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Karo-lina-s [1.5K]

3 years ago

11

A car travels at a constant 1km per hour in the positive direction down a straight line highway. What is the displacement of the

car 45 seconds later?

1 answer:

svlad2 [7]3 years ago

5 0

Explanation:

1 km/hr × (1000 m/km) × (1 hr / 3600 s) × 45 s = 12.5 m

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A Carnot engine operates between two heat reservoirs at temperatures THTH and TCTC. An inventor proposes to increase the efficie

Marat540 [252]

Answer:

e_12=1-Tc/Th

This is same as the original Carnot engine.

Explanation:

For original Carnot engine, its efficiency is given by

e = 1-Tc/Th

For the composite engine, its efficiency is given by

e_12=(W_1+W_2)/Q_H1

where Q_H1 is the heat input to the first engine, W_1 s the work done by the first engine and W_2 is the work done by the second engine.

But the work done can be written as

W= Q_H + Q_C with Q_H as the heat input and Q_C as the heat emitted to the cold reservoir. So.

e_12=(Q_H1+Q_C1+Q_H2+Q_C2)/Q_H1

But Q_H2 = -Q_C1 so the second and third terms in the numerator cancel

each other.

e_12=1+Q_C2/Q_H1

but, Q_C2/Q_H2= -T_C/T'

⇒ Q_C2 = -Q_H2(T_C/T')

= Q_C1(T_C/T')

(T1 is the intermediate temperature)

But, Q_C1 = -Q_H1(T'/T_H)

so, Q_C2 = -Q_H1(T'/T_H)(T_C/T') = Q_H1(T_C/T_H) So the efficiency of the composite engine is given by

e_12=1-Tc/Th

This is same as the original Carnot engine.

7 0

3 years ago

Which of the following occurs when light is reflected from a rough or unpolished surface

valentinak56 [21]

It gets blurred and you can't see the light very well.

6 0

3 years ago

g A rotating wheel requires 5.00 s to rotate 28.0 revolutions. Its angular velocity at the end of the 5.00-s interval is 96.0 ra

Setler [38]

Answer:

The angular acceleration of the wheel is 15.21 rad/s².

Explanation:

Given that,

Time = 5 sec

Final angular velocity = 96.0 rad/s

Angular displacement = 28.0 rev = 175.84 rad

Let $\alpha$ be the angular acceleration

We need to calculate the angular acceleration

Using equation of motion

$\theta=\omega_{i} t+\dfrac{1}{2}\alpha t^2$

Put the value in the equation

$175.84=\omega_{i}\times 5+\dfrac{1}{2}\times\alpha\times(5)^2$

$175.84=\omega_{i}\times 5+12.5\alpha$ ......(I)

Again using equation of motion

$\omega_{f}=\omega_{i}+\alpha t$

Put the value in the equation

$96.0=\omega_{i}+\alpha \times 5$

On multiply by 5 in both sides

$480=\omega_{i}\times 5+\alpha\times 25$ ....(II)

On subtract equation (I) from equation (II)

$480-175.84=\alpha(25-5)$

$304.16=\alpha\times20$

$\alpha=\dfrac{304.16}{20}$

$\alpha=15.21\ rad/s^2$

Hence, The angular acceleration of the wheel is 15.21 rad/s².

5 0

3 years ago

An athlete swings a 5.00-kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.800 m at an angular s

iogann1982 [59]

Answer:

0.32m/s2

Explanation:

obtained from a=wr2 where w=anular speed

r-radius

3 0

3 years ago

A baseball player hits a ball in such a way that it leaves bat with an initial speed of 40.0 m/s at 38.0 degrees.

Serggg [28]

Answer:

d

Explanation:

8 0

3 years ago