Answer:
Explanation:
According to a free body diagram the forces in the horizontal direction on body 1 would be:
F₁ = a₁*m₁ = -N
and on body 2:
F₂ = a₂*m₂ = N - F
N: normal force between the two blocks
F: frictional force on block 2
Since the two blocks are moving together, they need to have the same acceleration:
a₁ = a₂
This gives two equations with two unknown. Solving for a and N gives:
case A:
|a| = 0.96 m/s²
|N| = 2.9 N
case B:
|a| = 0.644 m/s²
|N| = 3.86 N
Answer:
The electric potential from a single charge is defined to be zero an infinite distance from the charge, and the electric potential associated with two charges is also defined to be zero when the charges are infinitely far apart.
Explanation:
Angular momentum is conserved, just before the clay hits and just after;
<span>mv(L/2) = Iw </span>
<span>I is the combined moment of inertia of the rod, (1/12)ML^2 , and the clay at the tip, m(L/2)^2 ; </span>
<span>I = [(1/12)ML^2 + m(L/2)^2] </span>
<span>Immediately after the collision the kinetic energy of rod + clay swings the rod up so the clay rises to a height "h" above its lowest point, giving it potential energy, mgh. From energy conservation in this phase of the problem; </span>
<span>(1/2)Iw^2 = mgh </span>
<span>Use the "w" found in the conservation of momentum above; and solve for "h" </span>
<span>h = mv^2L^2/8gI </span>
<span>Next, get the angle by noting it is related to "h" as; </span>
<span>h = (L/2) - (L/2)Cos() </span>
<span>So finally </span>
<span>Cos() = 1- 2h/L = 1 - mv^2L/4gI </span>
<span>m=mass of clay </span>
<span>M=mass of rod </span>
<span>L=length of rod </span>
<span>v=velocity of clay</span>
Answer:
the total force vector, on test charge is points from origin to point C( 1, 1 )
Explanation:
Given the data in the question, as illustrated in the image below;
from the Image, OA = 1, OB = AC = 1
so using Pythagoras theorem
a² = b² + c²
a = √( b² + c² )
so
OC = √( OB² + AC² )
we substitute
OC = √( OA² + AC² )
OC = √( 1² + 1² )
OC = √( 1 + 1 )
OC = √2
Coordinate of C( 1, 1 )
Hence, the total force vector, on test charge is points from origin to point C( 1, 1 )
