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baherus [9]
2 years ago
13

What are the transfer of energy for the 6 simple machine? Ex: inclined plane is gravitational potential energy to mechanical ene

rgy. What are the rest?
Engineering
1 answer:
lilavasa [31]2 years ago
8 0

Answer:

I. Inclined plane: GPE to KE.

II. Screw: KE to GPE.

III. Wheel and axle: GPE to KE.

IV. Lever: GPE to KE.

V. Wedge: GPE to KE.

VI. Pulley: GPE to KE.

Explanation:

A simple machine can be defined as a type of machine with no moving parts but can be used to perform work.

Basically, a simple machine allows for the transformation of energy into work. The six simple machines are;

I. <u>Inclined plane</u>: gravitational potential energy to mechanical energy (kinetic energy). It is set at an angle and then used to lift an object.

II. <u>Screw</u>: kinetic energy to potential energy. It is held in position and then used to tighten two or more objects together.

III. <u>Wheel and axle</u>: gravitational potential energy to mechanical energy (kinetic energy). It comprises of two circular objects which aids in the motion of an object such as a car.

IV. <u>Lever</u>: gravitational potential energy to mechanical energy (kinetic energy). It comprises of a pivoted bar at a fixed point (fulcrum) that allows it to be used for lifting heavy objects.

V. <u>Wedge</u>: gravitational potential energy to mechanical energy (kinetic energy). It can be used to split objects into halves.

VI. <u>Pulley</u>: gravitational potential energy to mechanical energy (kinetic energy). It is made up of a wheel and rope which allows force to be multiplied such as in a drawing well.

Energy can be defined as the ability (capacity) to do work. The two (2) main types of energy are;

a. Gravitational potential energy (GPE): it is an energy possessed by an object or body due to its position above the earth.

b. Kinetic energy (KE): it is an energy possessed by an object or body due to its motion.

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faltersainse [42]

Answer:

a) 0.489

b) 54.42 kg/s

c) 247.36 kW/s

Explanation:

Note that all the initial enthalpy and entropy values were gotten from the tables.

See the attachment for calculations

4 0
3 years ago
To put out a class D metal fire, you must _______ the fire.
gladu [14]

To put out a class D metal fire, you must smother the fire and eliminate the oxygen element in the fire.

<h3>What is a Class D fire?</h3>

A class D fire is a type of fire that cannot be extinguished by water. This is because adding water to it reacts with other elements in the fire intensifying the fire even more.

Smothering in this context involves adding a solution like carbon dioxide (CO2) into the fire, this results in a reduction of oxygen in the atmosphere surrounding the class D fire.

By so doing, smothering the fire eliminates the oxygen element in the fire, thereby extinguishing the fire.

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7 0
2 years ago
A cruise missile under test is moving horizontally at Ma =2 in the atmosphere at an elevation of 2000 m (Air temperature is 2 °C
telo118 [61]

Answer: the half-angle "alpha" of the Mach cone = 30⁰

Explanation:

To calculate the  half-angle "alpha" of the Mach cone.

we say ;

Sin∝ = 1 / Ma

given that Ma = 2

now we substitute

Sin∝ = 1 / 2

Sin∝ = 0.5

∝ = Sin⁻¹ 0.5

∝ = 30⁰

Therefore, the half-angle "alpha" of the Mach cone is 30⁰

3 0
3 years ago
A person walks into a refrigerated warehouse with head uncovered. Model the head as a 25- cm diameter sphere at 35°C with a surf
galina1969 [7]

Answer:

Hello some parts of your question is missing below is the missing part

Convection coefficient = 11 w/m^2. °c

answer : 44.83 watts

Explanation:

Given data :

surface emissivity ( ε )= 0.95

head ( sphere) diameter( D )  = 0.25 m

Temperature of sphere( T )  = 35° C

Temperature of surrounding ( T∞ )  = 25°C

Temperature of surrounding surface ( Ts ) = 15°C

б  = ( 5.67 * 10^-8 )

Determine the total rate of heat loss

First we calculate the surface area of the sphere

As = \pi D^{2}  

= \pi * 0.25^2 =  0.2 m^2

next we calculate heat loss due to radiation

Qrad = ε * б * As( T^{4} - T^{4} _{s} )  ---- ( 1 )

where ;

ε = 0.95

б = ( 5.67 * 10^-8 )

As = 0.2 m^2

T = 35 + 273 = 308 k

Ts = 15 + 273 = 288 k

input values into equation 1

Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )

         = 22.83  watts

Qrad ( heat loss due to radiation ) = 22.83 watts

calculate the heat loss due to convection

Qconv = h* As ( ΔT )

           = 11*0.2 ( 35 -25 )  = 22 watts

Hence total rate of heat loss

=  22 + 22.83

= 44.83 watts

5 0
3 years ago
Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa
Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air,  R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes  2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0
3 years ago
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