Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
Answer:
Explanation:
Given:
Steam Mass rate, ms = 1.5 kg/min
= 1.5 kg/min × 1 min/60 sec
= 0.025 kg/s
Air Mass rate, ma = 100 kg/min
= 100 kg/min × 1 min/60 sec
= 1.67 kg/s
A.
Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.
xf, quality = 0.9.
Tsat = 89.9°C
hf = 376.57 kJ/kg
hfg = 2283.38 kJ/kg
Using the equation for specific enthalpy,
hi = hf + (hfg × xf)
= 376.57 + (2283.38 × 0.9)
= 2431.552 kJ/kg
The specific enthalpy of the outlet, h2 = hf
= 376.57 kJ/kg
B.
Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy
= ms × (hi - h2)
= 0.025 × (2431.552 - 376.57)
= 0.025 × 2055.042
= 51.37455 kW
= 51.38 kW.
<span>The viewer travels through the electron cloud, then the nucleus, and then the electron cloud again.</span>
Answer:
(a)F= 3.83 * 10^3 N
(b)Altitude=8.20 * 10^5 m
Explanation:
On the launchpad weight = gravitational force between earth and satellite.
W = GMm/R²
where R is the earth radius.
Re-arranging:
WR² / GM = m
m = 4900 * (6.3 * 10^6)² / (6.67 * 10^-11 * 5.97 * 10^24) = 488 kg
The centripetal force (Fc) needed to keep the satellite moving in a circular orbit of radius (r) is:
Fc = mω²r
where ω is the angular velocity in radians/second. The satellite completes 1 revolution, which is 2π radians, in 1.667 hours.
ω = 2π / (1.667 * 60 * 60) = 1.05 * 10^-3 rad/s
When the satellite is in orbit at a distance (r) from the CENTRE of the earth, Fc is provided by the gravitational force between the earth and the satellite:
Fc = GMm/r²
mω²r = GMm / r²
ω²r = GM / r²
r³ = GM/ω² = (6.67 * 10^-11 * 5.97 * 10^24) / (1.05 * 10^-3)²
r³ = 3.612 * 10^20
r = 7.12 * 10^6 m
(a)
F = GMm/r²
F=(6.67 * 10^-11 * 5.97 * 10^24 * 488) / (7.12 * 10^6 )²
F= 3.83 * 10^3 N
(b) Altitude = r - R = (7.12 * 10^6) - (6.3 * 10^6) = 8.20 * 10^5 m
