Answer:
direction is given as
South of West
Explanation:
Part b)
displacement is given as
now we will have
total displacement is given as
direction is given as
South of West
Whenever an increasing temperature leads to a quicker movement of elementary particles and a collision, Thermal energy occurs. The hypothesis as per the given statement is provided below.
- Whenever the overall volume rises, subsequently heat is increased since a particular chemical solar radiation relies on one of its temperatures as well as critical or actual mass.
- Whenever subjected to about a similar quantity of infrared radiation, distinct masses completely modify the temperature at varying intervals. The reason would be that the mass may consume heat energy by things.
Learn more:
brainly.com/question/12706330
Because the average of one measurement would yield to itself, which provides nothing new to us. We need at least two for average to be purposeful.
Answer:
(a) 
(b) 
Explanation:
It is given that,
Force acting on the particle, F = 12 N
Displacement of the particle, 
Magnitude of displacement, 
(a) If the change in the kinetic energy of the particle is +30 J. The work done by the particle is given by :
is the angle between force and the displacement
According to work energy theorem, the charge in kinetic energy of the particle is equal to the work done.
So,
(b) If the change in the kinetic energy of the particle is (-30) J. The work done by the particle is given by :
Hence, this is the required solution.
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
