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3 years ago

12

In one cycle, a heat engine takes in 1000 J of heat from a high-temperature reservoir, releases 600 J of heat to a lower-tempera

ture reservoir, and does 400 J of work. What is its efficiency?

1 answer:

Talja [164]3 years ago

4 0

Answer:

η = 40 %

Explanation:

Given that

Qa ,Heat addition= 1000 J

Qr,Heat rejection= 600 J

Work done ,W= 400 J

We know that ,efficiency of a engine given as

$\eta=\dfrac{W(net)}{Q(heat\ addition)}$

Now by putting the values in the above equation ,then we get

$\eta=\dfrac{400}{1000}$

η = 0.4

The efficiency in percentage is given as

η = 0.4 x 100 %

η = 40 %

Therefore the answer will be 40%.

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Plz help it really easy

Mrac [35]

The answer is high to low.

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2 years ago

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17. Which of the following is the closest weight in newtons of a 7.0-kilogram

marusya05 [52]

Weight = (mass) x (gravity)

Weight = (7.0 kg) x (gravity)

On Earth, where (gravity) is roughly 10 N/kg . . .

Weight = (7.0 kg) x (roughly 10 N/kg)

Weight = roughly 70 Newtons

That's <em>B </em>on Earth.

It would be some other number on other bodies.

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3 years ago

A typical neutron star may have a mass equal to that of the Sun but a radius of only 20 km.(a) What is the gravitational acceler

Pepsi [2]

Answer:

$331665750000\ m/s^2$

3257806.62409 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Sun = $1.989\times 10^{30}\ kg$

r = Radius of Star = 20 km

u = Initial velocity = 0

v = Final velocity

s = Displacement = 16 m

a = Acceleration

Gravitational acceleration is given by

$g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2$

The gravitational acceleration at the surface of such a star is $331665750000\ m/s^2$

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 331665750000\times 16+0^2}\\\Rightarrow v=3257806.62409\ m/s$

The velocity of the object would be 3257806.62409 m/s

8 0

3 years ago

A glass tube 1.50 meters long and open at one end is weighted to keep it vertical and is then lowered to the bottom of a lake. W

Lilit [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The total pressure is $P_T = 10.79*10^{5} N/m^2$

The temperature at the bottom is $T_b = 284.2 \ K$

Explanation:

From the question we are told that

The length of the glass tube is $L = 1.50 \ m$

The length of water rise at the bottom of the lake $d = 1.33 \ m$

The depth of the lake is $h = 100 \ m$

The air temperature is $T_a = 27 ^oC = 27 +273 = 300 \ K$

The atmospheric pressure is $P_a = 1.01 *10^{5} N/m$

The density of water is $\rho = 998 \ kg/m^3$

The total pressure at the bottom of the lake is mathematically represented as

$P_T = P_a + \rho g h$

substituting values

$P_T = 1.01*10^{5} + 998 * 9.8 * 100$

$P_T = 10.79*10^{5} N/m^2$

According to ideal gas law

At the surface the glass tube not covered by water at surface

$P_a V_a = nRT_a$

Where is the volume of

$P_a *A * L = nRT_a$

At the bottom of the lake

$P_T V_b = nRT_b$

Where $V_b$ is the volume of the glass tube not covered by water at bottom

and $T_b$ i the temperature at the bottom

So the ratio between the temperature at the surface to the temperature at the bottom is mathematically represented as

$\frac{T_b}{T_a} = \frac{d * P_T}{P_a * h}$

substituting values

$\frac{T_b}{27} = \frac{0.133 * 10.79 *10^5}{1.01 *10^{5} * 1.5}$

=> $T_b = 284.2 \ K$

3 0

2 years ago

A Ferris wheel with a diameter of 35 m starts from rest and achieves its maximum operational tangential speed of 2.3 m/s in a ti

Butoxors [25]

Answer: 8.8e-3 rad/s², 0.0 m/s²

Explanation:

R = 35 / 2 = 17.5 m

a = 2.3 m/s / 15 s = 0.1533333... m/s²

α = a/R = 0.1533333/17.5 = 0.0087619... ≈ 8.8e-3 rad/s² ◄(a)

at maximum speed, no more acceleration in either tangential or angular

α = a = 0.0 rad/s² or m/s² ◄(b)

5 0

2 years ago