Answer:
for this problem, 2.5 = (5+2/2)-(5-2/2)erf (50×10-6m/2Dt)
It now becomes necessary to compute the diffusion coefficient at 750°C (1023 K) given that D0= 8.5 ×10-5m2/s and Qd= 202,100 J/mol.
we have D= D0exp( -Qd/RT)
=(8.5×105m2/s)exp(-202,100/8.31×1023)
= 4.03 ×10-15m2/s
Answer:
The magnitude of the velocity of the aircraft P relative to aircraft Q is zero
Explanation:
The velocity of the two aircraft, P & Q, v = 300 m/s
The angle of the direction between them, Ф = 90°
The magnitude of the velocity of aircraft P relative to aircraft Q is given by the formula
<em> V = v cos Ф
</em>
Substituting the values in the above equation
v = 300 x cos 90°
= 300 x 0
= 0
Since the aircraft are at right angles, the velocity of one aircraft relative to the other is zero.
Answer: 11 m/s
vinitial=2 m/s
time=3 s
acceleration = 3 m/s^2
vfinal = ?
The key here is that it is a constant acceleration, so we can use the constant acceleration equations. The easiest one to use would be:
vfinal=vinitial + a*t
We need vfinal, so algebraically we are ready to put in numbers into the equation:
vfinal=vinitial + a*t = 2 m/s + (3 m/s^2)*(3 s ) = 11 m/s is the final velocity
Answer:
The value is 
Explanation:
From the question we are told that
The mass of the block is 
The force constant of the spring is 
The amplitude is 
The time consider is 
Generally the angular velocity of this block is mathematically represented as
=> 
=> 
Given that the block undergoes simple harmonic motion the velocity is mathematically represented as
=> 
=>
