Answer:
A) 0 V
B) -117 kV
C) -0.468 J
Explanation:
q1=+2.00μC q2=−2.00μC q3 = -4.00 μC
A) The electric potential at point a due to q1 and q2 (
) is given as:
![V_a=k\Sigma \frac{q}{r_i} = k(\frac{q_1}{r_1}+\frac{q_2}{r_2} )\\ but\ r_2=r_1=d.\ Therefore\\V_a=k(\frac{q_1}{d}+\frac{q_2}{d} )=k(\frac{2}{d}-\frac{2}{d} )=0](https://tex.z-dn.net/?f=V_a%3Dk%5CSigma%20%5Cfrac%7Bq%7D%7Br_i%7D%20%3D%20k%28%5Cfrac%7Bq_1%7D%7Br_1%7D%2B%5Cfrac%7Bq_2%7D%7Br_2%7D%20%20%29%5C%5C%20but%5C%20r_2%3Dr_1%3Dd.%5C%20Therefore%5C%5CV_a%3Dk%28%5Cfrac%7Bq_1%7D%7Bd%7D%2B%5Cfrac%7Bq_2%7D%7Bd%7D%20%20%29%3Dk%28%5Cfrac%7B2%7D%7Bd%7D-%5Cfrac%7B2%7D%7Bd%7D%20%20%29%3D0)
B) he electric potential at point b due to q1 and q2 (
) is given as:
![V_b=k\Sigma \frac{q}{r_i} = k(\frac{q_1}{r_1}+\frac{q_2}{r_2} )\\ but\ r_2=4.5 cm=0.045m,\ r_1=\sqrt{0.045^2+0.045^2}= 0.0636.\ Therefore\\V_b= k(\frac{q_1}{r_1}+\frac{q_2}{r_2} )= 9*10^{9}(\frac{2*10^{-6}}{0.0636}-\frac{2*10^{-6}}{0.045} )=-117\ kV](https://tex.z-dn.net/?f=V_b%3Dk%5CSigma%20%5Cfrac%7Bq%7D%7Br_i%7D%20%3D%20k%28%5Cfrac%7Bq_1%7D%7Br_1%7D%2B%5Cfrac%7Bq_2%7D%7Br_2%7D%20%20%29%5C%5C%20but%5C%20r_2%3D4.5%20cm%3D0.045m%2C%5C%20r_1%3D%5Csqrt%7B0.045%5E2%2B0.045%5E2%7D%3D%200.0636.%5C%20Therefore%5C%5CV_b%3D%20k%28%5Cfrac%7Bq_1%7D%7Br_1%7D%2B%5Cfrac%7Bq_2%7D%7Br_2%7D%20%20%29%3D%209%2A10%5E%7B9%7D%28%5Cfrac%7B2%2A10%5E%7B-6%7D%7D%7B0.0636%7D-%5Cfrac%7B2%2A10%5E%7B-6%7D%7D%7B0.045%7D%20%20%29%3D-117%5C%20kV)
C) The work done on q3 by the electric forces exerted by q1 and q2 (W) is given by:
Answer:
The velocity is ![v = 6.66 \ m/s](https://tex.z-dn.net/?f=v%20%20%3D%20%206.66%20%5C%20%20m%2Fs%20)
Henrietta is at distance
from the under the window
Explanation:
From the question we are told that
The speed of Henrietta is ![v= 2.70 \ m/s](https://tex.z-dn.net/?f=v%3D%20%202.70%20%5C%20m%2Fs)
The height of the window from the ground is ![h = 36.5 \ m](https://tex.z-dn.net/?f=h%20%20%3D%20%2036.5%20%5C%20%20m)
Generally the time taken for the lunch to reach the ground assuming it fell directly under the window is
![t = \sqrt{\frac{2 * h }{g} }](https://tex.z-dn.net/?f=t%20%20%3D%20%20%5Csqrt%7B%5Cfrac%7B2%20%2A%20%20h%20%7D%7Bg%7D%20%7D)
=>
=>
Generally the time taken for the lunch to reach Henrietta is mathematically represented as
![T = t + t_1](https://tex.z-dn.net/?f=T%20%3D%20%20t%20%2B%20%20t_1)
Here
is the time duration that elapsed after Henrietta has passed below the window the value is given as 4 s
Now
=>
Generally the distance covered by Henrietta before catching her lunch is
![s= v * T](https://tex.z-dn.net/?f=s%3D%20%20v%20%20%2A%20%20T)
=> ![s= 2.70 * 6.73](https://tex.z-dn.net/?f=s%3D%20%202.70%20%20%2A%206.73)
=> ![s= 18.17 \ m](https://tex.z-dn.net/?f=s%3D%20%2018.17%20%5C%20%20m)
Generally the speed with which Bruce threw her lunch is mathematically represented as
![v = \frac{18.17}{2.73}](https://tex.z-dn.net/?f=v%20%20%3D%20%20%5Cfrac%7B18.17%7D%7B2.73%7D)
![v = 6.66 \ m/s](https://tex.z-dn.net/?f=v%20%20%3D%20%206.66%20%5C%20%20m%2Fs%20)
Answer:
τ = 1679.68Nm
Explanation:
In order to calculate the required torque you first take into account the following formula:
(1)
τ: torque
I: moment of inertia of the merry-go-round
α: angular acceleration
Next, you use the following formulas for the calculation of the angular acceleration and the moment of inertia:
(2)
(3) (it is considered that the merry-go-round is a disk)
w: final angular speed = 3.1 rad/s
wo: initial angular speed = 0 rad/s
M: mass of the merry-go-round = 432 kg
R: radius of the merry-go-round = 2.3m
You solve the equation (2) for α. Furthermore you calculate the moment of inertia:
![\alpha=\frac{\omega}{t}=\frac{3.1rad/s}{2.1s}=1.47\frac{rad}{s^2}\\\\I=\frac{1}{2}(432kg)(2.3)^2=1142.64kg\frac{m}{s}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cfrac%7B%5Comega%7D%7Bt%7D%3D%5Cfrac%7B3.1rad%2Fs%7D%7B2.1s%7D%3D1.47%5Cfrac%7Brad%7D%7Bs%5E2%7D%5C%5C%5C%5CI%3D%5Cfrac%7B1%7D%7B2%7D%28432kg%29%282.3%29%5E2%3D1142.64kg%5Cfrac%7Bm%7D%7Bs%7D)
Finally, you replace the values of the moment of inertia and angular acceleration in the equation (1):
![\tau=(1142.64kgm/s)(1.47rad/s^2)=1679.68Nm](https://tex.z-dn.net/?f=%5Ctau%3D%281142.64kgm%2Fs%29%281.47rad%2Fs%5E2%29%3D1679.68Nm)
The required torque is 1679.68Nm
Seems the both of us haven't done the project yet.