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puteri [66]
3 years ago
12

20 POINTS PLUS BRAINLIEST What work is done by a system on its environment when an ideal gas is maintained at a constant pressur

e of 200 newtons/meter^2 during an isobaric process, while the volume decreases by 0.2 cubic meters? Please explain briefly
Physics
2 answers:
Oliga [24]3 years ago
4 0

Say you have 60 cubic meters of an ideal gas at a pressure of 200 pascals. You heat the gas until it expands to a volume of 120 cubic meters. How much work does the gas do? All you have to do is plug in the numbers:

The gas does 12,000 joules of work as it expands under constant pressure.


kogti [31]3 years ago
4 0

The answer is C. -40 joules on plato, i got 100 on the test

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Answer:

q=1\times10^{-8}C

Explanation:

Let the charge on the ball bearing is q.

charge on glass bead, Q = 20 nC = 20 x 10^-9 C

Force between them, F = 0.018 N

Distance between them, d = 1 cm = 0.01 m

By use of Coulomb's law in electrostatics

F=\frac{KQq}{d^{2}}

By substituting the values

0.018=\frac{9\times10^{9}\times20\times10^{-9}q}{0.01^{2}}

q=1\times10^{-8}C

Thus, the charge on the ball bearing is q=1\times10^{-8}C

7 0
3 years ago
A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i
Mariana [72]

Answer:

451.13 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

c = 1500/[0.133(45-20)]

c = 1500/(0.133×25)

c = 1500/3.325

c = 451.13 J/kg.°C

4 0
3 years ago
Why does ice melt faster in water than in oil when both liquids are at the same temperature​
vekshin1
Water has a high specific heat capacity. Oil has a smaller heat capacity.
7 0
3 years ago
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A 2.61 g lead weight, initially at 11.1 ∘C, is submerged in 7.67 g of water at 52.6 ∘C in an insulated container. What is the fi
sleet_krkn [62]

Answer:

Equilibrium temperature will be T=52.2684^{\circ}C

Explanation:

We have given weight of the lead m = 2.61 gram

Let the final temperature is T

Specific heat of the lead c = 0.128

Initial temperature of the lead = 11°C

So heat gain by the lead = 2.61×0.128×(T-11°C)

Mass of the water m = 7.67 gram

Specific heat = 4.184

Temperature of the water = 52.6°C

So heat lost by water = 7.67×4.184×(T-52.6)

We know that heat lost = heat gained

So 2.61\times 0.128\times (T-11)=7.67\times 4.184\times (52.6-T)

0.334T-3.67=1688-32.031T

T=52.2684^{\circ}C

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3 years ago
List 3 additional real world examples that show work being done.
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