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
Hence for a mass of 3.82g 0r 0.00382kg we have the weight to be
To calculate the electric field,
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
Answer:
A major problem in all survey research is that respondents are almost always self-selected. Not everyone who receives a survey is likely to answer it, no matter how many times they are reminded or what incentives are offered.
Explanation:
<em><u>DISADVANTAGES</u></em>
Respondents may not feel encouraged to provide accurate, honest answers.
Respondents may not feel comfortable providing answers that present themselves in a unfavorable manner.
Respondents may not be fully aware of their reasons for any given answer because of lack of memory on the subject, or even boredom.
HAVE A GOOD DAY!
Answer:
Explanation:
A proton of charge
q=+1.609×10^-19C
Orbit a radius of 12cm
r=0.12m
Magnetic Field of 0.31T
Angle between velocity and field is 90°
a. Because the magnetic force F supplies the centripetal force Fc.
The magnitude of the magnetic force F on a charge q moving at a speed v in a magnetic field of strength B is given by
F = qvB sin θ
And the centripetal force is given as
Fc=mv²/r
Where m is mass of proton
m=1.673×10^-27kg
Then, F=Fc
qvB sin θ=mv²/r
qBSin90=mv/r
rqB=mv
Then, v=rqB/m
v=0.12×1.609×10^-19×0.31/1.673×10^-23
v=3577692.78m/s
v=3.58×10^6m/s
b. Since,
F=qVBSin90
F=1.609×10^-19×3.58×10^6×0.31
F=1.785×10^-13 N.
Answer:
a) a = 4.9 m / s², N = 16.97 N and b) F = 9.8 N
Explanation:
a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry
sin 30 = Wx / W
cos 30 = Wy / W
Wx = W sin30
Wy = W cos 30
Let's write the equations on each axis
X axis
Wx = ma
Y Axis
N- Wy = 0
N = Wy = mg cos 30
N = 2.0 9.8 cos 30
N = 16.97 N
We calculate the acceleration
a = Wx / m
a = mg sin 30 / m
a = g sin 30
a =9.8 sin 30
a = 4.9 m / s²
b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component
F -Wx = 0
F = Wx
F = m g sin 30
F = 2.0 9.8 sin 30
F = 9.8 N
Answer:
for 4.567 I can't tell if it x 
or x 
so:
answer using x = 0.0000644697N
answer using x = 0.00644697
Explanation:
use equation F = GMm/
