The electric force between the two particles are calculated through the equation,
F = kQ₁Q₂ / d²
where F is the force, k is a constant called Coulomb's law constant, Q₁ and Q₂ are the charges, and d is the distance. This equation is called the Coulomb's law.
It can be seen from the equation above that the electric forces between the objects are majorly affected by the substance's charges and distance.
The answer to this item is therefore letter A.
Answer:
No. 67
Peter Street
12th Road
Chennai
24th June 201_
Dear Amrish
I have come to know that since your school has closed for the Autumn Break you have plenty of free time at your disposal at the moment. I would like to tell you that even I am having holidays now.
It has been a long time since we have spent some time together. If you are free, I would welcome to have your company this weekend. Why don’t you come over to my house and spend a day or so with me?
I am anxiously waiting for your reply.
Yours affectionately
your name
Answer:
A) ω = 6v/19L
B) K2/K1 = 3/19
Explanation:
Mr = Mass of rod
Mb = Mass of bullet = Mr/4
Ir = (1/3)(Mr)L²
Ib = MbRb²
Radius of rotation of bullet Rb = L/2
A) From conservation of angular momentum,
L1 = L2
(Mb)v(L/2) = (Ir+ Ib)ω2
Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.
(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2
(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2
Divide each term by Mr;
vL/8 = (L²/3 + L²/16)ω2
vL/8 = (19L²/48)ω2
Divide both sides by L to obtain;
v/8 = (19L/48)ω2
Thus;
ω2 = 48v/(19x8L) = 6v/19L
B) K1 = K1b + K1r
K1 = (1/2)(Mb)v² + Ir(w1²)
= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)
= (1/8)(Mr)v²
K2 = (1/2)(Isys)(ω2²)
I(sys) is (Ir+ Ib). This gives us;
Isys = (19L²Mr/48)
K2 =(1/2)(19L²Mr/48)(6v/19L)²
= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152
Thus, the ratio, K2/K1 =
[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19
Answer:
<u>Inelastic collision:</u>
A collision in which there is a loss of Kinetic Energy due to internal friction of the bodies colliding.
<u>Characteristics of an inelastic collision:</u>
- <em>the momentum of the system is conserved</em>
- <em>the momentum of the system is conservedloss of kinetic energy</em><u> </u>
<em>I</em><em>n</em><em> </em><em>a perfectly elastic collision</em><em>, the two bodies </em><em>that</em><em> </em><em>collide with each other stick together.</em>
<u>Elastic </u><u>collision</u><u>:</u>
A collision in which the kinetic energy of the two bodies, before and after the collision, remains the same.
<u>Characteristic</u><u>s</u><u> </u><u>of</u><u> </u><u>elastic</u><u> </u><u>collision</u><u>:</u>
- <em>the</em><em> </em><em>momentum</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>system</em><em> </em><em>is</em><em> </em><em>conserved</em>
- <em>no</em><em> </em><em>loss</em><em> </em><em>o</em><em>f</em><em> </em><em>kinetic</em><em> </em><em>energy</em>
In everyday life, no collision is perfectly elastic.
__________________
ANSWER:
<u>Given examples:</u>
- Two cars colliding with each other form an example of inelastic collision.
<u>Reason:</u>
<em>(</em><em>T</em><em>hey</em><em> </em><em>lose</em><em> </em><em>kinetic</em><em> </em><em>energy</em><em> </em><em>and</em><em> </em><em>come</em><em> </em><em>to</em><em> </em><em>a</em><em> </em><em>stop</em><em> </em><em>after</em><em> </em><em>the</em><em> </em><em>collision</em><em>.</em><em>)</em>
- A ball bouncing after colliding with a surface is an example of elastic collision
<u>Reason:</u>
<em>(a very less amount of kinetic energy is lost)</em>
