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4 years ago

11

An engine absorbs 2150 J as heat from a hot reservoir and gives off 750 J as heat to a cold reservoir during each cycle. What is

the efficiency of the engine?

1 answer:

tia_tia [17]4 years ago

6 0

Answer:

65.2 %

Explanation:

Let Q1 = Heat absorbed by the system

Q2 = Heat released by the system

e= (1 - (Q2/Q1)) x 100

e= (1 - (750/2150)) x 100

e= (1 - 0.348) x 100

e= 0.652 x 100

e= 65.2 %

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What can you conclude about the pendulum at the instant shown by the graph below (just past 2 s)?

Semenov [28]

Below are the chocies
A. The pendulum is at its highest point.
<span>B. The pendulum is 0.9 m above its lowest point. </span>
<span>C. The pendulum is between its highest and lowest points. </span>
<span>D. The pendulum is at its lowest point.
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I think the answer is C..

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4 years ago

Assuming a roughly spherical shape and a density of 4000 kg/m3, estimate the diameter of an asteroid having the average mass of

vredina [299]

Explanation:

It is given that,

Density of asteroid, $\rho=4000\ kg/m^3$

Mass of asteroid, $m=10^{17}\ kg$

We need to find the diameter of the asteroid. The formula of density is given by:

$\rho=\dfrac{m}{V}$

V is the volume of spherical shaped asteroid, $V=\dfrac{4}{3}\pi r^3$

$r^3=\dfrac{3M}{4\pi \rho}$

$r^3=\dfrac{3\times 10^{17}\ kg}{4\pi \times 4000\ kg/m^3}$

$r^3=\sqrt{5.96\times 10^{12}}$

r = 2441311.12 m

Diameter = 2 × radius

d = 4882622.24 m

or

$d=4.88\times 10^6\ m$

Hence, this is the required solution.

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3 years ago

The weight of a girl with a mass of 40.0 kg is_N

faltersainse [42]

Answer: The weight of a girl with a mass of 40kg is 392.266 Newtons.

Explanation:

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3 years ago

Plants have organelles called

Natasha_Volkova [10]

Answer:

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4 years ago

A 70.0-kg skier is sliding at 4 m/s when they slide down a 2m high hill. At the bottom of the hill they run into a large 2800 N/

Katarina [22]

Answer:

$\Delta x=245\ mm$

Explanation:

Given:

mass of skier, $m=70\ kg$

initial velocity of skier, $u=4\ m.s^{-1}$

height of the hill, $h=2\ m$

spring constant, $k=2800\ N.m^{-1}$

<u>final velocity of skier before coming in contact of spring:</u>

Using eq. of motion:

$v^2=u^2+gh$

$v^2=4^2+9.8\times 2$

$v=5.9666\ m.s^{-1}$

<u>Now the time taken by the skier to reach down:</u>

$v=u+gt$

$5.9666=4+9.8\ t$

$t=0.2007\ s$

<u>Now we calculate force using Newton's second law:</u>

$F=\frac{dp}{dt}$

$F=\frac{m(v-u)}{t}$

$F=\frac{70\times(5.9666-4)}{0.2007}$

$F\approx686\ N$

<u>∴Compression in spring before the skier momentarily comes to rest:</u>

$\Delta x=\frac{F}{k}$

$\Delta x=\frac{686}{2800}$

$\Delta x=0.245\ m$

$\Delta x=245\ mm$

4 0

3 years ago