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soldi70 [24.7K]
3 years ago
12

You are walking from your math class to your science class. You are carrying books

Physics
1 answer:
kolezko [41]3 years ago
8 0

Answer:

1800J

Explanation:

Given parameters:

Weight of the book  = 20N

Total distance covered  = 45m + 15m + 30m  = 90m

Unknown:

Total work performed on the books  = ?

Solution:

To solve this problem we must understand that work done is the force applied to move a body through a certain distance.

So;

    Work done  = Force x distance

  Work done  = 20 x 90  = 1800J

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Two moles of helium are initially at a temperature of 21.0 ∘Cand occupy a volume of 3.30×10−2 m3 . The helium first expands at c
Anettt [7]

Answer:

(B) The total internal energy of the helium is 4888.6 Joules

(C) The total work done by the helium is 2959.25 Joules

(D) The final volume of the helium is 0.066 cubic meter

Explanation:

(B) ∆U = P(V2 - V1)

From ideal gas equation, PV = nRT

T1 = 21°C = 294K, V1 = 0.033m^3, n = 2moles, V2 = 2× 0.033=0.066m^3

P = nRT ÷ V = (2×8.314×294) ÷ 0.033 = 148140.4 Pascal

∆U = 148140.4(0.066 - 0.033) = 4888.6 Joules

(C) P2 = P1(V1÷V2)^1.4 =148140.4(0.033÷0.066)^1.4= 148140.4×0.379=56134.7 Pascal

Assuming a closed system

(C) Wc = (P1V1 - P2V2) ÷ 0.4 = (148140.4×0.033 - 56134.7×0.066) ÷ 0.4 = (4888.6 - 3704.9) ÷ 0.4 = 1183.7 ÷ 0.4 = 2959.25 Joules

(C) Final volume = 2×initial volume = 2×0.033= 0.066 cubic meter

6 0
3 years ago
What type of organization is used in a paragraph that lists similarities between two objects?
alekssr [168]
<span>A. Comparison 

</span>What type of organization is used in a paragraph that lists similarities between two objects? Comparison


NOT:
<span>B. Contrast 
C. Chronological order 
D. Cause and effect</span><span>
</span>
3 0
3 years ago
1. An electron travels 4.82 meters in 0.00360 seconds. What is its average speed?
vredina [299]

Answer:

speed =distance /time

speed =4.82/0.00360

speed =1338.8m/s

6 0
2 years ago
Help with these please someone?
DIA [1.3K]

Answer:

8. 2.75·10^-4 s^-1

9. No, too much of the carbon-14 would have decayed for radiation to be detected.

Explanation:

8. The half-life of 42 minutes is 2520 seconds, so you have ...

1/2 = e^(-λt) = e^(-(2520 s)λ)

ln(1/2) = -(2520 s)λ

-ln(1/2)/(2520 s) = λ ≈ 2.75×10^-4 s^-1

___

9. Reference material on carbon-14 dating suggests the method is not useful for time periods greater than about 50,000 years. The half-life of C-14 is about 5730 years, so at 65 million years, about ...

6.5·10^7/5.73·10^3 ≈ 11344

half-lives will have passed. Whatever carbon 14 may have existed at the time will have decayed completely to nothing after that many half-lives.

7 0
3 years ago
Suppose that during any period of ¼ second there is one instant at which the crests or troughs of component waves are exactly in
kolezko [41]
<h3><u>Answer;</u></h3>

A. 4

<h3><u>Explanation;</u></h3>
  • <em><u>The period of a wave or periodic time is the time taken for a complete oscillation to occur. </u></em>For example its is the time taken between two successive crests or troughs.
  • <em><u>The beats or oscillation that occur in one second represents the frequency. Frequency is the number of complete oscillations or beats in one second in a wave.</u></em>
  • Frequency, measured in Hertz is given by the reciprocal of the periodic time.
  • Thus; <u><em>Frequency or beats per second = 1/(1/4) = 4</em></u>
  • <u><em>Hence , 4 beats per second</em></u>

6 0
3 years ago
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