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3 years ago

what should be used to feed material into a machine? A.joy stick B.push stick C. your feet D. your hands

Engineering

1 answer:

fenix001 [56]3 years ago

5 0

Answer:

B. Push stick

Explanation:

A joy stick is used to play video games, and using either your hands or feet to feed material into a machine could be hazardous and cause injury.

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cricket20 [7]

Except the Table of Contents

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3 years ago

How are speed and acceleration related

erica [24]

R = distance

dr/dt speed or with a direction, velocity

d(dr/dt)/dt = the time derivative of the velocity is called acceleration.

Speed is a scalar. Acceleration is a vector.

6 0

3 years ago

Read 2 more answers

6. Find the heat flow in 24 hours through a refrigerator door 30.0" x 58.0" insulated with cellulose fiber 2.0" thick. The tempe

Ilia_Sergeevich [38]

Answer:

The heat flow in 24 hours through the refrigerator door is approximately 1,608.57 BTU

Explanation:

The given parameters are;

The duration of the heat transfer, t = 24 hours = 86,400 seconds

The area of the refrigerator door, A = 30.0" × 58.0" = 1,740 in.² = 1.122578 m²

The material of the insulator in the door = Cellulose fiber

The thickness of the insulator in the door, d = 2.0" = 0.0508 m

The temperature inside the fridge = 38° F = 276.4833 K

The temperature of the room = 78°F = 298.7056 K

The thermal conductivity of cellulose fiber = 0.040 W/(m·K)

By Fourier's law, the heat flow through a by conduction material is given by the following formula;

$\dfrac{Q}{t} = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d}$

$Q = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d} \times t$

Therefore, we have;

$Q = \dfrac{0.04 \times 1.122578 \times (298.7056 - 276.4833 ) }{0.0508} \times 86,400 =1,697,131.73522$

The heat flow in 24 hours through the refrigerator door, Q = 1,697,131.73522 J = 1,608.5705140685 BTU

7 0

3 years ago

An office building is served by an air-cooled chiller currently operating at 115 tons (404.5 kW). The measured chilled water sup

Andrei [34K]

Answer:

B.197 gpm and 12.4 L/s

Explanation:

Given that

Load Q = 404.5 KW

Water inlet temperature= 6.1 °C

Water outlet temperature= 13.9°C

We know that specific heat for water

$C_p=4.187\ \frac{KJ}{kg.K}$

Now from energy balance

$Q=\dot{m}C_p\Delta T$

by putting the values

$Q=\dot{m}C_p\Delta T$

$404.5=\dot{m}\times 4.187(13.9-6.1)$

$\dot{m}=12.38\ \frac{kg}{s}$ (1 Kg/s = 15.85 gal/min)

We can say that

$\dot{m}=196.31\ \frac{gal}{min}$

We know that

$\dot{m}=\rho\times volume\ flow\ rate$

12.38=1000 x volume flow rate

$volume\ flow\ rate\ = 12.38\times 10^{-3}\ \frac{m^3}{s}$

volume flow rate = 12.38 L/s

So the option B is correct.

8 0

3 years ago

Steam enters an adiabatic turbine at 10MPa and 500 C and leaves at 10 kPa with a quality of 90%. Neglecting the changes in kinet

Amanda [17]

Answer:

flow ( m ) = 4.852 kg/s

Explanation:

Given:

- Inlet of Turbine

P_1 = 10 MPa

T_1 = 500 C

- Outlet of Turbine

P_2 = 10 KPa

x = 0.9

- Power output of Turbine W_out = 5 MW

Find:

Determine the mass ow rate required

Solution:

- Use steam Table A.4 to determine specific enthalpy for inlet conditions:

P_1 = 10 MPa

T_1 = 500 C ---------- > h_1 = 3375.1 KJ/kg

- Use steam Table A.6 to determine specific enthalpy for outlet conditions:

P_2 = 10 KPa -------------> h_f = 191.81 KJ/kg

x = 0.9 -------------> h_fg = 2392.1 KJ/kg

h_2 = h_f + x*h_fg

h_2 = 191.81 + 0.9*2392.1 = 2344.7 KJ/kg

- The work produced by the turbine W_out is given by first Law of thermodynamics:

W_out = flow(m) * ( h_1 - h_2 )

flow ( m ) = W_out / ( h_1 - h_2 )

- Plug in values:

flow ( m ) = 5*10^3 / ( 3375.1 - 2344.7 )

flow ( m ) = 4.852 kg/s

3 0

3 years ago

what should be used to feed material into a machine? A.joy stick B.push stick C. your feet D. your hands​

what should be used to feed material into a machine? A.joy stick B.push stick C. your feet D. your hands