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2 years ago

11

A well insulated and perfectly sealed room is installed with 4 HP fan to circulate air for 1.5 hours. Determine the increase in

internal energy of the air answer is 1.61 x10^7 J

1 answer:

marissa [1.9K]2 years ago

7 0

"Increase in internal energy = energy consumed by fan " as the room is perfectly insulated and sealed.

energy consumed by fan = power x time
power of fan = 4HP = 4 x 745.7 W = 2982.8 W
time = 1.5 hours = 1.5 x 60 x 60 = 5400 s

energy consumed by fan = 2982.8 x 5400 = 16107120 J = 1.61x10^7 J

Increase in internal energy is 1.61x10^7 J

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two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0

2 years ago

A ship sends out a 1200 Hz sound wave, which has a wavelength of 120 cm in the water. What would happen if that ship sent out a

OlgaM077 [116]

Answer:

the wave length becomes doubled or becomes two times the initial wavelength = 240 cm

Explanation:

From wave,

v = λf................ Equation 1

Where v = velocity of the wave, λ = wavelength of the wave, f = frequency of the wave.

Given: f = 1200 Hz, λ = 120 cm = 1.2 m

Substitute into equation 1

v = 1200(1.2)

v = 1440 m/s.

When the ship sent out a 600 Hz sound wave,

make λ the subject of formula in equation 1

λ = v/f............. Equation 2

Given: f = 600 Hz, v = 1440 m/s

Substitute into equation 2

λ = 1440/600

λ = 2.4 m or 240 cm.

When the ship sent out a 600 Hz sound wave instead, the wave length becomes doubled or becomes two times the initial wavelength = 240 cm

3 0

2 years ago

How many calories are equal to one BTU? (One calorie = 4.186 J, one BTU = 1 054 J.)

I am Lyosha [343]

<h2>Option C is the correct answer.</h2>

Explanation:

We need to find how many calories is 1 BTU.

Given

1 BTU = 1054 J

1 calorie = 4.186 J

So we have

1 BTU = 4.186 x 251.79 J

1 BTU =251.79 calorie

1 BTU = 252 calorie.

Option C is the correct answer.

3 0

2 years ago

A bar 4.0m long weights 400 N. Its center of gravity is 1.5m from on end. A weight of 300N is attached at the light end. What ar

Debora [2.8K]

Answer:

The resulting, needed force for equilibrium is a reaction from a support, located at 2.57 meters from the heavy end. It is vertical, possitive (upwards) and 700 N.

Explanation:

This is a horizontal bar.

For transitional equilibrium, we just need a force opposed to its weight, thus vertical and possitive (ascendent). Its magnitude is the sum of the two weights, 400+300 = 700 N, since weight, as gravity is vertical and negative.

Now, the tricky part is the point of application, which involves rotational equilibrium. But this is quite simple if we write down an equation for dynamic momentum with respect to the heavy end (not the light end where the additional weight is placed). The condition is that the sum of momenta with respect to this (any) point of the solid bar is zero:

$0=\Sigma_{i}M=400\cdot1.5+300\cdot4-d\cdot700$

Where momenta from weights are possitive and the opposed force creates an oppossed momentum, then a negative term. Solving our unknown d:

$d=\frac{400\cdot1.5+300\cdot4}{700} =2.57 m$

So, the resulting force is a reaction from a support, located at 2.57 meters from the heavy end (the one opposed to the added weight end).

8 0

2 years ago

Anna applies a force of 19.5 newtons to push a book placed on a table. If the normal force of the book is 51.7 newtons, what is

GarryVolchara [31]

that would be given by

[email protected]

@ representing coefficient of kinetic friction.

thus 19.5/51.7 = 0.377

6 0

2 years ago