All categories

3 years ago

A Carnot engine has an efficiency of 0.537, and the temperature of its cold reservoir is 379 K.

Physics

1 answer:

katen-ka-za [31]3 years ago

8 0

Answer:

(A) Th = 818.6 K

(B) Qh = 14211.7 J

Explanation:

efficiency (n) = 0.537

temperature of cold reservoir (Tc) = 379 K

heat rejected (Qc) = 6580 J

(A) find the temperature of the hot reservoir (Th)

n = 1 - $\frac{Tc}{Th}$

0.537 = 1 - $\frac{379}{Th}$

$\frac{379}{Th}$ = 1 - 0.537 = 0.463

Th = $\frac{379}{0.463}$

Th = 818.6 K

(B) what amount of heat is put into the engine (Qh) ?

from $\frac{Tc}{Th} = \frac{Qc}{Qh}$

Qh = 6580 ÷ $\frac{379}{818.6}$

Qh = 14211.7 J

You might be interested in

What linear speed must an Earth satellite have to be in a circular orbit at an altitude of 203 km above Earth's surface

juin [17]

Answer:

7,790.38 m/s

Explanation:

Given the following :

What linear speed must an Earth satellite have to be in a circular orbit at an altitude of 203 km above Earth's surface

Altitude = 203 km

Using the formula :

V = √GM/r

Where G = gravitational constant =6.67×10^-11

Kindly check attached picture for detailed explanation.

3 0

3 years ago

How does air resistance affect an object’s speed?

tatuchka [14]

When air resistance acts, acceleration during a fall will be less than g because air resistance affects the motion of the falling objects by slowing it down. Air resistance depends on two important factors - thespeed of the object and its surface area. Increasing the surface area of an object decreases its speed.

4 0

3 years ago

A radar station detects an airplane approaching directly from the east. At first observation, the range to the plane is 375.0 m

Gekata [30.6K]

Answer:

819.78 m

Explanation:

Given:

OA = range of initial position of the airplane from the point of observation = 375 m
OB = range of the final position of the airplane from the point of observation = 797 m
$\theta$ = angle of the initial position vector from the observation point = $43^\circ$
$\alpha$ = angle of the final position vector from the observation point = $123^\circ$
$\vec{AB}$ = displacement vector from initial position to the final position

A diagram has been attached with the solution in order to clearly show the position of the plane.

$\vec{OA} = OA\cos \theta \hat{i}+OA \sin \theta \hat{j}\\\Rightarrow \vec{OA} = 375\ m\cos 43^\circ \hat{i}+375\ m\sin 43^\circ \hat{j}\\\Rightarrow \vec{OA} = (274.26\ \hat{i}+255.75\ \hat{j})\ m\\\vec{OB} = OB\cos \alpha \hat{i}+OB \sin \alpha \hat{j}\\\Rightarrow \vec{OB} = 797\ m\cos 123^\circ \hat{i}+797\ m\sin 123^\circ \hat{j}\\\Rightarrow \vec{OB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m$

Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:

$\vec{AB}=\vec{OB}-\vec{OA}\\\Rightarrow \vec{AB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m-(274.26\ \hat{i}+255.75\ \hat{j})\ m\\\Rightarrow \vec{AB} = (-708.34\ \hat{i}+412.67\ \hat{j})\ m$