Answer:
a. one line down one line to the right one live to the northwest from the object
b. t1=190 t2=310
Explanation:
When car is at the top of the hill its whole energy is stored in the form of gravitational potential energy
![U = mgh](https://tex.z-dn.net/?f=U%20%3D%20mgh)
so when height of the car becomes half then its potential energy is given as
![U_f = \frac{mgh}{2}](https://tex.z-dn.net/?f=U_f%20%3D%20%5Cfrac%7Bmgh%7D%7B2%7D)
so final potential energy when car falls down by half of the height will become half of the initial potential energy
So it is U = 50 MJ after falling down
Now by energy conservation we can say that final potential energy + final Kinetic energy must be equal to the initial potential energy of the car
So here at half of the height kinetic energy of car = 100 - 50 = 50 MJ
so we can say at this point magnitude of potential energy and kinetic energy will be same
<em>A. the same as the potential energy at that point.</em>
To solve this problem it is necessary to apply the concepts related to the described wavelength through frequency and speed. Mathematically it can be expressed as:
![\lambda = \frac{v}{f}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7Bv%7D%7Bf%7D)
Where,
Wavelength
f = Frequency
v = Velocity
Our values are given as,
![f = 2.8*10^3Hz](https://tex.z-dn.net/?f=f%20%3D%202.8%2A10%5E3Hz)
Speed of sound
Keep in mind that we do not use the travel speed of the ambulance because we are in front of it. In case it approached or moved away we should use the concepts related to the Doppler effect:
Replacing we have,
![\lambda = \frac{340}{2.8*10^3}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7B340%7D%7B2.8%2A10%5E3%7D)
![\lambda = 0.1214m](https://tex.z-dn.net/?f=%5Clambda%20%3D%200.1214m)
Therefore the frequency that you hear if you are standing in from of the ambulance is 0.1214m
Answer:
Explanation:
Given
Diameter of CD ![d=12.70\ cm](https://tex.z-dn.net/?f=d%3D12.70%5C%20cm)
radius ![r=6.35\ cm](https://tex.z-dn.net/?f=r%3D6.35%5C%20cm)
mass of CD ![m=16.30\ gm](https://tex.z-dn.net/?f=m%3D16.30%5C%20gm)
Tangential velocity ![v_t=1.150\ m/s](https://tex.z-dn.net/?f=v_t%3D1.150%5C%20m%2Fs)
music detected at ![r'=20.90\ mm=2.090\ cm](https://tex.z-dn.net/?f=r%27%3D20.90%5C%20mm%3D2.090%5C%20cm)
Moment of Inertia of disc ![I=\frac{mr^2}{2}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7Bmr%5E2%7D%7B2%7D)
![I=\frac{16.30\times 10^{-3}\times (6.35\times 10^{-2})^2}{2}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B16.30%5Ctimes%2010%5E%7B-3%7D%5Ctimes%20%286.35%5Ctimes%2010%5E%7B-2%7D%29%5E2%7D%7B2%7D)
![I=6.572\times 10^{-5} kg-m^2](https://tex.z-dn.net/?f=I%3D6.572%5Ctimes%2010%5E%7B-5%7D%20kg-m%5E2)
Work done is equal to change in kinetic Energy of CD
![W=\Delta K](https://tex.z-dn.net/?f=W%3D%5CDelta%20K)
![W=\frac{1}{2}I\omega_f^2-\frac{1}{2}I\omega_i^2](https://tex.z-dn.net/?f=W%3D%5Cfrac%7B1%7D%7B2%7DI%5Comega_f%5E2-%5Cfrac%7B1%7D%7B2%7DI%5Comega_i%5E2)
where
angular velocity
![v_t=\omega \times r](https://tex.z-dn.net/?f=v_t%3D%5Comega%20%5Ctimes%20r)
![W=\frac{1}{2}\times 6.572\times 10^{-5}\left ( \left ( \frac{1.15}{0.0209}\right )^2-0\right )](https://tex.z-dn.net/?f=W%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%206.572%5Ctimes%2010%5E%7B-5%7D%5Cleft%20%28%20%5Cleft%20%28%20%5Cfrac%7B1.15%7D%7B0.0209%7D%5Cright%20%29%5E2-0%5Cright%20%29)
![W=0.0994\ J](https://tex.z-dn.net/?f=W%3D0.0994%5C%20J)
Answer:
B) 4/5
Explanation:
The magnitude of the electric force between the two spheres is given by
![F=k\frac{q_1 q_2}{r^2}](https://tex.z-dn.net/?f=F%3Dk%5Cfrac%7Bq_1%20q_2%7D%7Br%5E2%7D)
where
k is the Coulombs' constant
q1 and q2 are the charges on the two spheres
r is the distance between the two spheres
Initially, we have
![q_1 = 5.0\mu C=5.0\cdot 10^{-6}C\\q_2 = 1.0 \mu C=1.0 \cdot 10^{-6}C\\r = L](https://tex.z-dn.net/?f=q_1%20%3D%205.0%5Cmu%20C%3D5.0%5Ccdot%2010%5E%7B-6%7DC%5C%5Cq_2%20%3D%201.0%20%5Cmu%20C%3D1.0%20%5Ccdot%2010%5E%7B-6%7DC%5C%5Cr%20%3D%20L)
So the force is
![F_1=k\frac{(5.0\cdot 10^{-6}C)(1.0\cdot 10^{-6}C)}{L^2}=(5.0 \cdot 10^{-12} C^2)\frac{k}{L^2}](https://tex.z-dn.net/?f=F_1%3Dk%5Cfrac%7B%285.0%5Ccdot%2010%5E%7B-6%7DC%29%281.0%5Ccdot%2010%5E%7B-6%7DC%29%7D%7BL%5E2%7D%3D%285.0%20%5Ccdot%2010%5E%7B-12%7D%20C%5E2%29%5Cfrac%7Bk%7D%7BL%5E2%7D)
Later, the two spheres are brought together so they are in contact: this means that the total charge will redistribute equally on the two spheres (because they are identical).
The total charge is
![Q=q_1 + q_2 = +4.0 \mu C=4.0\cdot 10^{-6}C](https://tex.z-dn.net/?f=Q%3Dq_1%20%2B%20q_2%20%3D%20%2B4.0%20%5Cmu%20C%3D4.0%5Ccdot%2010%5E%7B-6%7DC)
So each sphere will have a charge of
![q=\frac{Q}{2}=2.0\cdot 10^{-6} C](https://tex.z-dn.net/?f=q%3D%5Cfrac%7BQ%7D%7B2%7D%3D2.0%5Ccdot%2010%5E%7B-6%7D%20C)
So, the new force will be
![F_2=k\frac{(2.0\cdot 10^{-6}C)(120\cdot 10^{-6}C)}{L^2}=(4.0 \cdot 10^{-12} C^2)\frac{k}{L^2}](https://tex.z-dn.net/?f=F_2%3Dk%5Cfrac%7B%282.0%5Ccdot%2010%5E%7B-6%7DC%29%28120%5Ccdot%2010%5E%7B-6%7DC%29%7D%7BL%5E2%7D%3D%284.0%20%5Ccdot%2010%5E%7B-12%7D%20C%5E2%29%5Cfrac%7Bk%7D%7BL%5E2%7D)
And so the ratio of the two forces is
![\frac{F_2}{F_1}=\frac{4.0\cdot 10^{-12} C}{5.0\cdot 10^{-12} C}=\frac{4}{5}](https://tex.z-dn.net/?f=%5Cfrac%7BF_2%7D%7BF_1%7D%3D%5Cfrac%7B4.0%5Ccdot%2010%5E%7B-12%7D%20C%7D%7B5.0%5Ccdot%2010%5E%7B-12%7D%20C%7D%3D%5Cfrac%7B4%7D%7B5%7D)