsin%5Ctheta%7D%20%3D%20%5Cfrac%7B332%7D%7B356%7D)
so we have
![sin\theta = 0.358](https://tex.z-dn.net/?f=sin%5Ctheta%20%3D%200.358)
![\theta = 20.98 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2020.98%20degree)
Answer:
D. Sound Energy, Magnetic energy
Explanation:
Sound energy is in motion, and Magnetic energy is about to be in motion.
Answer:
I BELIEVE THE ANSWER A I REMEMBER THAT QUESTION WHEN I DID IT SO THATS THE BEST ONE OUT OF THE 3
Explanation:
Answer:
6 A
Explanation:
First of all, we need to calculate the equivalent resistance of the circuit. The three resistors are connected in parallel, so their equivalent resistance is given by:
![\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{120 \Omega}+\frac{1}{60 \Omega}+\frac{1}{40 \Omega}=\frac{3+2+1}{120 \Omega}=\frac{6}{120 \Omega}\\R_T = \frac{120}{6} \Omega](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_T%7D%3D%5Cfrac%7B1%7D%7BR_1%7D%2B%5Cfrac%7B1%7D%7BR_2%7D%2B%5Cfrac%7B1%7D%7BR_3%7D%3D%5Cfrac%7B1%7D%7B120%20%5COmega%7D%2B%5Cfrac%7B1%7D%7B60%20%5COmega%7D%2B%5Cfrac%7B1%7D%7B40%20%5COmega%7D%3D%5Cfrac%7B3%2B2%2B1%7D%7B120%20%5COmega%7D%3D%5Cfrac%7B6%7D%7B120%20%5COmega%7D%5C%5CR_T%20%3D%20%5Cfrac%7B120%7D%7B6%7D%20%5COmega)
And now we can use Ohm's law to find the current in the circuit:
![V=R_T II=\frac{V}{R_T}=\frac{120 V}{\frac{120}{6}\Omega}=6 A](https://tex.z-dn.net/?f=V%3DR_T%20II%3D%5Cfrac%7BV%7D%7BR_T%7D%3D%5Cfrac%7B120%20V%7D%7B%5Cfrac%7B120%7D%7B6%7D%5COmega%7D%3D6%20A)
Answer:
![E_y=1175510.2\ N.C^{-1}](https://tex.z-dn.net/?f=E_y%3D1175510.2%5C%20N.C%5E%7B-1%7D)
The Magnitude of electric field is in the upward direction as shown directly towards the charge
.
Explanation:
Given:
- side of a square,
![a=52.5\ cm](https://tex.z-dn.net/?f=a%3D52.5%5C%20cm)
- charge on one corner of the square,
![q_1=+45\times 10^{-6}\ C](https://tex.z-dn.net/?f=q_1%3D%2B45%5Ctimes%2010%5E%7B-6%7D%5C%20C)
- charge on the remaining 3 corners of the square,
![q_2=q_3=q_4=-27\times 10^{-6}\ C](https://tex.z-dn.net/?f=q_2%3Dq_3%3Dq_4%3D-27%5Ctimes%2010%5E%7B-6%7D%5C%20C)
<u>Distance of the center from each corners</u>![=\frac{1}{2} \times diagonals](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20diagonals)
![diagonal=\sqrt{52.5^2+52.5^2}](https://tex.z-dn.net/?f=diagonal%3D%5Csqrt%7B52.5%5E2%2B52.5%5E2%7D)
![diagonal=74.25\cm=0.7425\ m](https://tex.z-dn.net/?f=diagonal%3D74.25%5Ccm%3D0.7425%5C%20m)
∴Distance of center from corners, ![b=0.3712\ m](https://tex.z-dn.net/?f=b%3D0.3712%5C%20m)
Now, electric field due to charges is given as:
![E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Ctimes%20%5Cfrac%7Bq%7D%7Bb%5E2%7D)
<u>For charge
we have the field lines emerging out of the charge since it is positively charged:</u>
![E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}](https://tex.z-dn.net/?f=E_1%3D9%5Ctimes%2010%5E9%5Ctimes%20%5Cfrac%7B45%5Ctimes%2010%5E%7B-6%7D%7D%7B0.3712%5E2%7D)
<u>Force by each of the charges at the remaining corners:</u>
![E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}](https://tex.z-dn.net/?f=E_2%3DE_3%3DE_4%3D9%5Ctimes%2010%5E9%5Ctimes%20%5Cfrac%7B27%5Ctimes%2010%5E%7B-6%7D%7D%7B0.3712%5E2%7D)
<u> Now, net electric field in the vertical direction:</u>
![E_y=E_1-E_4](https://tex.z-dn.net/?f=E_y%3DE_1-E_4)
![E_y=1175510.2\ N.C^{-1}](https://tex.z-dn.net/?f=E_y%3D1175510.2%5C%20N.C%5E%7B-1%7D)
<u>Now, net electric field in the horizontal direction:</u>
![E_y=E_2-E_3](https://tex.z-dn.net/?f=E_y%3DE_2-E_3)
![E_y=0\ N.C^{-1}](https://tex.z-dn.net/?f=E_y%3D0%5C%20N.C%5E%7B-1%7D)
So the Magnitude of electric field is in the upward direction as shown directly towards the charge
.