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brilliants [131]
3 years ago
13

Claire is traveling home when her mother calls and asks her to go to her grandma's house. She is 300 meters north of her house w

hen she gets the call. She travels 100 meters south of her house to Grandma's house. Her entire trip tool 400 seconds. What was Claire's average velocity?
Physics
1 answer:
m_a_m_a [10]3 years ago
5 0

Answer:

you divide distance traveled by the time it takes to travel that distance. average speed. If your speed changes from 10 km/h to 6 km/h, you have a(n) ... 400 km. Suppose that the average speed your dog can run is 3 m/s. ... she drives her scooter 7 kilometres north. She stops for lunch and then drives 5 kilometres south.

i rlly dunno lma.o

Explanation:

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Answer:

Greatest Achievements of Mankind

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(please mark brainliest if correct/helped you <3)

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A light woman and a heavy man jump from an airplane at the same time and open their same-size parachutes at the same time. which
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It is B particle sizes
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How do I set up this equation?
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4 0
3 years ago
A particular heat engine has a mechanical power output of 4.00 kW and an efficiency of 26.0%. The engine expels 8.55 103 J of ex
Ivahew [28]

To develop the problem we will start by finding the energy taken by each cycle through the efficiency of the motor and the exhausted energy. Later the work will be found for the conservation of energy in which this is equivalent to the difference between the two calculated energy values. Finally the estimated time will be calculated with the work and the power given,

\text{Efficiency of the heat engine} = \eta = 26\% = 0.26

\text{Energy taken in by the heat engine during each cycle} = Q_h

\text{Energy exhausted by the heat engine in each cycle} = Q_c = 8.55*10^3 J

\eta = 1 - \frac{Q_{c}}{Q_{h}}

0.26 = 1 - \frac{8.55\ast 10^{3}}{Q_{h}}

\frac{8.55* 10^{3}}{Q_{h}} = 0.74

Q_h = \frac{8.55*10^3}{0.74}

Q_h = 11.554*10^3J

PART A)

Work done by the heat engine in each cycle = W

W = Q_h-Q_c

W = 11.554*10^3J-8.55*10^3J

W = 3004J

According to the value given we have that,

P = 4.0kW

P = 4000W

Power is defined as the variation of energy as a function of time therefore,

P = \frac{W}{t}

4000W = \frac{3004J}{t}

t = \frac{3004}{4000}

t = 0.75s

Therefore the interval for each cycle is 0.75s

5 0
3 years ago
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