A car travels from stop A to stop B with a speed of 30 km/h and then returns back to A with a speed of 50 km/h. Find

Air, considered an ideal gas, is contained in an insulated piston-cylinder assembly outfitted with a paddle wheel. It is initial

Maru [420]

Our data are,

State 1:

$P_1= 10psi=68.95kPa\\V_1 = 1ft^3=0.02831m^3\\T_1 = 100\°F = 310.93K$

State 2:

$P_2 =5psi=34.474kPa\\V_2 = 3ft^3=0.0899m^3$

We know as well that $3BTU=3.16kJ/K$

To find the mass we apply the ideal gas formula, which is given by

$P_1V_1=mRT_1$

Re-arrange for m,

$m= \frac{P_1V_1}{RT_1}\\m= \frac{68.95*0.02831}{(0.287)310.9}\\m=0.021893kg=0.04806lbm\\$

Because of the pressure, temperature and volume ratio of state 1 and 2, we have to

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

Replacing,

$T_2 = \frac{P_2V_2}{P_1V_1}T_1\\T_2 =\frac{34.474*0.0844}{68.95*0.02831}*310.93\\T_2 = 464.217K=375.5\°F$

For conservative energy we have, (Cv = 0.718)

$W = m C_v = 0.718 \Delta T +dw\\dw = W - mv\Delta T\\dw = 3.16-(0.0218*0.718)(454.127-310.93)\\dw = 0.765kJ=0.72BTU$

3 0

2 years ago

2. If a cyclist in the Tour de France traveled southwest a distance of 12,250 meters in one hour, what would the velocity of the

Luda [366]

Answer:

12,25 km/h

≈ 3,4 m/s

Explanation:

v = d/t

= 12250m/h

= 12,25km/h

or

v = d/t

= 12250m/h

1h = 60m×60s = 3600s

= 12250m/3600s

≈ 3,4 m/s

5 0

2 years ago

A light-year is a unit of a. time. c. mass. b. distance. d. density. please select the best answer from the choices provided a b

11111nata11111 [884]

Answer:

distance

Explanation:

it is the distance traveled by light in one year

3 0

1 year ago

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Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 65.3 ◦ . The velo

yanalaym [24]

Answer:

Plane will 741.6959 m apart after 1.7 hour

Explanation:

We have given time = 1.7 hr

So if we draw the vectors of a 2d graph we see that the difference in angles is = $102^{\circ}-65.3^{\circ}=36.7^{\circ}$

Speed of first plane = 730 m/h

So distance traveled by first plane = 730×1.7 = 1241 m

Speed of second plane = 590 m/hr

So distance traveled by second plane = 590×1.7 = 1003 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.

Using the law of cosine, $r^2$ representing the distance between the planes, we see that:

$r^2=1241^2+1003^2-2\times 1003\times 1241cos(36.7)=550112.8295$

r = 741.6959 m

3 0

3 years ago

A car's velocity as a function of time is given by vx(t)=α+βt2, where α=3.00m/s and β=0.100m/s3.

Mice21 [21]

1) Analyze the equation

Vₓ = α + β t²

Vₓ = 3.00 + 0.100 t²

That is a quadratic equation, so the graph is a parabola.

2) Therefore, take into account that the main points to draw a parabola are:

i) vertex

ii) concavity

iii) y-intercepts

iv) x - intercepts

Also, for all graphs you need the domain and the range.

3) Find the y-intercept (t = 0)

t = 0 ⇒ Vₓ = 3.00 + 0.100 (0)² = 3.00

4) Find the x-intercepts (Vₓ = 0)

Vₓ = 0 ⇒ 0 = 3.00 + 0.100 t²

⇒ t² = - 3.00 / 0.100 = -30.0. Since t² cannot be negative, there is not x-intercepts.

5) Concavity

Since, the coefficient of t² is positive, the parabola open upwards.

6) Vertex

It is the local minimum of the equation. You can find it by the first derivative

Vₓ' = 2×0.100 t = 0 ⇒ t = 0

Vₓ = 3.00 + 0.100 (0)² = 3.00 m/s

⇒ vertex = (0,3.00)

7) The domain is given t ∈ [0,5.00]

8) You can also build a table with several points in the domain

t =0; Vₓ = 3.00 + 0.100 (0)² = 0

t = 1; Vₓ = 3.00 + 0.100 (1)² = 3.10

t = 2; Vₓ = 3.00 + 0.100 (2)² = 3.40

t = 3; Vₓ = 3.00 + 0.100 (3)² = 3.90

t = 4; Vₓ = 3.00 + 0.100 (4)² = 4.60

t = 5; Vₓ = 3.00 + 0.100 (5)² = 5.50

9) Range: Vₓ ∈ [ 3.00, 5.50]

10) All that information permits you to put several points in a coordinate sysment and sketch the same graph as the shown in the figure attached.

4 0

3 years ago

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