Why is the recruitment and selection of staff the concern of all managers in an organization?
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The term Freebody diagram refer to the diagram that could be drawn to show the forces that act on an object.
<h3>What is a freebody diagram?</h3>The term Freebody diagram refer to the diagram that could be drawn to show the forces that act on an object. We must note that the forces that act on a body determine the direction in which the body moves.
This is because, force is a vector quantity. As such the magnitude and direction of a force are both responsible when we are trying to determine the force that acts on the object.
Attached is the image of the upward motion of a vector that shows the forces that act on the vector. This is what we refer to as a Freebody diagram.
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Answer
given,
length of rod = 21.5 cm = 0.215 m
mass of rod (m) = 1.2 Kg
radius, r = 1.50
mass of ball, M = 2 Kg
radius of ball, r = 6.90/2 = 3.45 cm = 0.0345 m
considering the rod is thin
I = 0.144 kg.m²
rotational kinetic energy of the rod is equal to
KE = 6.15 J
b) using conservation of energy
ω = 9.25 rad/s
c) linear speed of the ball
v = r ω
v = (L+R )ω
v = (0.215+0.0345) x 9.25
v =2.31 m/s
d) using equation of motion
v² = u² + 2 g h
v² = 0 + 2 x 9.8 x 0.248
v = √4.86
v =2.20 m/s
speed attained by the swing is more than free fall
% greater =
= 5 %
speed of swing is 5 % more than free fall
Answer:
C) 7.35*10⁶ N/C radially outward
Explanation:
- If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
- So, we can write the following equation:
- As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:
- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:
- As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.