Answer:
y = - (½ g / v₀²) x²
Explanation:
This is a projectile launch exercise where there is no acceleration on the x-axis so
x = v₀ₓ t
v₀ₓ = v₀ cos tea
y = 
t - ½ g t2
v_{oy} = v₀ sin θ
as the sphere is thrown horizontally, the angle is tea = 0º, so the initial velocity remains
v₀ₓ = v₀
v_{oy} = 0
we substitute in our equations
x = v₀ t
y = - ½ g t²
we eliminate the time from these equations, we substitute the first in the second
y = - ½ g (x / v₀)²
y = - (½ g / v₀²) x²
this is the equation of a parabola
Here in this question as we can see there is no air friction so we can use the principle of energy conservation
now here we know that
now plug in all values in above equation
divide whole equation by mass "m"
so height of the ball from ground will be 1.35 m
Answer:
M2 = 0.06404
P2 = 2.273
T2 = 5806.45°R
Explanation:
Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.
Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,
To1 = (1.008)*(1000) = 1008 ºR
R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)
F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga
For the air q = cp(To2– To1)
(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2
Table A.3 of steam table gives P/P* = 2.273,
T/T* = 0.2066,
To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =
F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit
750W 30mins...750x30x60 joules ... 75000x18 ... about 1,500,000j
D
Electrical generator which operates using a magnetic field. It is the beginning of modern dynamos.
