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anygoal [31]
3 years ago
13

During each cycle, a heat engine ejects 75 J of thermal energy for every 115 J of input thermal energy. This engine is used to l

ift a 345-kg load a vertical distance of 27.0 m at a steady rate of 52.5 mm/s. Part A How many cycles of the engine are needed to accomplish this task?
Physics
1 answer:
Nastasia [14]3 years ago
7 0

Answer:

no of cycle required is 1885.3

Explanation:

given data

thermal energy Q1 = 75 J

thermal energy Q2 = 115 J

load = 3454 kg

distance = 27 m

steady rate = 52.5 mm/s

to find out

How many cycles of the engine are need

solution

we find here first work done that is express as

work done W = Q2 - Q1     ..................1

put here all these value

W = 115 - 75

work done is 40 J

so energy required is

Energy required E = m×g×h     ..................2

put here value

E = 285 × 9.8 × 27

energy required = 75411 J

so here no of cycle is

no of cycle = energy required / work done

put here value

no of cycle required = 75411 / 40

so no of cycle required is 1885.3

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Answer: 2.74

Explanation:

We can solve this problem using the stopping distance formula:

d=\frac{(V_{o})^{2}}{2 \mu g}

Where:

d=132.1 m is the distance traveled by the car before it stops

V_{o}=30.7 m/s is the car's initial velocity

\mu is the coefficient of friction between the road and the tires

g=9.8 m/s^{2} is the acceleration due gravity

Isolating \mu:

\mu=\frac{2dg}{(V_{o})^{2}}

Solving:

\mu=\frac{2(132.1 m)(9.8 m/s^{2})}{(30.7 m/s)^{2}}

\mu=2.74 This is the coefficient of friction

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3 years ago
Why the temperature of a resistor increase when a current pass through it
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What are 1A, 3B, and 7A examples of on the periodic table?
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Estimate the kinetic energy of the Mars with respect to the Sun as the sum of the terms, that due to its daily rotation about it
kotykmax [81]

Answer:

K = 2.07 10³⁹ J

Explanation:

This problem must be solved using rotational kinematics.

Kinetic energy has the form

    K = ½ m v²

Linear velocity is related to angular velocity.

    v = w r

replace

    K = ½ m R² w²

With this equation we can find the total kinetic energy of Mars, formed by the rotation energy plus the translational energy

     K = Kr + Kt

Where Kr is the energy by the rotation on its axis and Kt is the energy by the rotation around the sun.

Let's reduce to SI units

Rotation         T₁ = 24.7 h (3600 s / 1 h) = 88920 s

                       R₁ = 3.4 10⁶ m

translation     T₂ = 686 day (24 h / 1 day) (3600 s / 1 h) = 5.927 10⁷ s

                      R₂ = 2.3 10⁸ km (1000 m / 1 km) = 2.3 10¹¹ m

The angular velocity is the angle rotated (radians) between the time taken, in this case as the order gives us the angle is 2pi rad (360º). Remember that the equations work only in radians

Rotation

    wr = 2π / T₁

    wr = 2 π / 88920

    wr = 7.066 10⁻⁵ rad / s

Translation

    wt = 2 π / T₂

    wt = 2 π / 5,927 10⁷

    wt = 1.06 10⁻⁷-7 rad/s

Let's explicitly write the equation of kinetic energy and calculate

    K = ½ m R₁² wr² + ½ m R₂² wt²2

    K = ½ m (R₁² wr² + R₂² wt²)

    K = ½ 6.4 10²³ [(3.4 10⁶)² 7.066² + (2.3 10¹¹)² (1.06 10⁻⁷)²]

    K = 3.2 10²³ [61.49 10¹² + 6.414 10¹⁵]

    K = 3.2 10²³ [61.49 10¹² + 6414 10¹²]

    K = 3.2 10²³ (6475 10¹²)

    K = 2.07 10³⁹ J

5 0
3 years ago
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