Answer:
Magnification= -image distance/object distance
.253=image distance/33.5
image distance= 8.48 cm
Answer:
Answer:
Explanation:
Given that
K=8.98755×10^9Nm²/C²
Q=0.00011C
Radius of the sphere = 5.2m
g=9.8m/s²
1. The electric field inside a conductor is zero
εΦ=qenc
εEA=qenc
net charge qenc is the algebraic sum of all the enclosed positive and negative charges, and it can be positive, negative, or zero
This surface encloses no charge, and thus qenc=0. Gauss’ law.
Since it is inside the conductor
E=0N/C
2. Since the entire charge us inside the surface, then the electric field at a distance r (5.2m) away form the surface is given as
F=kq1/r²
F=kQ/r²
F=8.98755E9×0.00011/5.2²
F=36561.78N/C
The electric field at the surface of the conductor is 36561N/C
Since the charge is positive the it is outward field
3. Given that a test charge is at 12.6m away,
Then Electric field is given as,
E=kQ/r²
E=8.98755E9 ×0.00011/12.6²
E=6227.34N/C
Answers is F=7.84 N
Friction force resists the effect of horizontal force and trying to approch to a limiting force.
we have formula for limiting friction force between block and floor
F=Ц N
where N=mg
putting values we get answer.
<span>An object roating at one revokution per second has an angular velocity of 360 degrees per second or 2pi radians per second. This is found by taking the number of revolutions over a period of time and than dividing by the chosen period of time to get the velocity. There are 360 degrees or 2pi radians in one revolution.</span>
