A 10-kg block A is released from rest 2 m above the 5-kg plate P, which can slide freely along the smooth vertical guides BC and
DE. Determine the velocity of the block and plate just after impact. The coefficient of restitution between the block and the plate is e = 0.75. Also, find the maximum compression of the spring due to impact.The spring has an unstretched length of 600 mm
The spring with the 5 kg plate are stationary before impact at 450mm. The 600mm unstretch length of the spring is without the 5 kg plate
The spring with the 5 kg plate are stationary before impact at 450mm. The 600mm unstretch length of the spring is without the 5 kg plate
1 answer:
Answer:
Check the explanation
Explanation:
<u>Given data </u>
Mass of the block A is Ma = 10kg
Mass of the plate is Mp = 5kg
Coefficient of restitution between the block and that of the plate is e=0.75
Unstretched length of the spring is 0.6m
From the conservation of energy, we have four motion of block A from (1) to (2)
Then accordingly, we will have the step by step calculation in the attached image below .
You might be interested in
Answer:
maximum allowable electrical power=4.51W/m
critical radius of the insulation=13mm
Explanation:
Hello!
To solve this heat transfer problem we must initially draw the wire and interpret the whole problem (see attached image)
Subsequently, consider the heat transfer equation from the internal part of the tube to the external air, taking into account the resistance by convection, and conduction as shown in the attached image
to find the critical insulation radius we must divide the conductivity of the material by the external convective coefficient
Answer:
(a) The mean time to fail is 9491.22 hours
The standard deviation time to fail is 9491.22 hours
(b) 0.5905
(c) 3.915 × 10⁻¹²
(d) 2.63 × 10⁻⁵
Explanation:
(a) We put time to fail = t
∴ For an exponential distribution, we have f(t) =
Where we have a failure rate = 10% for 1000 hours, we have(based on online resource);
e^(1000·λ) - 0.1·e^(1000·λ) = 1
0.9·e^(1000·λ) = 1
1000·λ = ㏑(1/0.9)
λ = 1.054 × 10⁻⁴
Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours
The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours
b) Here we have to integrate from 5000 to ∞ as follows;
(c) The Poisson distribution is presented as follows;
p(x = 3) = 3.915 × 10⁻¹²
d) Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours
The Cumulative Distribution Function is given as follows;
p( t ≤ 1/4) .
Answer:
a) 0.4393 mm
b) -0.141 mm
Explanation:
Cylindrical bar : Diameter = 20.3 mm , Length = 205 mm
Force that deforms bar of metal = 46400 N
elastic modulus = 66.9 GPa
Poisson's ratio ( u ) = 0.32
A) Determine the amount by which this specimen will elongate in the direction of applied stress
dl =
where P = 46400 N
L = 205 mm
A = = 323.65
E = 66.9 GPa
dl = ( 46400 * 205 ) / ( 323.65 * 66.9 * 10^3 ) = 0.4393 mm
B) determine the change in diameter of the specimen
change in diameter( compressed due to elongation in length )
= - u * dl
= - 0.32 * 0.4393
= -0.141 mm