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Amiraneli [1.4K]
2 years ago
5

An eagle is flying around and its velocity v as a function of time t is given in the graph below where

Physics
1 answer:
ahrayia [7]2 years ago
8 0
I honestly don’t even know
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In order to determine the mass moment of inertia of a flywheel of radius 600 mm, a 12-kg block is attached to a wire that is wra
shtirl [24]

Answer:

Explanation:

Given that,

When Mass of block is 12kg

M = 12kg

Block falls 3m in 4.6 seconds

When the mass of block is 24kg

M = 24kg

Block falls 3m in 3.1 seconds

The radius of the wheel is 600mm

R = 600mm = 0.6m

We want to find the moment of inertia of the flywheel

Taking moment about point G.

Then,

Clockwise moment = Anticlockwise moment

ΣM_G = Σ(M_G)_eff

M•g•R - Mf = I•α + M•a•R

Relationship between angular acceleration and linear acceleration

a = αR

α = a / R

M•g•R - Mf = I•a / R + M•a•R

Case 1, when y = 3 t = 4.6s

M = 12kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 4.6²

3 × 2 = 4.6²a

a = 6 / 4.6²

a = 0.284 m/s²

M•g•R - Mf = I•a / R + M•a•R

12 × 9.81 × 0.6 - Mf = I × 0.284/0.6 + 12 × 0.284 × 0.6

70.632 - Mf = 0.4726•I + 2.0448

Re arrange

0.4726•I + Mf = 70.632-2.0448

0.4726•I + Mf = 68.5832 equation 1

Second case

Case 2, when y = 3 t = 3.1s

M= 24kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 3.1²

3 × 2 = 3.1²a

a = 6 / 3.1²

a = 0.6243 m/s²

M•g•R - Mf = I•a / R + M•a•R

24 × 9.81 × 0.6 - Mf = I × 0.6243/0.6 + 24 × 0.6243 × 0.6

141.264 - Mf = 1.0406•I + 8.99

Re arrange

1.0406•I + Mf = 141.264 - 8.99

1.0406•I + Mf = 132.274 equation 2

Solving equation 1 and 2 simultaneously

Subtract equation 1 from 2,

Then, we have

1.0406•I - 0.4726•I = 132.274 - 68.5832

0.568•I = 63.6908

I = 63.6908 / 0.568

I = 112.13 kgm²

8 0
3 years ago
Pls heplp 70 points!!!!!
Rasek [7]

Answer

the answer is d for sure

5 0
3 years ago
efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.
PolarNik [594]

Answer:

Inlet : v_i=0.0646\frac{m}{s}

Outlet:  v_o=0.171\frac{m}{s}

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

\dot{m}=\frac{\dot{V}}{\upsilon}, where \dot{V} represent the flow rate and \upsilon the specific volume at the pressure and temperature given.

A_i=0.5m^2 is the inlet area

P_i=600Kpa pressure at the inlet area

T_i=70C temperature at the inlet area

A_o=1m^2 is the outlet area

P_o=100Kpa pressure at the outlet area

T_o=C temperature at the outlet area

\dot{m}=0.75\frac{kg}{s} represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

\upsilon_i =0.04304\frac{kg}{m^3} and the entropy is h_i=1.0645\frac{KJ}{KgK}=h_o

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy h_o=1.0645\frac{KJ}{KgK}

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}

h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}

Our interest value would be given using interpolation like this:

\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that \dot{m}=\frac{\dot{V}}{\upsilon}, but since \dot{V}=Av we have this:

\dot{m}=\frac{Av}{\upsilon}

If we solve from the velocity v we have this:

v=\frac{\upsilon \dot{m}}{A}   (*)

And now we just need to replace the values into equation (*)

For the inlet case:

v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}

For the oulet case:

v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}

7 0
3 years ago
HELPPPPPPP PLEASE!!! does anybody have the answer key to edge. physics lab report : mechanical equivalent to heat? one page of d
Karo-lina-s [1.5K]

Answer:

the mechanical equivalent of heat states that motion and heat are mutually interchangeable and that in every case

Explanation:

6 0
2 years ago
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Which term is used to describe the variety of inheritable traits in a species?
choli [55]
Genetic i think is it right?
7 0
3 years ago
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