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emmainna [20.7K]
3 years ago
5

Air enters into the hollow propeller tube at A with a mass flow of 4 kg/s and exits at the ends B and C with a velocity of 400 m

/s, measured relative to the tube. If the tube rotates at 1500 rev/min, determine the frictional torque M on the tube.
Physics
1 answer:
mash [69]3 years ago
6 0

Answer:

643N.m

Explanation:

From this question we have:

Mass flow = 4kg/s

Velocity V = 400m/s

Rotation N = 1500rev/min

We get the relative velocity at exit to be:

V2 = V - r2w

400-0.5x [(2*π*1500)/60]

= 400-78.5

= 321.5m/s

Then we have to calculate the frictional torque My

Mt = Mr2 x V2

= 4x0.5x321.5

= 643Nm

From the calculations above, we get the frictional torque M on the tube to be 643Nm.

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A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t
Charra [1.4K]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

                       v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time  

                       V = at + U

using equation  v - u = at to get line equation for the graph of the motion of the train on the incline plane

                       V_{x} = mt + V_{o}      where m is the slope

Comparing equation (1) and (2)

V = V_{x}

a = m    

U = V_{o}

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

          a = -  1.40 m/s²

Sunstituting a = -  1.40 m/s² and  u = 22 m/s

                        V_{x} = -1.40t + 22

                            V_{x} = -1.40(7.30) + 22

                             V_{x} = -10.22 + 22

                             V_{x} = 11. 78 m/s

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

           Area of triangle +  Area of rectangle

          [\frac{1}{2} * (22 - 11.78) * (7.30)]  + [(11.78 - 0) * (7.30)]

                           = 37.303 + 85.994

                           = 123. 297 m

                           ≈ 123. 30 m

                 

4 0
3 years ago
Read 2 more answers
A hockey puck moving at 0.4600 m/s collides with another puck that was at rest. The pucks have equal mass. The first puck is def
Sladkaya [172]

Answer:

Speed =0.283m/ s

Direction = 47.86°

Explanation:

Since it is a two dimensional momentum question with pucks having the same mass, we derive the momentum in xy plane

MU1 =MU2cos38 + MV2cos y ...x plane

0 = MU2sin38 - MV2sin y .....y plane

Where M= mass of puck, U1 = initial velocity of puck 1=0.46, U2 = final velocity of puck 1 =0.34, V2 = final velocity of puck 2, y= angular direction of puck2

Substitute into equation above

.46 = .34cos38 + V2cos y ...equ1

.34sin38 = V2sin y...equ2

.19=V2cos Y...x

.21=V2sin Y ...y

From x

V2 =0.19/cost

Sub V2 into y

0.21 = 0.19(Sin y/cos y)

1.1052 = tan y

y = 47.86°

Sub Y in to x plane equ

.19 = V2 cos 47.86°

V2=0.283m/s

7 0
3 years ago
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