Answer:
The net charge is ![67.89 \mu C](https://tex.z-dn.net/?f=67.89%20%5Cmu%20C)
Solution:
As per the question:
Mass of the plastic bag, m = 12.0 g = ![12.0\times 10^{-3}\ kg](https://tex.z-dn.net/?f=12.0%5Ctimes%2010%5E%7B-3%7D%5C%20kg)
Magnitude of electric field, E = ![10^{3}\ N/C](https://tex.z-dn.net/?f=10%5E%7B3%7D%5C%20N%2FC)
Angle made by the string, ![\theta = 30^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2030%5E%7B%5Ccirc%7D)
Now,
To calculate the net charge, Q on the ball:
Vertical component of the tension in the string, ![T = Tcos\theta](https://tex.z-dn.net/?f=T%20%3D%20Tcos%5Ctheta)
Horizontal component of the tension in the string, ![T = Tsin\theta](https://tex.z-dn.net/?f=T%20%3D%20Tsin%5Ctheta)
Now,
Balancing the forces in the x-direction:
![Tsin\theta = QE](https://tex.z-dn.net/?f=Tsin%5Ctheta%20%3D%20QE)
(1)
Balancing the forces in the y-direction:
![Tcos\theta = mg](https://tex.z-dn.net/?f=Tcos%5Ctheta%20%3D%20mg)
where
g = acceleration due to gravity = ![9.8\ m/s^{2}](https://tex.z-dn.net/?f=9.8%5C%20m%2Fs%5E%7B2%7D)
Thus
![T = \frac{mg}{cos\theta }](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7Bmg%7D%7Bcos%5Ctheta%20%7D)
![T = \frac{12.0\times 10^{-3}\times 9.8}{cos30^{\circ}} = 0.1357\ N](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B12.0%5Ctimes%2010%5E%7B-3%7D%5Ctimes%209.8%7D%7Bcos30%5E%7B%5Ccirc%7D%7D%20%3D%200.1357%5C%20N)
Use T = 0.1357 N in eqn (1):
![Q = {0.1357\times sin30^{\circ}}{10^{3}} = 6.789\times 10^{- 5}\ C](https://tex.z-dn.net/?f=Q%20%3D%20%7B0.1357%5Ctimes%20sin30%5E%7B%5Ccirc%7D%7D%7B10%5E%7B3%7D%7D%20%3D%206.789%5Ctimes%2010%5E%7B-%205%7D%5C%20C)
![Q = 67.89\times 10^{- 5}\ C = 67.89\mu C](https://tex.z-dn.net/?f=Q%20%3D%2067.89%5Ctimes%2010%5E%7B-%205%7D%5C%20C%20%3D%2067.89%5Cmu%20C)
Answer:
The answer to your question is: letter D.
Explanation:
a.The mass that a mole of substance has, measured in grams per mole. Density is not measure in moles, so this is not the correct answer.
b.The amount of substance dissolved in a liquid, measured in moles per liter. The substance dissolved in a liquid must be measure in grams not in moles, so this answer is incorrect.
c.The mass of substance dissolved in a liquid, measured in grams per milliliter. I think that this definition is correct but is incomple, so this answer is wrong.
d.The ratio of a substance's mass to its volume, measured in grams per milliliter and also equivalent to grams per cubic centimeter. This is the right description to density, so this is the correct answer.
Answer:
3.004A
Explanation:
The voltage across an inductor is expressed as:
![V=IX_L\\V=I(2\pi fL)](https://tex.z-dn.net/?f=V%3DIX_L%5C%5CV%3DI%282%5Cpi%20fL%29)
Given
f = 270Hz
V = 48V
L =0.053 H
Get the current
270 = I (2*3.14*270*0.053)
270 = 89.8668I
I = 270/89.8668
I = 3.004A
Hence the current in the inductor is 3.004A
Answer:
Velocity of Gulf Stream=
west of north at a speed of ![2.03\ \rm m/s](https://tex.z-dn.net/?f=2.03%5C%20%5Crm%20m%2Fs)
Explanation:
Given
- speed of ship relative to Gulf stream=
west of north at a speed of 4 m/s
- speed of ship relative to earth=
west of north at a speed of 4 .8 m/s
Let
be the velocity of the ship relative to the earth and
be the velocity of the ship with respect to the Gulf stream
Let the west direction be negative x axis and north direction be positive x axis
Now
![V_{sg}=-4\sin25^\circ \vec i+4\cos25^\circ \vec j\\V_s-V_g=-4\sin25^\circ \vec i+4\cos25^\circ \vec j\\-4.8\sin5^\circ \vec i+4.8\cos5^\circ \vec j-V_g=-4\sin25^\circ \vec i+4\cos25^\circ \vec j\\Vg=-1.68\vec\ i+1.156\vec\ j\ \rm m/s\\](https://tex.z-dn.net/?f=V_%7Bsg%7D%3D-4%5Csin25%5E%5Ccirc%20%5Cvec%20i%2B4%5Ccos25%5E%5Ccirc%20%5Cvec%20j%5C%5CV_s-V_g%3D-4%5Csin25%5E%5Ccirc%20%5Cvec%20i%2B4%5Ccos25%5E%5Ccirc%20%5Cvec%20j%5C%5C-4.8%5Csin5%5E%5Ccirc%20%5Cvec%20i%2B4.8%5Ccos5%5E%5Ccirc%20%5Cvec%20j-V_g%3D-4%5Csin25%5E%5Ccirc%20%5Cvec%20i%2B4%5Ccos25%5E%5Ccirc%20%5Cvec%20j%5C%5CVg%3D-1.68%5Cvec%5C%20i%2B1.156%5Cvec%5C%20j%5C%20%5Crm%20m%2Fs%5C%5C)
magnitude of the velocity is given by
![V_g=\sqrt{(1.68^2+1.156^2)}\\V_g=2.03\ \rm m/s\\](https://tex.z-dn.net/?f=V_g%3D%5Csqrt%7B%281.68%5E2%2B1.156%5E2%29%7D%5C%5CV_g%3D2.03%5C%20%5Crm%20m%2Fs%5C%5C)
Let the velocity of gulf stream makes an angle
with the positive y axis we have
![\tan\theta=\dfrac{1.156}{1.68}\\\theta=34.5^\circ](https://tex.z-dn.net/?f=%5Ctan%5Ctheta%3D%5Cdfrac%7B1.156%7D%7B1.68%7D%5C%5C%5Ctheta%3D34.5%5E%5Ccirc)
Velocity of Gulf Stream
west of north at a speed of ![2.03\ \rm m/s](https://tex.z-dn.net/?f=2.03%5C%20%5Crm%20m%2Fs)
Use kinematic equations to solve:
1) yf = yo + vo*t + 1/2at²
yf = final height
yo = initial height
vo = initial velocity
a = acceleration
t = time
yf - yo = vo*t + 1/2at²
yf - yo = h
vo = 0
Thus,
h = 1/2at²
h = 1/2(9.8)(12)² = 705.6 m
2) vf = vo + at
vo = 0
Thus,
vf = at
vf = (9.8)(12) = 117.6 m/s