Time before projectile hits wall
= 88.2 m / 29.4 m/s = 3 seconds
Vertical velocity of projectile after three seconds
= 3*9.8 = 29.4 m/s
Horizontal velocity of projectile after three seconds, assuming no air resistance
= 29.4 m/s (given)
Conclusion:
velocity of projectile when it hits the wall
= < 29.4, -29.4> m/s
= sqrt(29.4^2+29.4^2) m/s east-bound at 45 degrees below horizontal
= 41.58 m/s east-bound at 45 degrees below horizontal.
Answer:
pu = 1260.9kg/m^3
the density of the unknown liquid is 1260.9kg/m^3
Explanation:
The density of a liquid is inversely proportional to the volume (height) of object submerged in it.
High density liquid possess higher buoyant force preventing objects from submerging.
p ∝ 1/V ∝ 1/h
since V = Ah
pu/pw = hw/hu
pu = pwhw/hu
Where;
p = density
h = height submerged
pu and pw is the density of unknown liquid and water respectively
hu and hw is the height of object submerged in unknown liquid and water respectively
pw = 1000kg/m^3
hu = 4.6cm = 0.046m
hw = 5.8cm = 0.058m
Substituting the given values;
pu = 1000×0.058/0.046
pu = 1260.9kg/m^3
the density of the unknown liquid is 1260.9kg/m^3
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
<h2>
Person must have 8.18 m/s to catch the ball</h2>
Explanation:
Consider the vertical motion of ball
We have equation of motion s = ut + 0.5at²
Initial velocity, u = 12 m/s
Acceleration, a = -9.81 m/s²
Displacement, s = -25 m
Substituting
-25 = 12 x t + 0.5 x -9.81 x t²
4.905 t² -12t - 25 = 0
t = 3.79 sec
Ball hits ground after 3.79 seconds.
So person need to cover 31 m in 3.79 seconds
Consider the horizontal motion of person
We have equation of motion s = ut + 0.5at²
Initial velocity, u = ?
Acceleration, a = 0 m/s²
Displacement, s = 31 m
Time, t = 3.79 seconds
Substituting
31 = u x 3.79 + 0.5 x 0 x 3.71²
u = 8.18 m/s
Person must have 8.18 m/s to catch the ball
The answer is 1.01 x 10^(-11) N. I arrived to this answer through calculating the GPEs of both balls. Bjorn's ball has a GPE of 1.402 x 10^(-11) N. Billie Jean's ball has a GPE of <span>2.503 x 10^(-11) N. I subtracted the two and I found that Billie Jean's tennis ball has a GPE of 1.01 x 10^(-11) more than Bjorn's tennis ball.</span>
