The kinetic energy of an object is given by:
KE = 0.5mv²
KE = kinetic energy, m = mass, v = speed
Given values:
KE = 0.161J, v = 2.33m/s
Plug in and solve for m:
0.161 = 0.5m(2.33)²
m = 0.059kg
Answer:
θ = 66º
Explanation:
This exercise of Newton's second law must be solved in part, let's start by finding the slowing down acceleration of the ball
a = v² / r
the radius of the circle is
sin θ = r / L
r = L sin θ
we substitute
a = v² /L sin θ
now let's write Newton's second law
vertical axis
T_y -W = 0
T_y = W
radial axis
Tₓ = m a (1)
let's use trigonometry for the components of the string tension
cos θ = T_y / T
sin θ = Tₓ / T
Tₓ = T sin θ
we substitute in 1
T sin θ =
T L sin² θ = m v²
we write our system of equations
T cos θ = m g
T L sin ² tea = m v²
we divide the two equations
L = v² / g
(1 -cos²)/ cos θ =
1 - cos² θ = cos θ
cos² θ + 0.97044 cos θ -1 = 0
we change variable cos θ = x
x² + 0.97044 x - 1 =0
x=
since the square root is imaginary there is no real solution to the problem, suppose that the radius is 1 m r = 1 m
T sin θ =
T cos θ = m g
resolved
tan θ =
θ = tan⁻¹ ( 4.75²/ 1 9.81)
θ = 66º
Given Information:
Inlet Temperature of hot air= Th₁ = 450K
Exit Temperature of hot air = Th₂ = 350K
Inlet Temperature of cold air = Tc₁ = 300K
Volume flow rate of hot air = vh = 0.02 m³/s
Volume flow rate of cold air = vc = 1 m³/s
Required Information:
Exit Temperature of cold air= Tc₂ = ?
Answer:
Exit Temperature of cold air = Tc₂ = 302 °C
Explanation:
In a heat exchanger, the cold air absorbs heat that is lost by the hot air,
Heat absorbed by cold air = Heat lost by hot air
Therefore, the exit temperature of the cold air is 302 °C or 575K
Note:
m = ρv
Where ρ is density of air and v is the volume flow rate and m is the mass flow rate.
cp is the specific heat capacity of air.