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3 years ago

How to calculate overall heat transfer coefficient for cross flow Air Preheater as need to design Air preheater?

Physics

2 answers:

JulijaS [17]3 years ago

5 0

Answer:

here lol......................

............

Irina-Kira [14]3 years ago

3 0

answer:

write number easy

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A cannonball is launched from the ground at an angle of 30 degrees above the horizontal and a speed of 30 m/s. Ideally (no air r

NikAS [45]

As we know that here no air resistance while ball is moving in air

So here we will say that

initial total energy = final total energy

$KE_i + U_i = KE_f + U_f$

here we know that

$Ui = U_f = 0$ (as it will be on ground at initial and final position)

so we will say

$KE_i = KE_f$

since mass is always conserved

so we will say that final speed of the ball must be equal to the initial speed of the ball

so we have

$v_f = v_i = 30 m/s$

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3 years ago

Bi Name two processes by which a hot drinks cool

7nadin3 [17]

Wind,cold water, ice

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3 years ago

Question 2:

Vesna [10]

Explanation:

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3 years ago

Read 2 more answers

Sam is pulling a box up to the second story of his apartment via a string. The box weighs 92.6 kg and starts from rest on the gr

grin007 [14]

Answer: 2.7 seconds

Explanation:

We only want to answer: How long until the box reaches a height of 13.1 m?

Then we only must integrate the movement equations.

We know that the velocity of the box increases by 3.6 m/s every second, then the acceleration is constant, and can be written as:

a(t) = 3.6m/s^2

Now, for the velocity, we should integrate over time, and because we know that the box starts from rest, the initial velocity (the constant of integration) will be zero.

v(t) = (3.6m/s^2)*t

For the position equation we should integrate again over time, and if we define the position 0 as the ground, we know that the box starts at the ground, then the initial position (the constant of integration) will be zero.

p(t) = (1/2)*(3.6m/s^2)*t^2.

Now we want to find how long will take until the height of the box is equal to 13.1m

Then we must solve:

p(t) = 13.1m = (1/2)*(3.6m/s^2)*t^2

Let's solve this for t.

13.1m = (1/2)*(3.6m/s^2)*t^2

13.1m*2 = (3.6m/s^2)*t^2

26.2m/(3.6m/s^2) = t^2

7.27<u>7</u>.... s^2 = t^2

√(7.27<u>7</u>.... s^2) = t = 2.7 seconds.

So it will take 2.7 seconds for the box to reach the height of 13.1m

4 0

3 years ago

There is a spot of paint on the front wheel of the bicycle. Take the position of the spot at time t=0 to be at angle θ=0 radians

mario62 [17]

Here is the missing information.

An exhausted bicyclist pedal somewhat erraticaly when exercising on a static bicycle. The angular velocity of the wheels takes the equation ω(t)=at − bsin(ct) for t≥ 0, where t represents time (measured in seconds), a = 0.500 rad/s2 , b = 0.250 rad/s and c = 2.00 rad/s .

Answer:

0.793 rad

Explanation:

From the given question:

The angular velocity of the wheel is expressed by the equation:

$\omega (t) =\dfrac{d\theta}{dt}$

The angular velocity of the wheels takes the description of the equation ω(t)=at−bsin(ct)

SO;

$\dfrac{d \theta}{dt} = at - b \ sin \ ct$

dθ = at dt - (b sin ct) dt

Taking the integral of the above equation; we have:

$\int \limits^{\theta}_{0} \ d \theta = \int \limits ^{t=2}_{0} at \ dt - (b \ sin \ ct) \dt$

$[\theta] ^{\theta}_{0} = a \bigg [\dfrac{t^2}{2} \bigg]^2_0 - \bigg[ -\dfrac{b}{c} \ cos \ ct \bigg] ^2_0$

where;

a = 0.500 rad/s2 ,

b = 0.250 rad/s and

c = 2.00 rad/s

$\theta = (0.500 \ rad/s^2 ) \bigg [\dfrac{(2s)^2}{2} \bigg] - \bigg[ -\dfrac{0.250 \ rad/s}{2.00 \ rad/s} \ cos \ (2.00 \ rad/s )( 2.00 \ s) \bigg] - \bigg [ \dfrac{0.250 \ rad/s}{2.00 \ rad/s}\bigg ] cos 0^0$

$\mathbf{\theta = 0.793 \ rad}$

Hence, the angular displacement after two seconds = 0.793 rad

3 0

3 years ago

How to calculate overall heat transfer coefficient for cross flow Air Preheater as need to design Air preheater?​

How to calculate overall heat transfer coefficient for cross flow Air Preheater as need to design Air preheater?