Answer:
a) 31557600 seconds b) 3.168x10⁻⁸ years
Explanation:
1 hour is equivalent to 3600 seconds, that multiplied by 24 hours (1 day) gives 86.400 seconds
<em>a) how many seconds there are in 1.00 year?</em>
⇒ 31557600 s
So in 1 year there are a total of 31557600 seconds.
<em>b) how many years are there in 1.00 second?</em>
Using the quantity from part a) it is get:
⇒ 3.168x10⁻⁸ years
So in 1 second there are a total of 3.168x10⁻⁸ years.
Answer:
35870474.30504 m
Explanation:
r = Distance from the surface
T = Time period = 24 h
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
m = Mass of the Earth = 5.98 × 10²⁴ kg
Radius of Earth =
The gravitational force will balance the centripetal force
From Kepler's law we have relation
Distance from the center of the Earth would be
Answer:
a) F = 64.30 N, b) θ = 121.4º
Explanation:
Forces are vector quantities so one of the best methods to add them is to decompose each force and add the components
let's use trigonometry
Force F1
sin 170 = F_{1y} / F₁
cos 170 = F₁ₓ / F₁
F_{1y} = F₁ sin 170
F₁ₓ = F₁ cos 170
F_{1y} = 100 sin 170 = 17.36 N
F₁ₓ = 100 cos 170 = -98.48 N
Force F2
sin 30 = F_{2y} / F₂
cos 30 = F₂ₓ / F₂
F_{2y} = F₂ sin 30
F₂ₓ = F₂ cos 30
F_{2y} = 75 sin 30 = 37.5 N
F₂ₓ = 75 cos 30 = 64.95 N
the resultant force is
X axis
Fₓ = F₁ₓ + F₂ₓ
Fₓ = -98.48 +64.95
Fₓ = -33.53 N
Y axis
F_y = F_{1y} + F_{2y}
F_y = 17.36 + 37.5
F_y = 54.86 N
a) the magnitude of the resultant vector
let's use Pythagoras' theorem
F = Ra Fx ^ 2 + Fy²
F = Ra 33.53² + 54.86²
F = 64.30 N
b) the direction of the resultant
let's use trigonometry
tan θ’= F_y / Fₓ
θ'=
θ'= tan⁻¹ (54.86 / (33.53)
θ’= 58.6º
this angle is in the second quadrant
The angle measured from the positive side of the x-axis is
θ = 180 -θ'
θ = 180- 58.6
θ = 121.4º
Answer:
the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg
Explanation:
To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables
Mathematically this can be determined as
Where
Temperature at inlet of turbine
Temperature at exit of turbine
Pressure at exit of turbine
Pressure at exit of turbine
The steady flow Energy equation for an open system is given as follows:
Where,
m = mass
m(i) = mass at inlet
m(o)= Mass at outlet
h(i)= Enthalpy at inlet
h(o)= Enthalpy at outlet
W = Work done
Q = Heat transferred
v(i) = Velocity at inlet
v(o)= Velocity at outlet
Z(i)= Height at inlet
Z(o)= Height at outlet
For the insulated system with neglecting kinetic and potential energy effects
Using the relation T-P we can find the final temperature:
From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK
the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg