Ask question

Ask question

All categories

2 years ago

14

Calculate the mass of the air contained in a room that measures 2.50 m x 5.50 m x 3.00 m if the density of air is 1.29 g/dm3.53.

2 g3.13 x 10-3 g3.20 x 104 g5.32 x 104 g5.32 x 107 g

1 answer:

Law Incorporation [45]2 years ago

6 0

Answer:

$5.32\cdot 10^4 g$

Explanation:

First of all, we need to find the volume of the room, which is given by

$V=2.50 m \cdot 5.50 m \cdot 3.00 m =41.3 m^3$

Now we can find the mass of the air by using

$m=dV$

where

$d=1.29 g/dm^3$ is the density of the air

$V=41.3 m^3 = 41,300 dm^3$ is the volume of the room

Substituting,

$m=(1.29)(41300)=5.32\cdot 10^4 g$

You might be interested in

Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o

sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

$a_{x} = -\omega^{2} x$

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders is given as:

$I=\frac{1}{2}MR^{2}$ -----(1)

Newton's second law for angular motion is:

∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

∑Fₓ = Maₓ ------ (3)

By definition of torque:

τ = RF --------(4)

Put (4) and (1) in (2)

$RF=\frac{1}{2}MR^{2}\alpha$

$RF=\frac{1}{2}MR^{2}\alpha$

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

$f_{s}=\frac{1}{2}MR\alpha$ ------ (5)

As angular acceleration is related to linear by:

$a= R\alpha$

Eq (5) becomes

$f_{s}=\frac{1}{2}Ma_{x}$ ---- (6)

aₓ shows displacement in horizontal direction

From (3)

∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

$\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

$a_{x} = \frac{2k}{3M} x$ --- (9)

For spring mass system

$a= -\omega^{2} x$ ----- (10)

Equating (9) and (10)

$\omega^{2} = \frac{2k}{3M}$

$\omega = \sqrt{ \frac{2k}{3M}}$

then (9) becomes

$a_{x} = - \omega^{2}x$

(The minus sign says that x and aₓ have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

$T=\frac{2\pi }{\omega}$

$T = 2\pi \sqrt{\frac{3M}{2k}$

5 0

2 years ago

Question 8 (4 points)

katrin2010 [14]

Answer:

The answer to your question is:

Explanation:

Data

Duane Albert

d = 5 m ; v = 3 m/s v = 4.2 m/s

a) b)

Duane's Albert's

d = 5 + (3)t d = 4.2t

d = 5 + 3t

c) 5 + 3t = 4.2t

4.2t - 3t = 5

1.2t = 5

t = 4.17 s

d)

Duane's

d= 5 + 3(4.17)

d = 17.51 m

Alberts

d = 4.2(4.17)

d = 17.51 m

4 0

2 years ago

Read 2 more answers

A battery charges a parallel-plate capacitor fully and then is removed. The plates are then slowly pulled apart. What happens to

White raven [17]

Answer:

<h2>The potential difference increases </h2>

Explanation:

from the relation $E= \frac{V}{d}$

where E= electric field (force per coulomb)

V= voltage

d= distance

Hence the voltage is going to be V= E×d.

Therefore this means that increasing the distance increases the voltage.

3 0

2 years ago

A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvat

Stolb23 [73]

Answer: f=150cm in water and f=60cm in air.

Explanation: Focal length is a measurement of how strong light is converged or diverged by a system. To find the variable, it can be used the formula:

$\frac{1}{f}$ = (nglass - ni)( $\frac{1}{R1}$ - $\frac{1}{R2}$ ).

nglass is the index of refraction of the glass;

ni is the index of refraction of the medium you want, water in this case;

R1 is the curvature through which light enters the lens;

R2 is the curvature of the surface which it exits the lens;

Substituting and calculating for water (nwater = 1.3):

$\frac{1}{f}$ = (1.5 - 1.3)( $\frac{1}{10}$ - $\frac{1}{15}$ )

$\frac{1}{f}$ = 0.2( $\frac{1}{30}$ )

f = $\frac{30}{0.2}$ = 150

For air (nair = 1):

$\frac{1}{f}$ = (1.5 - 1)( $\frac{1}{10}$ - $\frac{1}{15}$ )

f = $\frac{30}{0.5}$ = 60

In water, the focal length of the lens is f = 150cm.

In air, f = 60cm.

5 0

2 years ago

Read 2 more answers

A bicyclist of mass 68 kg rides in a circle at a speed of 3.9 m/s. If the radius of the circle is 6.5 m, what is the centripetal

ASHA 777 [7]

Data:
Centripetal Force = ? (Newton)
m (mass) = 68 Kg
s (speed) = 3.9 m/s
R (radius) = 6.5 m

Formula:
$F_{centripetal\:force} = \frac{m*s^2}{R}$

Solving:
$F_{centripetal\:force} = \frac{m*s^2}{R}$
$F_{centripetal\:force} = \frac{68*3.9^2}{6.5}$
$F_{centripetal\:force} = \frac{68*15.21}{6.5}$
$F_{centripetal\:force} = \frac{1034.28}{6.5}$
$\boxed{\boxed{F_{centripetal\:force} = 159.12\:N}}$
Answer:
<span>B.159 N</span>

3 0

2 years ago