A cannonball has just been shot out of a cannon aimed 45∘ above the horizontal rightward direction. Drag forces cannot be neglec
1 answer:
Answer:
The free-body diagram of the cannonball is found in the attachment below
<em>Note The question is incomplete. The complete question is as follows:</em>
<em>A cannonball has just been shot out of a cannon aimed 45∘ above the horizontal rightward direction. Drag forces cannot be neglected.</em>
<em>Draw the free-body diagram of the cannonball.</em>
Explanation:
Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation.
In order to construct free-body diagrams, it is important to know the various types of forces acting on the object in that situation. Then, the direction in which each of the forces is acting is determined. Finally the given object is drawn using any given representation, usually a box, and the direction of action of the forces are represented using arrows.
In the given situation of a cannonball which has just been shot out of a cannon aimed 45∘ above the horizontal rightward direction., the forces acting on it are:
F = force exerted by the cannon acting in the direction of angle of projection
Fdrag = drag force. The drag force acts in a direction opposite to the force exerted by the cannon
Fw = weight of the cannonball acting in a downward direction
The free body diagram is as shown in the attachment below.
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Answer:
a)KE=878.8 J
b)W=2636.4 J
Explanation:
Given that
mass ,m = 65 kg
Initial speed ,u = 5.2 m/s
a)
We know that kinetic energy KE is given as follows
m=mass
u=velocity
Now by putting the values in the above equation we get
KE=878.8 J
b)
We know that
Work done by all forces = Change in the kinetic energy
The final velocity , v= 2 u = 2 x 5.2 m/s
v= 10.4 m/s
Now by putting the values in the above equation we get
W=2636.4 J
a)KE=878.8 J
b)W=2636.4 J
Answer:
(a) A. Uniform line of charge and B. Uniformly charged sphere
(b) To three digits of precision:
λ = 1.50 * 10^-10 C/m
p = 2.81 * 10^-4 C/m^3
Explanation:
