As soon as you let go of it it is at its max speed because gravity is constantly pulling it down
Answer
Time period T = 1.50 s
time t = 40 s
r = 6.2 m
a)
Angular speed ω = 2π/T
=
= 4.189 rad/s
Angular acceleration α = 
= 
= 0.105 rad/s²
Tangential acceleration a = r α = 6.2 x 0.105 = 0.651 m/s²
b)The maximum speed.
v = 2πr/T
= 
= 25.97 m/s
So centripetal acceleration.
a = 
= 
= 108.781 m/s^2
= 11.1 g
in combination with the gravitation acceleration.


<span>The particles through which compressional waves travel move in the same direction as the wave. This may be observed by fixing one end of a large spring and then compressing and extending the other end. The wave travels from one end to the other and the spring's parts move in the same direction.</span>
Answer:
Temperature at the exit = 
Explanation:
For the steady energy flow through a control volume, the power output is given as

Inlet area of the turbine = 
To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.
Assuming Argon behaves as an Ideal gas, we have the specific volume 
as


for Ideal gasses, the enthalpy change can be calculated using the formula

hence we have


<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>
evaluating the above equation, we have 
Hence, the temperature at the exit = 
Answer:
2.45 J
Explanation:
The following data were obtained from the question:
Mass (m) = 0.5 kg
Height (h) = 1 m
Kinetic energy (KE) =?
Next, we shall determine the velocity of the rock after it has fallen half way. This can be obtained as follow:
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 1/2 = 0.5 m
Final velocity (v) =?
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 0.5)
v² = 9.8
Take the square root of both side
v = √9.8
v = 3.13 m/s
Finally, we shall determine the kinetic energy of the rock after it has fallen half way. This can be obtained as follow:
Mass (m) = 0.5 kg
Velocity (v) = 3.13 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 0.5 × 3.13²
KE = 0.25 × 9.8
KE = 2.45 J
Therefore, the kinetic energy of the rock after it has fallen half way is 2.45 J