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puteri [66]
3 years ago
11

List 6 different mechanisms in the Rube Goldberg cartoon and predict the purpose of each. Does the mechanism change speed, force

, torque, direction, or type of movement?
Engineering
1 answer:
vodka [1.7K]3 years ago
7 0

Answer: Rotary - because it has to

Around in a circle

Explanation:

i hope this helped u

You might be interested in
Which one of the following is not an economic want?
Nostrana [21]

The available options are:

a. Want for a television set

b. Want for medical attention

c. Want for friendship

d. Want for new clothing

Answer:

Want for friendship

Explanation:

Given that economic want is what humans desire to have or possess such that they pay money to acquire them.

Hence, from the available options "want for friendship " is not economic want because it can't be bought with money, while other options can be bought with money or monetary transaction.

8 0
3 years ago
A controller on an electronic arcade game consists of a variable resistor connected across the plates of a 0.227 μF capacitor. T
anyanavicka [17]

Answer:

R min = 28.173 ohm

R max = 1.55 × 10^{4}  ohm

Explanation:

given data

capacitor = 0.227 μF

charged to 5.03 V

potential difference across the plates =  0.833 V

handled effectively = 11.5 μs to 6.33 ms

solution

we know that resistance range of the resistor is express as

V(t) = V_o \times e^{t\RC}    ...........1

so R will be

R = \frac{t}{C\times ln(\frac{V_o}{V})}    ....................2

put here value

so for t min 11.5 μs

R = \frac{11.5}{0.227\times ln(\frac{5.03}{0.833})}

R min = 28.173 ohm

and

for t max 6.33 ms

R max = \frac{6.33}{11.5} \times 28.173  

R max = 1.55 × 10^{4}  ohm

4 0
3 years ago
A man weighs 145 lb on earth.Part ASpecify his mass in slugs.Express your answer to three significant figures and include the ap
adell [148]

Answer:

<em>a) 4.51 lbf-s^2/ft</em>

<em>b) 65.8 kg</em>

<em>c) 645 N</em>

<em>d) 23.8 lb</em>

<em>e) 65.8 kg</em>

<em></em>

Explanation:

Weight of the man on Earth = 145 lb

a) Mass in slug is...

32.174 pound = 1 slug

145 pound = x slug

x = 145/32.174 = <em>4.51 lbf-s^2/ft</em>

b) Mass in kg is...

2.205 pounds = 1 kg

145 pounds = x kg

x = 145/2.205 = <em>65.8 kg</em>

c) Weight in Newton = mg

where

m is mass in kg

g is acceleration due to gravity on Earth = 9.81 m/s^2

Weight in Newton = 65.8 x 9.81 = <em>645 N</em>

d) If on the moon with acceleration due to gravity of 5.30 ft/s^2,

1 m/s^2 = 3.2808 ft/s^2

x m/s^2 = 5.30 ft/s^2

x = 5.30/3.2808 = 1.6155 m/s^2

weight in Newton = mg = 65.8 x 1.6155 = 106

weight in pounds = 106/4.448 = <em>23.8 lb</em>

e) The mass of the man does not change on the moon. It will therefore have the same value as his mass here on Earth

mass on the moon = <em>65.8 kg</em>

3 0
3 years ago
P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea- level standard air through a converging nozzle whose throat diame
eimsori [14]

Answer:

a)  m=0.17kg/s

b)  Ma=0.89

Explanation:

From the question we are told that:

Pressure P=60kPa

Diameter d=3cm

Generally at sea level

T_0=288k\\\\\rho_0=1.225kg/m^3\\\\P_0=101350Pa\\\\r=1.4

Generally the Power series equation for Mach number is mathematically given by

\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}

\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}

Ma=0.89

Therefore

Mass flow rate

\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}

\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}

\rho=0.848kg/m^3

Generally the equation for Velocity at throat is mathematically given by

V=Ma(r*T_0\sqrt{T_e})

Where

T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}

T_e=248

Therefore

V=0.89(1.4*288\sqrt{248})\\\\V=284

Generally the equation for Mass flow rate is mathematically given by

m=\rho*A*V

m=0.84*\frac{\pi}{4}*3*10^{-2}*284

m=0.17kg/s

6 0
3 years ago
A four-cylinder, four-stroke internal combustion engine operates at 2800 RPM. The processes within each cylinder are modeled as
Ulleksa [173]

Answer:

1) 287760.4 Hp

2) 18410899.5 kPa

Explanation:

The parameters given are;

p₁ = 14.7 lbf/in² = 101325.9 Pa

v₁ = 0.0196 ft³ = 0.00055501 m³

T₁ = 80°F = 299.8167 K

k = 1.4

Assumptions;

1) Air standard conditions are appropriate

2) There are negligible potential and kinetic energy changes

3) The air behaves as an ideal gas and has constant specific heat capacities of temperature and pressure

1) Process 1 to 2

Isentropic compression

T₂/T₁ = (v₁/v₂)^(1.4 - 1) = 10^0.4

p₂/p₁ = (v₁/v₂)^(1.4)

p₂ = p₁×10^0.4 =  101325.9*10^0.4 = 254519.153 Pa

T₂ = 299.8167*10^0.4 = 753.106 K

p₃ = 1080 lbf/in² = 7,446,338 Pa

Stage 2 to 3 is a constant volume process

p₃/T₃ = p₂/T₂

7,446,338/T₃ =   254519.153/753.106

T₃ = 7,446,338/(254519.153/753.106) = 22033.24 K

T₃/T₄ = (v₁/v₂)^(1.4 - 1) = 10^0.4

T₄ = 22033.24/(10^0.4) = 8771.59 K

The heat supplied, Q₁ = cv(T₃ - T₂) = 0.718*(22033.24 -753.106) = 15279.14 kJ

The heat rejected = cv(T₄ - T₁) = 0.718*(8771.59 - 299.8167) = 6082.73 kJ

W(net) = The heat supplied - The heat rejected = (15279.14 - 6082.73) = 9196.41 kJ

The power = W(net) × RPM/2*1/60 = 9196.41 * 2800/2*1/60 = 214582.9 kW

The power by the engine = 214582.9 kW = 287760.4 Hp

2) The mean effective pressure, MEP  = W(net)/(v₁ - v₂)

v₁ = 0.00055501 m³

v₁/v₂ = 10

v₂ = v₁/10 = 0.00055501/10 = 0.000055501

MEP  = 9196.41/(0.00055501 -  0.000055501) = 18410899.5 kPa

4 0
3 years ago
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