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sergeinik [125]

1 year ago

15

an auditorium measures 30.0 m ✕ 15.0 m ✕ 5.0 m. the density of air is 1.20 kg/m3. (a) what is the volume of the room in cubic fe

et? 79458 correct: your answer is correct. ft3 (b) what is the weight of air in the room in pounds?

1 answer:

Fynjy0 [20]1 year ago

5 0

dimension = 30.0 m ✕ 15.0 m ✕ 5.0 m.

density = 1.20 kg/m3

(a)volume = lenght * breadth * height

= 30 * 15 * 5

= 2250 metre cube = 2.25 cubic meter

(b) mass of air = density * volume

mass of air = 1.2 * 2250

mass of air = 2700kg

weight = mass * 9.8

= 2700 * 9.8

= 26,460 N

The definition of Density is the amount of matter in a given space, or volume
Density = mass/volume
units for density kg/m^3
Density of water 1g/ml
Salt water is denser that is why don't sink as easily.

To know more about density visit : brainly.com/question/15164682

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In our Solar System, the inner planets are rocky because Choose one: A. warm temperatures in the inner disk caused the inner pla

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The inner planets are rocky because The warm temperatures in the inner disk caused the inner planetesimals to be formed of mostly rocky material.

What are rocky planets?

Rocky planets are the planet in which constituents are mostly silicate rocks or metal. They are also regarded as a planet with a solid surface.
The formation of rocky planets is said to have occurred billions of years ago and its process of formation is termed accretion. Through accretion are its constituents formed as the more it goes bigger, the higher the rising temperature and pressure in its core and the elements which have to undergo accreted heat up, melt, and spread. Through this process, heavier elements go deeper into the core of the planet and lighter elements float toward the surface.
In the formation of rocky planets, the inner portions of the disk are said to be warm from the protostar thereby resulting in the production of the heavy elements that stay there.

Examples of rocky planets are Earth or Mars

Hence, from the above, we can say that,

The warm temperatures in the inner disk caused the inner planetesimals to be formed of mostly rocky material.

Here,

Option A is correct.

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3 0

2 years ago

1. A student lifts a box of books that weighs 185 N. The box is

aksik [14]

1) 148 J

When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:

$W = \Delta U = (mg) \Delta h$

where

mg is the weight of the object

$\Delta h$ is the change in height

For the box in this problem,

mg = 185 N

$\Delta h = 0.800 m$

Substituting into the equation, we find:

$W=(185)(0.800)=148 J$

2) (a) 28875 J

The work done by a force applied parallel to the direction of motion of the object is given by

$W=Fd$

where

F is the magnitude of the force

d is the displacement

In this problem,

F = 825 N is the force applied by the two students together

d = 35 m is the displacement of the car

Substituting,

$W=(825)(35)=28875 J$

2) (b) 57750 J

As seen previously, the equation that gives the work done by the force is

$W=Fd$

We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.

The work done previously was

W = 28875 J

Now the force is doubled, so the new work done will be

$W' = 2(28875)=57750 J$

3) 4.4 J

In this case, the force acting on the ball is the force of gravity, whose magnitude is:

$F = mg$

where

m = 0.180 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

Solving the equation,

$F=(0.180)(9.8)=1.76 N$

Now we find the work done by gravity using the same formula applied before:

$W=Fd$

where d = 2.5 m is the displacement of the ball. We can apply this version of the formula since the force is parallel to the displacement. Substituting,

$W=(1.76)(2.5)=4.4 J$

4) 595.2 kg

In this case, we have the work done on the box:

W = 7.0 kJ = 7000 J

And we also know the change in height of the box:

$\Delta h = 1.2 m$

As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:

$W=mg \Delta h$

Solving for m, we find

$m=\frac{W}{g \Delta h}$

And substituting the numerical values, we find the mass of the box:

$m=\frac{7000}{(9.8)(1.2)}=595.2 kg$

5) They do the same work

In fact, the net work done by each person on the box is equal to the change in gravitational potential energy of the box:

$W=mg \Delta h$

Where $\Delta h$ is the difference in height between the final position and the initial position of the box.

This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of $\Delta h$ is the same for both of them, so the work they have done is exactly the same.

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3 years ago

A century ago, swords were made from various metals, especially steel. The property that makes steel a good choice for sword mak

Nikolay [14]

The answer is a solid

5 0

3 years ago

An 87.6 g lead ball is dropped from rest from a height of 7.00 m. The collision between the ball and the ground is totally inela

bija089 [108]

Taking specific heat of lead as 0.128 J/gK = c

We have energy of ball at 7.00 meter height = mgh = $87.6*10^{-3}*9.81*7$

When leads gets heated by a temperature ΔT energy needed = mcΔT

= $87.6*10^{-3}*0.128*10^3$ ΔT

Comparing both the equations

$87.6*10^{-3}*9.81*7$ = $87.6*10^{-3}*0.128*10^3$ ΔT

ΔT = 0.536 K

Change in temperature same in degree and kelvin scale

So ΔT = 0.536 $^0C$

7 0

3 years ago

The pressure in a compressed air storage tank is 1200 kPa. What is the tank’s pressure in (a) kN and m units; (b) kg, m, and s u

elixir [45]

Explanation:

Given data

Pressure P=1200 kPa

To find

Pressure in

(a) kN/m²

(b) kg/m.s²

(c) kg/km.s²

Solution

For Part (a)

$P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )\\P=1200kN/m^{2}$

For Part (b)

$P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )(\frac{1000kg.m/s^{2} }{1kN} )\\P=1,200,000kg/m.s^{2}$

For Part (c)

$P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )(\frac{1000kg.m/s^{2} }{1kN} )(\frac{100m}{1km} )\\P=1,200,000,000kg/km.s^{2}$

8 0

3 years ago