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:
Breh seriously. Ugh fine.
1.B
2.D
3.C
4.C
5.D,A and B
6.A,C and D
Answer:
18.2145 meters
Explanation:
Using the conservation of momentum, we have that:
m1 = m1' is the mass of the astronaut, m2=m2' is the mass of the satellite, v1 and v2 are the inicial speed of the astronaut and the satellite (v1 = v2 = 0), and v1' and v2' are the final speed of the astronaut and the satellite. Then we have that:
The negative sign of this speed just indicates the direction the astronaut goes, which is the opposite direction of the satellite.
If the astronaut takes 7.5 seconds to come into contact with the shuttle, their initial distance is:
Step 2: Use the slope to find<span> the y-intercept. </span>Line<span> is </span>parallel<span> so use m = 2/5. </span>6<span>. </span>Find<span>the </span>equation<span> of a </span>line passing through the point<span> (8, –</span>9<span>) perpendicular to the </span>line<span> 3x + 8y = 4.</span>
