Answer:
raise the board to a higher angle
Explanation:
Static friction is the force opposite to the applied force.
Static friction is dependent on the angle of inclination, it means as the angle of incline increases, the force of friction will increases as normal force will decrease.
So, if the board will be raised to a higher angle, it will increase the angle of incline and will overcome the static friction and block will be able slide.
Hence, the correct option is "raise the board to a higher angle".
It is 1,000kg/m³ i hope this helps you with something
[H3O+] is just the same with [H+]. There are quite a few
relationships between [H+] and [OH−]
ions. And because there is a large range of number between 10 to 10-15
M, the pH is used. pH = -log[H+] and pOH = -log[OH−]. In aqueous solutions, [H+
][OH- ] = 10-14.
Answer:
Explanation:
See the attached figure . See the forces acting on man pulling up the box .
Man is stationary so net force acting on man is zero .
T + R = Wman
R is the reaction force of the ground of second floor .
R = Wman - T
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
