My guesses would be g. or F
Answer:
c
Explanation:
It's c the last one u see
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
Answer:
250,000
Explanation:
<h2> </h2>
<h2>formula = ( F=ma </h2>
- F=1500N
- a=6m/s^2
- F= ma
- m=?
- 1500/6 = m
- m=250 kg
- 1kg =1000gm so 250kg =250,000gm
- m =250×10^3 gm
Answer:
A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if
the dispersion is great
