How are elements and compounds similar

Ford used to make a 7.0 liter engine. Calculate how many many cubic inches would be the same size engine.

Anestetic [448]

Answer : The volume of engine in cubic inches is, $427.166\text{ inch}^3$

Explanation :

As we are given that the volume of engine is, 7.0 liters. Now we have to calculate the volume of engine in cubic inches.

Conversion used :

$1\text{ liter}=1000cm^3$

So, $7\text{ liter}=7000cm^3$

$1\text{ inch}=2.54cm$

or,

$1cm=\frac{1}{2.54}\text{ inch}$

$1cm^3=(\frac{1}{2.54}\text{ inch})^3$

As,

$1cm^3=(\frac{1}{2.54}\text{ inch})^3$

So,

$7000cm^3=\frac{7000cm^3}{1cm^3}\times (\frac{1}{2.54}\text{ inch})^3=427.166\text{ inch}^3$

Thu, the volume of engine in cubic inches is, $427.166\text{ inch}^3$

3 0

3 years ago

High rate of speed?

defon

<em>velocity over time I think</em>

6 0

3 years ago

Read 2 more answers

Learning Goal: How do 2 ordinary waves build up a "standing" wave? A very generic formula for a traveling wave is: y1(x,t)=Asin(

zheka24 [161]

Answer:

Explanation:

=Asin(kx−ωt). This general mathematical form can represent the displacement of a string, or the strength of an electric field, or the height of the surface of water, or a large number of other physical waves!

Part C Find ye(x) and yt(t). Remember that yt(t) must be a trig function of unit amplitude. Express your answers in terms of A, k, x, ω, and t. Separate the two functions with a comma. Use parentheses around the argument of any trig functions.

Part E At the position x=0, what is the displacement of the string (assuming that the standing wave ys(x,t) is present)? Part G From

Part F we know that the string is perfectly straight at time t=π2ω. Which of the following statements does the string's being straight imply about the energy stored in the strJHJMNMMUJJHTGGHing?

a.There is no energy stored in the string: The string will remain straight for all subsequent times.

b.Energy will flow into the string, causing the standing wave to form at a later time.

c.Although the string is straight at time t=π2ω, parts of the string have nonzero velocity. Therefore, there is energy stored in the string.

d.The total mechanical energy in the string oscillates but is constant if averaged over a complete cycle.

3 0

3 years ago

Does anyone know this?

e-lub [12.9K]

Explanation: The first one
Source: it literally has fusion in the name

5 0

3 years ago

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In a car lift used in a service station, compressed air exerts a force on a small piston of circular cross section having a radi

Vitek1552 [10]

Answer:

(a) 1,569.63 N

(b) 195,933.99 Pa

(c) As pressure and volume are equal for each piston, workdone must also be equal

(d) 1,647.47 kg

Explanation:

Let the cross-sectional area (CSA) of small piston = A₁

Let the cross-sectional area (CSA) of the bigger piston = A₂

Let the Force applied at the smaller piston = F₁

Let the Force applied at the bigger piston = F₂

The principle of hydraulic lift assumes the that the fluid is in-compressible, resulting to a constant pressure system.

F₁/A₁ =F₂/A₂----------------------------------------------------------- (1)

(a) F₁= F₂ xA₁ /A₂

F₂ = 13,300 N

A₁ = π r₁²

=π x (0.0505)²

= 0.008011 m²

A₂= π r₂²

=π x (0.147)²

= 0.06788 m²

Substituting into (1)

F₁ = 13,300 x 0.008011/0.06788

= 1,569.6272

≈ 1,569.63 N

(b) Air pressure = Force/Area

= F₁/A₁

= 1,569.6272/ 0.008011

= 195,933.99 Pa

(c) The pressure is constant for both pistons according to Pascal Law.

Workdone = force x distance----------------------------------------- (2)

force = pressure × area

distance = volume/area from

Substituting into (2)

Workdone = pressure × volume.

As pressure and volume are equal for each piston, work must also be equal

(d) F₂ = F₁ x A₂/ A₁---------------------------------------------------- (3)

A₁ = π r₁²

=π x (0.079)²

= 0.01960 m²

A₂= π r₂²

=π x (0.353)²

= 0.3914 m²

Substituting into (2)

F₂= 825 x 0.3914/ 0.01960

= 16,474.7448

≈ 16,474.74 N

= 1,647.47 kg

3 0

3 years ago