Answer:
<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong):</em>
a) a×b = 34.27k
b) a·b = 128.43
c) (a + b)·b = 305.17
d) The component of a along the direction of b = 9.66
Explanation:
<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong)</em> we can proceed as follows:
a) The vectorial product, a×b is:
![a \times b = (8.6*9.5 - 5.1*9.3)k = 34.27k](https://tex.z-dn.net/?f=%20a%20%5Ctimes%20b%20%3D%20%288.6%2A9.5%20-%205.1%2A9.3%29k%20%3D%2034.27k%20)
b) The escalar product a·b is:
![a\cdot b = (8.6*9.3) + (5.1*9.5) = 128.43](https://tex.z-dn.net/?f=%20a%5Ccdot%20b%20%3D%20%288.6%2A9.3%29%20%2B%20%285.1%2A9.5%29%20%3D%20128.43%20)
c) <u>Asumming (a</u><u> </u><u>+ b)·b</u> <em>instead a+b·b</em> we have:
![(a + b)\cdot b = [(8.6 + 9.3)i + (5.1 + 9.5)j]\cdot (9.3i + 9.5j) = (17.9i + 14.6j)\cdot (9.3i + 9.5j) = 305.17](https://tex.z-dn.net/?f=%28a%20%2B%20b%29%5Ccdot%20b%20%3D%20%5B%288.6%20%2B%209.3%29i%20%2B%20%285.1%20%2B%209.5%29j%5D%5Ccdot%20%289.3i%20%2B%209.5j%29%20%3D%20%2817.9i%20%2B%2014.6j%29%5Ccdot%20%289.3i%20%2B%209.5j%29%20%3D%20305.17)
d) The component of a along the direction of b is:
![a*cos(\theta) = \frac{a\cdot b}{|b|} = \frac{128.43}{\sqrt{9.3^{2} + 9.5^{2}}} = 9.66](https://tex.z-dn.net/?f=%20a%2Acos%28%5Ctheta%29%20%3D%20%5Cfrac%7Ba%5Ccdot%20b%7D%7B%7Cb%7C%7D%20%3D%20%5Cfrac%7B128.43%7D%7B%5Csqrt%7B9.3%5E%7B2%7D%20%2B%209.5%5E%7B2%7D%7D%7D%20%3D%209.66%20)
I hope it helps you!
They are right the answer is A true
Answer:
2271.16N/C upward
Explanation:
The diagram well illustrate all the forces acting on the mass. The weight is acting downward and the force is acting upward in other to balance the weight.since the question says it is motionless, then indeed the forces are balanced.
First we determine the downward weight using
![W=mg\\g=9.81m/s^{2}](https://tex.z-dn.net/?f=W%3Dmg%5C%5Cg%3D9.81m%2Fs%5E%7B2%7D)
Hence for a mass of 3.82g 0r 0.00382kg we have the weight to be
![W=0.00382kg*9.81m/s^{2}](https://tex.z-dn.net/?f=W%3D0.00382kg%2A9.81m%2Fs%5E%7B2%7D)
![W=0.0375N](https://tex.z-dn.net/?f=W%3D0.0375N)
To calculate the electric field,
![E=f/q\\E=0.0375/16.5*10^{-6} \\E=2271.16N/C](https://tex.z-dn.net/?f=E%3Df%2Fq%5C%5CE%3D0.0375%2F16.5%2A10%5E%7B-6%7D%20%5C%5CE%3D2271.16N%2FC)
Since the charge on the mass is negative, in order to generate upward force, there must be a like charge below it that is repelling it, Hebce we can conclude that the electric field lines are upward.
Hence the magnitude of the electric force is 2271.16N/C and the direction is upward
Newton's 2nd law:
Fnet = ma
Fnet is the net force acting on an object, m is the object's mass, and a is the acceleration.
The electric force on a charged object is given by
Fe = Eq
Fe is the electric force, E is the electric field at the point where the object is, and q is the object's charge.
We can assume, if the only force acting on the proton and electron is the electric force due to the electric field, that for both particles, Fnet = Fe
Fe = Eq
Eq = ma
a = Eq/m
We will also assume that the electric field acting on the proton and electron are the same. The proton and electron also have the same magnitude of charge (1.6×10⁻¹⁹C). What makes the difference in their acceleration is their masses. A quick Google search will provide the following values:
mass of proton = 1.67×10⁻²⁷kg
mass of electron = 9.11×10⁻³¹kg
The acceleration of an object is inversely proportional to its mass, so the electron will experience a greater acceleration than the proton.