Answer:
a) 3.5 m
b) 14 secs
c) 1.4 secs
Explanation:
<u>a) Determine the distance the particle will travel</u>
given velocity ( final velocity ) = 5 m/s
v^2 = u^2 + 2as
s = ( v^2 - u^2 ) / 2a
= ( 5^2 - 8^2 ) / 2 ( -0.5 * 5^3/2 )
= 3.5 m
<u>b) Determine the time when v = 1m/s</u>
V = u + at
1 = 8 + ( -0.5 * 1^3/2 ) * t
∴ t = 14 secs
c) Determine the time required for particle to travel 8 m
<em>we will employ both equations above </em>
V^2 = u^2 + 2as
s = 8 m , V = unknown , u = 8 m/s back to equation
V^2 = 8^2 + 2 ( - 1/2 * V^3/2 ) * 8
∴ V^2 + 8V^3/2 - 64 = 0
resolving the above equation
V = 3.478 m/s
now using the second equation
V = u + at
3.478 = 8 + ( - 1/2 * 3.478^3/2 ) * t
hence : t = 1.4 secs
Answer:
88.18 W
Explanation:
The weight of the boy is given as 108 lb
Change to kg =108*0.453592= 48.988 kg = 49 kg
The slope is given as 6% , change it to degrees as
6/100 =0.06
tan⁻(0.06)= 3.43°
The boy is travelling at a constant speed up the slope = 7mi/hr
Change 7 mi/h to m/s
7*0.44704 =3.13 m/s
Formula for power P=F*v where
P=power output
F=force
v=velocity
Finding force
F=m*g*sin 3.43°
F=49*9.81*sin 3.43° =28.17
Finding the power out
P=28.17*3.13 =88.18 W
The Lamborghini SCV12 has 830 horse power.
Answer:
The radius of a wind turbine is 691.1 ft
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m
Explanation:
Given;
power generation potential (PGP) = 1000 kW
Wind speed = 5 mph = 2.2352 m/s
Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³
Radius of the wind turbine r = ?
Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²
Wind energy per unit mass of air = 2.517 J/kg
PGP = mass flow rate * energy per unit mass
PGP = ρ*A*V*e
r = 210.64 m = 691.1 ft
Thus, the radius of a wind turbine is 691.1 ft
PGP = CVᵃ
For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)
Let C = 0.4
PGP = Cvᵃ
take log of both sides
ln(PGP) = a*ln(CV)
a = ln(PGP)/ln(CV)
a = ln(1000)/ln(0.4 *2.2352) = 7.73
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m
Answer: OHMMETER & MEGOHMMETER:
Explanation: The ohmmeter measures circuit resistance; the megohmmeter measures the high resistance of insulation. A meter used to measure electric current. It is connected as part of a circuit.
