Answer:
xcritical = d− m1
/m2
( L
/2−d)
Explanation: the precursor to this question will had been this
the precursor to the question can be found online.
ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)
. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces
smallest possible value of x such that the bar remains stable (call it xcritical)
∑τA = 0 = m2g(d− xcritical)− m1g( −d)
xcritical = d− m1
/m2
( L
/2−d)
The best and most correct answer among the choices provided by the question is the first choice, larger.
Rankine is Fahrenheit + 460 , while Kelvin is Celsius + 273. We all know that Fahrenheit has larger number compared to kelvin , thus rankine is much larger.
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.
Answer:
Explanation:
Given
mass flow rate=0.3 kg/s
diameter of pipe=5 cm
length of pipe=10 m
Inside temperature=22
Pipe surface =100
Temperature drop=30
specific heat of vapor(c)=2190 J/kg.k
heat supplied 
Heat due to convection =hA(100-30)




19,710=122.538 h

Answer:
F = 316.22 N
Explanation:
Given that,
The wind blows a jay bird south with a force of 300 Newtons.
The jay bird flies north, against the wind, with a force of 100 newtons.
Both the forces are acting perpendicular to each other. The net force is given by the resultant of forces as follows :

Hence, the net force on the jay bird is 316.22 N.
Answer:
Given that
T= 0.43 s
Radius of the ball path's , r=2.1 m
a)
We know that
f= 1/T
Here f= frequency
T= Time period
Now by putting the values
f= 1/T
T= 0.43 s
f= 1/0.43
f=2.32 Hz
b)
We know that
V= ω r
ω = 2 π f
ω=Angular speed
V= Linear speed
ω = 2 π f=ω = 2 x π x 2.32 =14.60 rad/s
V= ω r= 14.60 x 2.1 = 30.66 m/s
c)
Acceleration ,a
a =ω ² r
a= 14.6 ² x 2.1 = 447.63 m/s²
We know that g = 10 m/s²
So
a= a/g= 447.63/10 = 44.7 g m/s²
a= 44.7 g m/s²