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Vlad1618 [11]
3 years ago
8

Starting from rest, a 6.79 kg block slides 2.82 m down a rough 20.7 ◦ incline. The coefficient of kinetic friction between the b

lock and the incline is 0.425. The acceleration of gravity is 9.8 m/s 2 . Find the work done by the force of gravity. Answer in units of J.
Physics
1 answer:
Veronika [31]3 years ago
8 0

Answer:

23.52092 J

Explanation:

m = Mass of block = 6.79 kg

s = Sliding distance = 2.82 m

\theta = Angle of slide = 20.7°

\mu = Coefficient of kinetic friction = 0.425

g = Acceleration due to gravity = 9.8 m/s²

Work done by the force of gravity is given by

W=mgsin\theta\\\Rightarrow W=6.79\times 9.8\times sin20.7\\\Rightarrow W=23.52092\ J

The work done by the force of gravity is 23.52092 J

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Plant's produce oxygen, to explain this you would state that the plant uses CO2 from the air and removes the carbon and then release breathable O2 into the atmosphere. To test this you could use a small fern and put it in a sealed container of some kind big enough to hold it, then add a burning candle in the container with the plant, since we know fire burms O2 and creates CO2 the plant should use the CO2 and create more O2 keeping the fire from burning out from lack of oxygen
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A mole of a monatomic ideal gas at point 1 (101 kPa, 5 L) is expanded adiabatically until the volume is doubled at point 2. Then
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Answer:

(a). Check attachment.

(b). 280.305 J.

(c). 31.81 kpa; 38.26K.

(d). 24.05K.

(e). 24.05k; 40kpa.

(f). -138.6J.

Explanation:

(a). Kindly check the attached picture for the diagram showing the four process.

1 - 2 = adiabatic expansion process.

2 - 3 = Isochoric process.

3 - 4 = isothermal process.

4 - 1 = isochoric process.

(b). Recall that the process from 1 to is an adiabatic expansion process.

NB: b = 5/3 for a monoatomic gas.

Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].

= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.

Thus, the workdone = 280.305 J.

(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.

T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.

(d). The process 2 - 3 is an Isochoric process, then;

T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.

(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.

The pressure can be determine as below;

P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.

(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J

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Assume you are given an int variable named nElements and a two-dimensional array that has been created and assigned to a2d. Writ
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Answer:

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an object weighing 15 newtons is lifted from the ground to a height of 0.22 meter what is the increase in the object's gravitati
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I hope this helps ~~Charlotte~~
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