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inn [45]
3 years ago
13

An engine uses 5280 j of input heat to do 702 j of work.how much heat is rejected into the air

Physics
1 answer:
navik [9.2K]3 years ago
6 0

Answer:

4578 J

Explanation:

INPUT ENERGY = WORK + ENERGY REJECTED

or  ENERGY REJECTED = INPUT ENERGY - WORK

                                         = 5280-702

                                          = 4578J

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5. A car accelerates from 0 to 72 km/hour in 8.0 seconds. What is the car's acceleration?
MA_775_DIABLO [31]

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2.5 m/s

Explanation:

There are calculators online that can help you easily calculate the accerlation.

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2 years ago
If a baseball is in space and someone hits it with a bat what does the ball do
Darina [25.2K]

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It's will fly like float

Explanation:

Since in space there is no gravity that everything can float

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3 years ago
The Atomic number tells us the number of ____ in an atom.
yawa3891 [41]

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Protons

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5 0
3 years ago
Read 2 more answers
The PE of the box that is on a 2.0 m high self is 1600 J. What is the power expelled to lift the box to this height in 10.0 seco
Pani-rosa [81]

Answer:

  160 W

Explanation:

Power is the ratio of work to time:

  (1600 J)/(10 s) = 160 J/s = 160 W

7 0
3 years ago
A,b, e are complete. Help on the others would be so appreciated!!
bixtya [17]

Answer:

serie Ceq=0.678 10⁻⁶ F  and the charge Q = 9.49 10⁻⁶ C

Explanation:

Let's calculate all capacity values

a) The equivalent capacitance of series capacitors

    1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3 + 1 / C4 + 1 / C5

    1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 1 / 6.2 + 1 / 6.2

    1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 2 / 6.2

    1 / Ceq = 0.666 + 0.3030 +0.1818 +0.3225

    1 / Ceq = 1,147

    Ceq = 0.678 10⁻⁶ F

b) Let's calculate the total system load

   Dv = Q / Ceq

   Q = DV Ceq

   Q = 14 0.678 10⁻⁶

   Q = 9.49 10⁻⁶ C

In a series system the load is constant in all capacitors, therefore, the load in capacitor 5.5 is Q = 9.49 10⁻⁶ C

c) The potential difference

   ΔV = Q / C5

   ΔV = 9.49 10⁻⁶ / 5.5 10⁻⁶

   ΔV = 1,725 ​​V

d) The energy stores is

    U = ½ C V²

    U = ½ 0.678 10-6 14²

    U = 66.4 10⁻⁶ J

e) Parallel system

   Ceq = C1 + C2 + C3 + C4 + C5

   Ceq = (1.5 +3.3 +5.5 +6.2 +6.2) 10⁻⁶

   Ceq = 22.7 10⁻⁶ F

f) In the parallel system the voltage is maintained

   Q5 = C5 V

   Q5 = 5.5 10⁻⁶ 14

   Q5 = 77 10⁻⁶ C

g) The voltage is constant V5 = 14 V

h) Energy stores

   U = ½ C V²

   U = ½ 22.7 10-6 14²

   U = 2.2 10⁻³ J

8 0
3 years ago
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