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N76 [4]
3 years ago
12

A system absorbed 44 joules of heat from its surroundings. After doing work, the increase in the internal energy of the system i

s 32 joules. How much energy was used by the system to do useful work?
Physics
2 answers:
Minchanka [31]3 years ago
7 0

12 j is correct on edge

Lelu [443]3 years ago
4 0
We can solve the problem by using the first law of thermodynamics:
\Delta U = Q-W
where
\Delta U is the variation of internal energy of the system
Q is the heat absorbed by the system
W is the work done by the system

In this problem, the heat absorbed by the system is Q=+44 J, and the increase in internal energy is \Delta U = +32 J. So we can rearrange the equation to calculate the work done:
W=Q-\Delta U=44 J-32 J=+12 J
and the work is positive, which means that it is done by the system on the surrounding.
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What investigations are best for demonstrating cause and effect relationship
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What is the identify atom shown
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3 years ago
Help pleasee
agasfer [191]

Answer:

i/f = i/o + i/i       f = focal, o = object, i = image

1 / i = 1 / f - 1 / o  =    (o - f) / o f

i = o * f / ( o - f)      image distance

i = 12.5 * 22 / (12.5 - 22) = -28.9 cm

Image is real

Image is 28.9 cm to left of lens

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8 0
3 years ago
A 0.950 kg block is attached to a spring with spring constant 16.0 N/m . While the block is sitting at rest, a student hits it w
Bumek [7]

Answer:

Explanation:

Given that,

Mass attached m = 0.95kg

Spring constant k = 16N/m

Instantaneous speed v = 36cm/s = 0.36m/s

Amplitude A=?

When x = 0.7A

Using conservation of energy

∆K.E + ∆P.E = 0

K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0

At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.

Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0

So, the equation becomes

— K.E(initial) + P.E(final) = 0

K.E(initial) = P.E(final)

½mv² = ½kA²

mv² = kA²

0.95 × 0.36² = 16×A²

0.12312 = 16•A²

A² = 0.12312/16

A² = 0.007695

A = √0.007695

A = 0.088 m

A = 8.8cm

B. Speed at x = 0.7A

Using the same principle above

K.E(initial) = P.E(final)

½mv² = ½kA²

Where A = 0.7A = 0.7 × 0.088 = 0.0614m

Then,

½× 0.95 × v² = ½ × 16 × 0.0614²

0.475v² = 0.0310644

v² = 0.0310644/0.475

v² = 0.0635

v = √0.0635

v = 0.252 m/s

v = 25.2 cm/s

8 0
3 years ago
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