Explanation:
Given that,
Angle by the normal to the slip α= 60°
Angle by the slip direction with the tensile axis β= 35°
Shear stress = 6.2 MPa
Applied stress = 12 MPa
We need to calculate the shear stress applied at the slip plane
Using formula of shear stress
Put the value into the formula
Since, the shear stress applied at the slip plane is less than the critical resolved shear stress
So, The crystal will not yield.
Now, We need to calculate the applied stress necessary for the crystal to yield
Using formula of stress
Put the value into the formula
Hence, This is the required solution.
Answer:
no ... hahahha! but I know every boys wait for the day when their heart beat is faster than normal ever in life
Answer
given,
discharge rate from pipe = 1000 gallons/minutes
now,
flow rate in cubic meters per second
1 gallon = 0.00378541 m³
1 min = 60 s
Q = 
Q = 0.063 m³/s
flow rate in liters per minute
1 gallon = 3.78541 L
Q = 
Q = 3785.41 m³/min
flow rate in cubic feet per second
1 gallon = 0.133681 ft³
1 min = 60 s
Q = 
Q = 2.23 ft³/s
Answer:
x(t) = - 6 cos 2t
Explanation:
Force of spring = - kx
k= spring constant
x= distance traveled by compressing
But force = mass × acceleration
==> Force = m × d²x/dt²
===> md²x/dt² = -kx
==> md²x/dt² + kx=0 ------------------------(1)
Now Again, by Hook's law
Force = -kx
==> 960=-k × 400
==> -k =960 /4 =240 N/m
ignoring -ve sign k= 240 N/m
Put given data in eq (1)
We get
60d²x/dt² + 240x=0
==> d²x/dt² + 4x=0
General solution for this differential eq is;
x(t) = A cos 2t + B sin 2t ------------------------(2)
Now initially
position of mass spring
at time = 0 sec
x (0) = 0 m
initial velocity v= = dx/dt= 6m/s
from (2) we have;
dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)
put t =0 and dx/dt = v(0) = -6 we get;
-2A sin 2(0)+2Bcos(0) =-6
==> 2B = -6
B= -3
Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get
x(t) = - 6 cos 2t
==>
Answer:
α = τ/I = 0.77 / (0.70(0.30²)) = 12.22222... = 12 rad/s²
Explanation:
