The speed of a car travelling over a hill that has a radius of curvature should not exceed a certain speed other it will topple. This speed is related to the radius of curvature and the gravitational acceleration as shown below:
V^2 = Rg, where V = maximum speed, R = Radius of curvature, g = gravitational acceleration.
Substituting;
V = Sqrt (Rg) = Sqrt (120*9.81) = 34.31 m/s
Answer:
3.26 secs
Explanation:
Diameter of sphere ( D )= 10 mm
T1 = 75°C
P = 1 atm
T∞ = 23°C
T2 = 35°c
Velocity = 10 m/s
<u>Determine how long it will take to cool the sphere to 35°C</u>
<em>Using the properties of copper and air given in the question</em>
Nu = 2 + (Re)^0.8 (Pr)^0.33
hd / k = 2 + ( vd/v )^0.8 (Pr)^0.33
∴ h ≈ 2594.7 W/m^2k
Given that :
(T2 - T∞) / ( T1 - T∞ ) = exp [ ( -hA / pv CP ) t ]
( 35 - 23 ) / ( 75 - 23 ) = exp [ - 2594.7 * 6 * t / 8933 * 387 * 10 * 10^-3 ]
= ln ( 12/52 ) = -1.466337069 = - 0.45032919 * t
∴ t ≈ 3.26 secs ( -1.466337069 / -0.45032919 )
Answer:
0.86 m/s
Explanation:
A = cross-sectional area of the duct = 900 cm² = 900 x 10⁻⁴ m²
v = speed of air in the duct
t = time period of circulation = 40 min = 40 x 60 sec = 2400 sec
V = Volume of the air in the room = volume of room = 7 x 11 x 2.4 = 184.8 m³
Volume of air in the room is given as
V = A v t
inserting the values
184.8 = (900 x 10⁻⁴) (2400) v
v = 0.86 m/s
Answer:
Kinetic energy is the energy an object has because of its motion. If we want to accelerate an object, then we must apply a force. Applying a force requires us to do work. After work has been done, energy has been transferred to the object, and the object will be moving with a new constant speed.
