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3 years ago

A certain heat engine does 30.2 kJ of work and dissipates 9.14 kJ of waste heat in a cyclical process.

Physics

1 answer:

valina [46]3 years ago

4 0

Answer:

a) $H_{in}=39.34 kJ$

b) Efficiency=76.77%

Explanation:

In order to solve this problem, we can use the following formula:

$H_{in}=H_{out}+W$

the problem provides us with all the necessary information so we can directly use the formula:

$H_{in}=9.14kJ+30.2kJ$

$H_{in}=39.34 kJ$

b) In order to find the efficiency, we can use the following formula:

$Efficiency=\frac{W}{H_{in}}*100\%$

so we get:

$Efficiency=\frac{30.2kJ}{39.34kJ}*100\%$

Efficiency=76.77%

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A solid 0.4550 kg ball rolls without slipping down a track toward a vertical loop of radius R = 0.6750 m . What minimum translat

Talja [164]

Answer:

u = 3.35 m/s

Explanation:

given,

mass , m = 0.455 kg

R = 0.675 m

Height of Loop = 1.021 m

the speed required at the top of loop be v

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$m g =\dfrac{mv^2}{r}$

$9.81 =\dfrac{v^2}{0.675}$

v² = 6.622

v = 2.57 m/s

Let the initial speed of ball be u

using conservation of energy

$\dfrac{1}{2}mu^2 + \dfrac{1}{2}I\omega^2 + m g h = \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2+ m g (2 R)$

where, $I =\dfrac{2}{5}mr^2$

$\dfrac{1}{2}mu^2 + \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{u}{r})^2 + m g h = \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{v}{r})^2 + \dfrac{1}{2}mv^2+ m g (2 R)$

$0.7 u^2 + g H = 0.7 v^2 + g(2R)$

$0.7 u^2 +9.81 \times 1.021= 0.7\times 2.57^2 + 9.81 \times 2\times 0.675)$

0.7 u² = 7.85092

u² = 11.2156

u = 3.35 m/s

the initial speed is 3.35 m/s

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3 years ago

A stone is dropped from the upper observation deck of a tower, "500" m above the ground. (Assume g = 9.8 m/s2.) (a) Find the dis

Yuri [45]

Answer:

h₍₁₎ = 495,1 meters

h₍₂₎ = 480,4 m

h₍₃₎ = 455,9 m

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Explanation:

The exercise is "free fall". t = $\sqrt{\frac{2h}{g} }$

Solving with this formula you find the time it takes for the stone to reach the ground (T) = 102,04 s

The heights (h) according to his time (t) are found according to the formula:

h(t) = 500 - 1/2 * g * t²

Remplacing "t" with the desired time.

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