Answer:
a) frequency = 0.1724 Hz
b) Period = 5.8 sec
c) speed = 7.04 m/s
d) acceleration = 7.62 m/s²
Explanation:
Given that;
radius = 6.5m
time period = 5.8 sec every circle
a) the frequency
frequency is the number of rotation in unit time
frequency = 1 / time period = 1/5.8
frequency = 0.1724 Hz
b) the period
period is time taken in one rotation
period = total time / rotation = 5.8 / 1
Period = 5.8 sec
c) the speed
speed = distance/time = circumference/time period = 2πr / t = (2π×6.5) / 5.8
speed = 7.04 m/s
d) acceleration
To find the acceleration we take the linear velocity squared divided by the radius of the circle.
so
acceleration = v² / r = (7.04)² / 6.5 = 49.5616 / 6.5
acceleration = 7.62 m/s²
Answer:
a) vB = 10.77 ft/s
b) W = 11.30 lb*ft
Explanation:
a) W = 8 lb ⇒ m = W/g = 8 lb/32.2 ft/s² = 0.2484 slug
vA <em>lin</em> = 5 ft/s
rA = 2 ft
v <em>rad</em> = 4 ft/s
vB = ?
rB = 1 ft
W = ?
We can apply The law of conservation of angular momentum
L<em>in</em> = L<em>fin</em>
m*vA*rA = m*vB*rB ⇒ vB = vA*rA / rB
⇒ vB = (5 ft/s)*(2 ft) / (1 ft) = 10 ft/s (tangential speed)
then we get
vB = √(vB tang² + vB rad²) ⇒ vB = √((10 ft/s)² + (4 ft/s)²)
⇒ vB = 10.77 ft/s
b) W = ΔK = K<em>B</em> - K<em>A</em> = 0.5*m*vB² - 0.5*m*vA²
⇒ W = 0.5*m*(vB² - vA²) = 0.5*0.2484 slug*((10.77 ft/s)²-(5 ft/s)²)
⇒ W = 11.30 lb*ft
If the type of friction is sliding friction, it may be increased by increasing the roughness of the surface and may be decreased by smoothing the surface or applying a lubricant.
W=F*S
W - Work
F - Force
S - Distance (from latin word 'spatium)
so...
S= 5 (m)
W=60 (J)
W=F*S
F=W/S
F=60/5=12 J/m = 12 N (Newtons)
Answer:
(a) 83475 MW
(b) 85.8 %
Explanation:
Output power = 716 MW = 716 x 10^6 W
Amount of water flows, V = 1.35 x 10^8 L = 1.35 x 10^8 x 10^-3 m^3
mass of water, m = Volume x density = 1.35 x 10^8 x 10^-3 x 1000
= 1.35 x 10^8 kg
Time, t = 1 hr = 3600 second
T1 = 25.4° C, T2 = 30.7° C
Specific heat of water, c = 4200 J/kg°C
(a) Total energy, Q = m x c x ΔT
Q = 1.35 x 10^8 x 4200 x (30.7 - 25.4) = 3 x 10^12 J
Power = Energy / time
Power input = 
Power input = 83475 MW
(b) The efficiency of the plant is defined as the ratio of output power to the input power.
Thus, the efficiency is 85.8 %.
