To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as
Where,
F = Force/Weight
A = Area
= Compression
= Original Length
According to the values given we have to
Replacing this values at our previous equation we have,
Therefore the Weight of the object is 3.82kN
Answer:
a.
Explanation:
a. Find the probability that exactly one photo detector of a pair is acceptable:
Let photo is accepted and the probability
is defected.
Therefore:
#The probability of exactly one photo detector of a pair is accepted is 7/25
b.Find the probability that both photo detectors in a pair are defective,P(D1D2):
Hence, from out tree diagram,the probability that both photo detectors in a pair are defective is 6/25
Answer:
Speed of ball A after collision is 3.7 m/s
Speed of ball B after collision is 2 m/s
Direction of ball A after collision is towards positive x axis
Total momentum after collision is m×4·21 kgm/s
Total kinetic energy after collision is m×8·85 J
Explanation:
<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>Let the mass of each ball be m kg
v be the velocity of ball A along positive x axis
v be the velocity of ball A along positive y axis
u be the velocity of ball B along positive y axis
Conservation of momentum along x axis
m×3·7 = m× v
∴ v = 3.7 m/s along positive x axis
Conservation of momentum along y axis
m×2 = m×u + m× v
2 = u + v → equation 1
∴ relative velocity of approach = relative velocity of separation
-2 = v - u → equation 2
By adding both equations 1 and 2 we get
v = 0
∴ u = 2 m/s along positive y axis
Kinetic energy before collision and after collision remains constant because it is an elastic collision
Kinetic energy = (m×2² + m×3·7²)÷2
= 8·85×m J
Total momentum = m×√(2² + 3·7²)
= m× 4·21 kgm/s
