Ask question

Ask question

All categories

3 years ago

9

The engine oil at 150 degree Celsius is cooled to 80 degree Celsius in a parallel flow heat exchanger by water entering at 25 de

gree Celsius and leaving at 60 degree Celsius. Estimate the number of transfer units If you think you require the specific heats for water and oil, use 4.18 and 2.20 respectively

1 answer:

Setler [38]3 years ago

6 0

Answer:

Explanation:

Attach is the solution

You might be interested in

Examine the scenario. Object A has 5 protons and 5 electrons. Object B has 5 protons and 7 electrons. Which option most accurate

Travka [436]

<span>Last choice on the list:
Object A has a net charge of 0 because the positive and negative
charges are balanced.
Object B has a net charge of –2 because there is an imbalance of
charged particles (2 more negative electrons than positive protons).</span>

4 0

3 years ago

A 3.5 x 10-6 C charge is located 0.28 m from a 2.8 x 10-6 C charge. What is the magnitude of the force being exerted on the smal

Margarita [4]

The electrostatic force between two charges is given by Coulomb's law:
$F=k_e \frac{q_1 q_2}{r^2}$
where
ke is the Coulomb's constant
q1 is the first charge
q2 is the second charge
r is the separation between the two charges

By substituting the data of the problem into the equation, we can find the magnitude of the force between the two charges:
$F=(8.99 \cdot 10^{9} Nm^2C^{-2} ) \frac{(3.5 \cdot 10^{-6} N/C)(2.8 \cdot 10^{-6}N/C)}{(0.28m)^2}=1.2 N$

7 0

3 years ago

A speeder passes a parked police car at a constant speed of 23.3 m/s. At that instant, the police car starts from rest with a un

KonstantinChe [14]

Answer:

32s

Explanation:

We must establish that by the time the police car catches up to the speeder, both have travelled a certain distance during the same amount of time. However, the police car experiences accelerated motion whereas the speeder travels at a constant velocity. Therefore we will establish two formulas for distance starting with the speeder's distance:

$x=vt=23.3\frac{m}{s}t$

and the police car distance:

$x=vt+\frac{at^{2}}{2}=0+\frac{2.75\frac{m}{s^{2}} t^{2}}{2}=0.73\frac{m}{s^{2}}$

Since they both travel the same distance x, we can equal both formulas and solve for t:

$0 = 0.73\frac{m}{s^{2}}t^{2}-23.3\frac{m}{s} t\\\\0=t(0.73\frac{m}{s^{2}}t-23.3\frac{m}{s} )\\\\$

Two solutions exist to the equation; the first one being $t=0$

The second solution will be:

$0.73\frac{m}{s^{2}}t=23.3\frac{m}{s}\\\\t=\frac{23.3\frac{m}{s}}{0.73\frac{m}{s^{2}}}=32s$

This result allows us to confirm that the police car will take 32s to catch up to the speeder

7 0

3 years ago

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

3 years ago

At t = 0 the components of a radio-controlled car's velocity are vx = 0.5 m/s and vy = 1.2 m/s. Find the components of the car's

FinnZ [79.3K]

Answer:

x-component of velocity = 5.7 m/s

y-component of velocity = -1.4 m/s

Explanation:

Use first equation of motion to find components of velocity at a given time:

$v = u + at$

where, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.

Given:

$u_x= 0.5 m/s\\u_y=1.2 m/s\\a_x=2 m/s^2\\a_y=-1m/s^2\\t = (2.6-0)s =2.6 s$

$v_x=u_x+a_xt\\\Rightarrow v_x=0.5+2\times 2.6\\\Rightarrow v_x=5.7 m/s$

$v_y=u_y+a_yt\\\Rightarrow v_y=1.2-1\times 2.6\\\Rightarrow v_y=-1.4 m/s$

5 0

3 years ago