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A mass of gas has a volume of 4m3, a temperature of 290k, and an absolute pressure of 475 kpa. When the gas is allowed to expand

motikmotik

Given:

V1 = 4m3

T1 = 290k

P1 = 475 kpa = 475000 Pa

V2 = 6.5m3

T2 = 277K

Required:

P

Solution:

n = PV/RT

n = (475000 Pa)(4m3) / (8.314 Pa-m3/mol-K)(290k)

n = 788 moles

P = nRT/V

P = (788 moles)(8.314 Pa-m3/mol-K)(277K)/(6.5m3)

P = 279,204 Pa or 279 kPa

4 0

3 years ago

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A snowball starting at rest rolls down a hill and reaches 5 m/s. If the hill is

Lostsunrise [7]

Answer:

The acceleration of the snowball is 0.3125

Explanation:

The initial speed of the snowball up the hill, u = 0

The speed the snowball reaches, v = 5 m/s

The length of the hill, s = 40 m

The equation of motion of the snowball given the above parameters is therefore;

v² = u² + 2·a·s

Where;

a = The acceleration of the snowball

Plugging in the values, we have;

5² = 0² + 2 × a × 40

∴ 2 × 40 × a = 5² = 25

80 × a = 25

a = 25/80 = 5/16

a = The acceleration of the snowball = 5/16 m/s².

The acceleration of the snowball = 5/16 m/s² = 0.3125 m/s² .

4 0

3 years ago

Four vehicles approach an intersection with a 4 way stop at the same time. Car B is ahead of Car A and both are in the same lane

garik1379 [7]

Answer: car D

Explanation:

4 0

3 years ago

A box is being dragged with a horizontal force of 65 N for 12 meters. If there is a force of friction acting on it

asambeis [7]

Answer:

A. 780 J

B. 120 J

C. 660 J

Explanation:

From the question given above the following data were obtained:

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

A. Determination of the work done by the dragging force.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Workdone (Wd) by dragging force =?

Wd = Fₔ × s

Wd = 65 × 12

Wd = 780 J

Therefore, the work done by the dragging force is 780 J

B. Determination of the work done by friction.

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Workdone (Wd) by friction =?

Wd = Fբ × s

Wd = 10 × 12

Wd = 120 J

Therefore, the work done by friction is 120 J

C. Determination of the net work done on the box.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Net work done (Wd) =?

Next, we shall determine the net force acting on the box. This can be obtained as follow:

Dragging force (Fₔ) = 65 N

Force of friction (Fբ) = 10 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fբ

Fₙ = 65 – 10

Fₙ = 55 N

Thus, the net force acting on the box is 55 N

Finally, we shall determine the net work done on the box as follow:

Distance (s) = 12 m

Net force (Fₙ) = 55 N

Net work done (Wd) =?

Wd = Fₙ × s

Wd = 55 × 12

Wd = 660 J

Therefore, the net work done on the box is 660 J

4 0

3 years ago

Introduction: The specific heat capacity of a substance is the amount of energy needed to change the temperature of that substan

ki77a [65]

Answer:

A) 8,368 J

B) ) 0.893 J/gºC

Explanation:

A)

The heat gained by the water can be obtained solving the following equation:

$q_{g} = c_{w} * m * \Delta T (1)$

where cw = specific heat of water = 4.184 J/gºC
m= mass of water = 1,000 g
ΔT = 2ºC
Replacing these values in (1) we get:

$q_{g} = c_{w} * m * \Delta T = 4.184 J/gºC*1,000 g* 2ºC = 8,368 J (2)$

B)

Assuming that the heat energy gained by the water is equal to the one lost by the aluminum, we can use the same equation, taking into account that the energy is lost by the aluminum, so the sign is negative: -8,368 J.
Replacing by the mass of aluminum (125 g), and the change in temperature (-74.95ºC), in (1), we can solve for the specific heat of aluminum, as follows:

$q_{l} = c_{Al} * m_{Al} * \Delta T (3)$

⇒ $-8,368 J = c_{Al}* 125 g * (-74.95ºC) (4)$

$c_{Al} = \frac{-8,368J}{125g*(-74.95ºC} = 0.893 J/gºC (5)$

which is pretty close to the Aluminum's accepted specific heat value of 0.900 J/gºC.

8 0

3 years ago