A sphere is charged with electrons to −9 × 10−6 C. The value given is the total charge of all the electrons present in the sphere. To calculate the number of electrons in the sphere, we divide the the total charge with the charge of one electron.
N = 9 × 10−6 C / 1.6 × 10−19 C
N = 5.6 x 10^13
Answer:
![v_0 = 3.53~{\rm m/s}](https://tex.z-dn.net/?f=v_0%20%3D%203.53~%7B%5Crm%20m%2Fs%7D)
Explanation:
This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.
The initial velocity is in the x-direction, and there is no acceleration in the x-direction.
On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.
Applying the equations of kinematics in the x-direction gives
![x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}](https://tex.z-dn.net/?f=x%20-%20x_0%20%3D%20v_%7Bx_0%7D%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7Da_x%20t%5E2%5C%5C63%20%5Ctimes%2010%5E%7B-3%7D%20%3D%20v_0t%20%2B%200%5C%5Ct%20%3D%20%5Cfrac%7B63%5Ctimes%2010%5E%7B-3%7D%7D%7Bv_0%7D)
For the y-direction gives
![v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t](https://tex.z-dn.net/?f=v_y%20%3D%20v_%7By_0%7D%20%2B%20a_y%20t%5C%5Cv_y%20%3D%200%20-9.8t%5C%5Cv_y%20%3D%20-9.8t)
Combining both equation yields the y_component of the final velocity
![v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}](https://tex.z-dn.net/?f=v_y%20%3D%20-9.8%28%5Cfrac%7B63%5Ctimes%2010%5E%7B-3%7D%7D%7Bv_0%7D%29%20%3D%20-%5Cfrac%7B0.61%7D%7Bv_0%7D)
Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.
![\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%29%20%3D%20%5Cfrac%7Bv_y%7D%7Bv_x%7D%5C%5C%5Ctan%28177.2%5E%5Ccirc%29%20%3D%20-0.0489%20%3D%20%5Cfrac%7Bv_y%7D%7Bv_0%7D%20%3D%20%5Cfrac%7B-0.61%2Fv_0%7D%7Bv_0%7D%20%3D%20-%5Cfrac%7B0.61%7D%7Bv_0%5E2%7D%5C%5Cv_0%20%3D%203.53~%7B%5Crm%20m%2Fs%7D)
The value of parameter C for the function in the figure is 2.
<h3>What is amplitude of a wave?</h3>
The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.
f(x) = Acos(x - C)
where;
- A is amplitude of the wave
- C is phase difference of the wave
<h3>What is angular frequency of a wave?</h3>
Angular frequency is the angular displacement of any element of the wave per unit time.
From the blue colored graph; at y = 1, x = -2 cm
1 = cos(2 - C)
(2 - C) = cos^(1)
(2 - C) = 0
C = 2
Thus, the value of parameter C for the function in the figure is 2.
Learn more about phase angle here: brainly.com/question/16222725
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Solve for "x"
X=force
18/6=x/9
cross multiply
162=6x
x=27
Hope this helps