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QveST [7]
3 years ago
13

Consider that 168.0 J of work is done on a system and 305.6 J of heat is extracted from the system. In the sense of the first la

w of thermodynamics, What is the value (including algebraic sign) of W, the work done by the system?
Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
3 0

To solve this problem we must resort to the Work Theorem, internal energy and Heat transfer. Summarized in the first law of thermodynamics.

dQ = dU + dW

Where,

Q = Heat

U = Internal Energy

By reference system and nomenclature we know that the work done ON the system is taken negative and the heat extracted is also considered negative, therefore

W = -168J \righarrow  Work is done ON the system

Q = -305.6J \rightarrow Heat is extracted FROM the system

Therefore the value of the Work done on the system is -158.0J

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Why are a 12 ounce hard seltzer and 1.5 ounces of liquor both standard drinks? alchohol edu
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12 ounce hard seltzer and 1.5 ounces of liquor are standard drinks because they contain the <u>same amount</u><u> </u><u>of </u><u>pure alcohol</u>

<h3>What is standard drink?</h3>

This is  a term used to refer to the measure of alcoholic content of drinks  such that the drink should have 14 grams of pure alcohol or 0.6 fluid ounces of pure alcohol. This concept of standard drink is applicable in the United States of America.

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2 years ago
Janice has just measured the density of an object. Which value is possible? (Density: D = )
babunello [35]
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3 years ago
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A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car ent
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Answer:

0.8712 m/s²

Explanation:

We are given;

Velocity of first car; v1 = 33 m/s

Distance; d = 2.5 km = 2500 m

Acceleration of first car; a1 = 0 m/s² (constant acceleration)

Velocity of second car; v2 = 0 m/s (since the second car starts from rest)

From Newton's equation of motion, we know that;

d = ut + ½at²

Thus,for first car, we have;

d = v1•t + ½(a1)t²

Plugging in the relevant values, we have;

d = 33t + 0

d = 33t

For second car, we have;

d = v2•t + ½(a2)•t²

Plugging in the relevant values, we have;

d = 0 + ½(a2)t²

d = ½(a2)t²

Since they meet at the next exit, then;

33t = ½(a2)t²

simplifying to get;

33 = ½(a2)t

Now, we also know that;

t = distance/speed = d/v1 = 2500/33

Thus;

33 = ½ × (a2) × (2500/33)

Rearranging, we have;

a2 = (33 × 33 × 2)/2500

a2 = 0.8712 m/s²

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