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3 years ago

12

A kettle full of water is brought to a boil in a room with temperature 20Â°c. after 15 minutes the temperature of the water has

decreased from 100Â° to 75Â°c. find the temperature after another 5 min

1 answer:

kogti [31]3 years ago

4 0

Answer: The temperature after another 5 minutes is 68.5Â°c

Explanation: Please see the attachments below

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(eText prob. 4.25 with some values changed) Air enters a diffuser of a jet engine operating at steady state at 2.65 psia, 389◦R,

krek1111 [17]

Answer:

$V_2 = 45.44m/s$

Explanation:

We have to many data in different system, so we need transform everything to SI, that is

$P_1 = 2.65 Psi = 18.271 kPa\\T_1= 389\°R = 216 K\\V_1 = 869ft/s = 264m/s\\T_2 = 450\°R = 250K$

When we have all this values in SI apply a Energy Balance Equation,

$\dot{Q}_{cv}-\dot{W}_{cv}+\dot{m}[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0$

Solving for V_2

$V_2 = \sqrt{V_1^2+2(h_1-h_2)}$

From the table of gas properties we calculate for $T_1 = 216K$ and $T_2 = 250K$

$h_1 = 209.97+(219.97-209.97)(\frac{216-210}{220-210})$

$h_1 = 215.97kJ/kg$

For T_2;

$h_2 = 250.05kJ/kg$

Substituting in equation for V_2

$V_2 = \sqrt{V_1^2+2(h_1-h_2)}$

$V_2 = \sqrt{265^2+2(215.97-250.05)*10^3}\\V_2 = 45.44m/s$

4 0

3 years ago

5. A man pulls down on a rope to hoist a sail on a sailboat. This is an example of a machine

tino4ka555 [31]

B. changing the direction over which a force is exerted

5 0

2 years ago

A small block is released from rest at the top of a frictionless incline. The distance from the top of the incline to the bottom

ehidna [41]

Complete Question

A small block is released from rest at the top of a frictionless incline. The distance from the top of the incline to the bottom, measured along the incline, is 3.40 m. The vertical distance from the top of the incline to the bottom is 1.09 m. If $g=9.8 \ m/s^2$ , what is the acceleration of the block as it slides down the incline plane

Answer:

The acceleration is $a = 3.142 m/s^2$

Explanation:

From the question we are told that

The distance from top to bottom of the inclined plane measured along the incline is $d = 3.40 \ m$

The distance from top to bottom of the inclined plane measured along the vertical axis is

$D = 1.90 \ m$

According to the SOHCAHTOA rule

$sin \theta = \frac{D}{d}$

=> $\theta = sin ^{-1} [\frac{D}{d} ]$

substituting values

=> $\theta = sin ^{-1} [\frac{1.09}{3.40} ]$

$\theta = 18.699^o$ T

The acceleration of a block on a frictionless inclined plane is mathematically represented as

$a = gsin \theta$

substituting values

$a = 9.8 * sin(18.699)$

$a = 3.142 m/s^2$

8 0

3 years ago

Which is a convex lens? (The blue arrows represent the direction of light as it interacts with the lenses.)

liraira [26]

Answer:

Image (c) is a convex lens

Explanation:

The lenses are of two types i.e. concave lens and convex lens. Figure (3) shows a convex lens. The blue arrows represent the direction of light as it interacts with the lenses. It is also known as converging lens. It is because, it converge all the rays of light that are coming parallel to principal axis at focal point (F).

The image formed by convex lens is real and inverted always. But when the object is placed between the focus and the optical center of the lens, the formed image is virtual and erect.

So, the correct option is (c).

3 0

3 years ago

Read 2 more answers

Suppose I have an infinite plane of charge surrounded by air. What is the maximum charge density that can be placed on the surfa

gregori [183]

Answer:

$53.1\mu C/m^2$

Explanation:

We are given that

Electric field,E= $3\times 10^6V/m$

We have to find the value of maximum charge density that can be placed on the surface of the plane before dielectric breakdown of the surrounding air occurs.

We know that

$E=\frac{\sigma}{2\epsilon_0}$

Where $\epsilon_0=8.85\times 10^{-12}$

Using the formula

$3\times 10^6=\frac{\sigma}{2\times 8.85\times 10^{-12}}$

$\sigma=3\times 10^6\times 2\times 8.85\times 10^{-12}$

$\sigma=5.31\times 10^{-5}C/m^2$

$\sigma=53.1\times 10^{-6}C/m^2=53.1\mu C/m^2$

$1\mu C=10^{-6} C$

3 0

3 years ago