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saw5 [17]
2 years ago
9

Select the correct answer. Which of the following devices is a simple machine? A.

Engineering
2 answers:
tatyana61 [14]2 years ago
7 0

Answer:

A Pulley would be a simple machine.

Explanation:

A pulley has only has one kind of movement

astra-53 [7]2 years ago
6 0

Answer:

a pulley

Explanation:

letter B.

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In order to produce a certain semiconductor, a process called doping is performed in which phosphorus is diffused into germanium
Nimfa-mama [501]

Answer:

The diffusivity is given as 8.064\times 10^{-16} m^2/s

Explanation:

From the given data as in the attached question found via the search (because the values were not clear in this one)

D_o=2.0\times10^{-4}m^2/s\\Q_d=240.6 kJ/mol=2.406 \times 10^5 J/mol\\Gas Constant=R=8.314 J/mol K\\Temperature =830 C =830+273 =1103 K

So  the Diffusion coefficient is given as

D=D_oe^{\frac{-Q_d}{RT}}\\D=(2.0\times 10^{-4})e^{\frac{-2.406\times 10^5}{8.314\times 1103}}\\D=8.0640\times 10^{-16} m^2/s

So the diffusivity is given as 8.064\times 10^{-16} m^2/s

4 0
4 years ago
How can I solve 23.5 million Nona meters to millimeters using no calculator because I have to show my work
lozanna [386]

Answer:

its so simple. u must mind some formulas.

Explanation:

mili->10^(-3)

micro->10^(-6)

nano->10^(-9)

so write the exact number and move "." to left or right depend question.

in this one:

23.5 is 23500000.0 nano with a default dot at the end

for turning to mili u must move the dot 6 steps to left so it will be: 23.5 milimeter.

6 0
3 years ago
A train which is traveling at 70 mi/hr applies its brakes as it reaches point A and slows down with a constant deceleration. Its
Ugo [173]

Answer:

a) 0 mi/s^2

b) 52 mi/s

Explanation:

Assuming the crossing is 1/2 mile past point A and that point B is near point A (it isn't clear in the problem)

The train was running at 70 mi/h at point A and with constant deceleration reachesn the crossing 1/2 mile away with a speed of 52 mi/h

The equation for position under constant acceleration is:

X(t) = X0 + V0 * t + 1/2 * a * t^2

I set my reference system so that the train passes point A at t=0 and point A is X = 0, so X0 = 0.

Also the equation for speed under constant acceleration is:

V(t) = V0 + a * t

Replacing

52 = 70 + a * t

Rearranging

a * t = 52 - 70

a = -18/t

I can then calculate the time it will take it to reach the crossing

1/2 * a * t^2 + V0 * t  - X(t) = 0

Replacing

1/2 (-18/t) * t^ + 70 * t - 1/2 = 0

-9 * t + 70 * t = 1/2

61 * t = 1/2

t = (1/2)/61 = 0.0082 h = 29.5 s

And the acceleration is:

a = -18/0.0082 = -2195 mi/(h^2)

To beath the train the car must reach the crossing in 29.5 - 4.3 = 25.2 s

X(t) = X0 + V0 * t + 1/2 * a * t^2

52 mi/h = 0.0144 mi/s

1/2 = 0 + 0.0144 * 25.2 + 1/2 * a * 25.2^2

1/2 = 0.363 + 317.5 * a

317.5 * a = 0.5 - 0.363

a = 0.137/317.5 = 0.00043 mi/s^2 (its almost zero)

The car should remain at about constant speed.

It will be running at the same speed.

4 0
3 years ago
Insulating concrete forms (ICFs): Select one: a. place the insulation at the core of the concrete wall. b. can be filled with co
Hitman42 [59]

Answer:

Insulating Concrete forms (ICFs) a. place the insulation at the core of the concrete wall. This insulation is like a framework made from materials like polystyrene foam or cement bonded wood fiber etc and is placed at the core of the wall. These concrete walls have the long life time, strength and also provide protection from fire.

3 0
4 years ago
An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,
tester [92]

Answer:

percentage change in volume is 2.60%

water level rise is 4.138 mm

Explanation:

given data

volume of water V = 500 L

temperature T1 = 20°C

temperature T2 = 80°C

vat diameter = 2 m

to find out

percentage change in volume and how much water level rise

solution

we will apply here bulk modulus equation that is ratio of change in pressure   to rate of change of volume to change of pressure

and we know that is also in term of change in density also

so

E = -\frac{dp}{dV/V}  ................1

And -\frac{dV}{V} = \frac{d\rho}{\rho}   ............2

here ρ is density

and we know ρ  for 20°C = 998 kg/m³

and ρ  for 80°C = 972 kg/m³

so from equation 2 put all value

-\frac{dV}{V} = \frac{d\rho}{\rho}

-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}

dV = 0.0130 m³

so now  % change in volume will be

dV % = -\frac{dV}{V}  × 100

dV % = -\frac{0.0130}{500*10^{-3} }  × 100

dV % = 2.60 %

so percentage change in volume is 2.60%

and

initial volume v1 = \frac{\pi }{4} *d^2*l(i)    ................3

final volume v2 = \frac{\pi }{4} *d^2*l(f)    ................4

now from equation 3 and 4 , subtract v1 by v2

v2 - v1 =  \frac{\pi }{4} *d^2*(l(f)-l(i))

dV = \frac{\pi }{4} *d^2*dl

put here all value

0.0130 = \frac{\pi }{4} *2^2*dl

dl = 0.004138 m

so water level rise is 4.138 mm

8 0
3 years ago
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