Answer:
0.6
Explanation:
The volume of a sphere = 
Therefore 
r of the disc = 
Using conservation of angular momentum;
The 
of the sphere = 
of the disc = 
= 0.6
The Young modulus is given by:
where
F is the force applied
is the initial length of the wire
is the cross-sectional area of the wire
is the stretch of the wire
The wire in the problem stretches by
of its length, this means
We can also calculate the area of the wire; its radius is in fact half the diameter:
and so the area is
We know the force applied to the wire, F=20 N, so now we have everything to calculate the Young modulus:
Hot combustion gases are accelerated in a 92% efficient
adiabatic nozzle from low velocity to a specified velocity. The exit velocity
and the exit temp are to be determined.
Given:
T1 = 1020 K à
h1 = 1068.89 kJ/kg, Pr1 = 123.4
P1 = 260 kPa
T1 = 747 degrees Celsius
V1 = 80 m/s ->nN = 92% -> P2
= 85 kPa
Solution:
From the isentropic relation,
Pr2<span> = (P2 / P1)PR1 = (85
kPa / 260 kPa) (123.4) = 40.34 = h2s = 783.92 kJ/kg</span>
There is only one inlet and one exit, and thus, m1 =
m2 = m3. We take the nozzle as the system, which is a
control volume since mass crosses the boundary.
h2a = 1068.89 kJ/kg – (((728.2 m/s)2 –
(80 m/s)2) / 2) (1 kJ/kg / 1000 m2/s2) =
806.95 kJ/kg\
From the air table, we read T2a = 786.3 K
Answer:
they meet from point o at distance 50.46 m and time taken is 11.6 seconds
Explanation:
given data
acceleration = 0.75 m/s²
speed B = 6 m/s
time B = 20 s
to find out
when and where the vehicles passed each other
solution
we consider here distance = x , when they meet after o point
and time = t for meet point z
we find first Bus B distance for 20 s ec
distance B = velocity × time
distance B = 6 × 20
distance B = 120 m
so
B take time to meet is calculate by distance formula
distance = velocity × time
120 - x = 6 × t
x = 120 - 6t .................1
and
distance of A when they meet by distance formula
distance = ut + 1/2 × at²
here u is initial speed = 0 and t is time
x = 0 + 1/2 × 0.75 × t²
x = 0.375 × t² .............2
so from equation 1 and 2
0.375 × t² = 120 - 6t
t = 11.6
so time is 11.6 second
and
distance from point o from equation 2
x = 0.375 (11.6)²
x = 50.46
so distance from point o is 50.46 m
Answer:
1 greater distances fallen in successive seconds
Explanation:
When a body falls freely it is subjected to the action of the force of gravity, which gives an acceleration of 9.8 m / s2, consequently, we are in an accelerated movement
If we use the kinematic formula we can find the position of the body
Y = Vo t + ½ to t2
Where the initial velocity is zero or constant and the acceleration is the acceleration of gravity
Y = - ½ g t2 = - ½ 9.8 t2 = -4.9 t2
Let's look for the position for successive times
t (s) Y (m)
1 -4.9
2 -19.6
3 -43.2
The sign indicates that the positive sense is up
It can be clearly seen that the distance is greatly increased every second that passes
