Answer:
153.2 J
Explanation:
Let’s first identify our known variables:
mass (m) of the block = 10 kg
distance slid down (i.e., displacement) = 2 m
coefficient of kinetic friction (μk) = 0.2
In the following diagram, if we analyze the force component directed along the displacement, we find
Fcos 40°
100 (cos 40°)
76.60 N
The work done by the frictional force is then calculated as:
W = 
× displacement
W = 76.60 × 2
W = 153.2 J
Therefore, the work done by the frictional force equals 153.2 J
Answer:
a) v₂ = 26.6 ft/s
b) v₂ = 31.9 ft/s
Explanation:
a) Applying the principle of energy conservation for the soccer arm from the point of impact to its lowest point:
The ball's velocity in the tangential direction is given by:
v₂*sinθ = v₁*sin30
With the values:
if v₁ = 0
θ = 0º
The restitution coefficient is:
where O=θ
The aggregate momentum equation is:
, where O = θ
When we incorporate gravitational acceleration:
Isolating v₂ results in:
v₂ = 26.6 ft/s
b) To find the ball's velocity, we utilize:
v₂ * sinθ = v₁ * sin30
if v₁ = 10 ft/s
then v₂ * sinθ = 10 * sin30
thus sinθ = 5/v₂
The restitution coefficient remains:
where O = θ
Solving for v₂ yields:
v₂ = 31.9 ft/s
That statement is incorrect!
The claim is untrue
Answer:
The change in entropy of the steam is 2.673 kJ/K
Explanation:
The mass of the liquid-vapor mixture is 1.5 kg
The mass in the liquid phase is calculated as 3/4 × 1.5 kg = 1.125 kg
The mass in the vapor phase is calculated as 1.5 - 1.125 = 0.375 kg
According to the steam tables
At a pressure of 200 kPa (200/100 = 2 bar), the specific entropy of steam is found to be 7.127 kJ/kgK
The entropy of steam can be calculated as specific entropy multiplied by mass = 7.127 × 0.375 = 2.673 kJ/K
Answer:
B) P1 would have to increase to sustain the flow rate (correct)
C) Resistance would rise (correct)
Explanation:
Flow rate is measured at 10 liters per minute
Driving pressure (P1) stands at 20 cm H2O
Fixed downstream pressure (P2) is 5 cm H2O
The accurate statements when the lumen is pinched in the center of the tube are: P1 will increase to maintain the flow rate, and resistance will rise. This occurs because pinching the lumen decreases its diameter, leading to higher resistance, which is linearly related to pressure, thus P1 will also increase.
The incorrect statement is: the flow would decrease.
