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4 years ago

Calculate the mass of air in a room of floor dimensions =10M×12M and height 4m(Density of air =1.26kg/m cubic

Physics

1 answer:

Alinara [238K]4 years ago

7 0

The volume of the room is the product of its dimensions:

$10\times 12 \times 4 = 480\text{ m}^3$

Now, from the equation

$d=\dfrac{m}{V}$

where d is the density, m is the mass and V is the volume, we deduce

$m=dV$

So, multiply the density and the volume to get the mass of air in the room.

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An object of known mass M with speed v0 travels toward a wall. The object collides with it and bounces away from the wall in the

Bingel [31]

Neither side of the equation may be used because there are too many unknown quantities before, during, and after the collision

Explanation:

The impulse theorem states that the change in momentum of an object is equal to the impulse, which is the product between the average force applied and the duration of the collision:

$\Delta p = F \Delta t$

where

$\Delta p$ is the change in momentum

F is the average force

$\Delta t$ is the duration of the collision

In this problem, neither side of the equation can be used to measure the change in momentum. In fact:

- The change in momentum (left side) is given by

$\Delta p = m(v-u)$

where

m is the mass of the object

u is the initial velocity

v is the final velocity

Here the final velocity is not known, so it's not possible to use this side of the equation

- The impulse (right side) is given by

$F\Delta t$

here the average force is known, however the duration of the collision is not known, so it's not possible to use this side of the equation.

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4 years ago

Which of the following best describes gravitational potential energy,

s2008m [1.1K]

The energy stored in an object when you lift it,

8 0

3 years ago

Explain how bar graphs can show comparisons between 2 topics.

Pachacha [2.7K]

Answer:

A bar chart is typically used for the comparison of informations obtained from a data collection by representing each value (information) with rectangular bars respectively.

Explanation:

A bar chart also known as a bar graph can be defined as the graphical representation of data (informations) by using vertical columns or rectangular bars. When a data set is observed, a bar chart can be used to graphically represent or record the total or overall amount of information contained therein. Generally, bar charts are used for the comparison of informations obtained from a data collection by representing each value (information) with rectangular bars respectively.

The bar graph has a x-axis and a y-axis used to represent the various categories of data collected.

In this scenario, the x-axis is used to denote the points earned by the teams while the y-axis is used to denote the seasons as seen in the image attached.

<em>Hence, point scores by team (Team A, B and C) per seasons (2002, 2003, 2004 and 2005) can best be compared through the use of a bar chart (graph).</em>

8 0

3 years ago

Read 2 more answers

a 3.00 kg block moving 2.09 m/s right hits a 2.22 kg block moving 3.92 m/s left. afterward, the 3.00 kg block moves 1.11 m/s lef

LuckyWell [14K]

The momentum of the 2.22 kg afterwards is 0.91 kg m/s to the right

Explanation:

We can solve this problem by applying the law of conservation of momentum.

In fact, the total momentum before and after the collision must be conserved. So, we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 3.00 kg$ is the mass of the first block

$u_1 = 2.09 m/s$ is the initial velocity of the first block (we take the right as positive direction)

$v_1 = -1.11 m/s$ is the final velocity of the first block (to the left)

$m_2 = 2.22 kg$ is the mass of the second block

$u_2 = -3.92$ is the initial velocity of the second block

$v_2$ is the final velocity of the second block

Re-arranging the equation and substituting the values, we find: the final velocity of the second block:

$v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(3.00)(2.09)+(2.22)(-3.92)-(3.00)(-1.11)}{2.22}=0.41 m/s$

(to the right, since it is positive)

And so, the momentum of the 2.22 kg block afterwards is:

$p_2 = m_2 v_2 = (2.22)(0.41)=0.91 kg m/s$ (to the right)

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3 years ago

The smallest unit of charge is − 1.6 × 10 − 19 C, which is the charge in coulombs of a single electron. Robert Millikan was able

vovangra [49]

Answer:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

Explanation:

<u>Charge of an Electron</u>

Since Robert Millikan determined the charge of a single electron is

$q_e=-1.6\cdot 10^{-19}\ C$

Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is $5\times -1.6\cdot 10^{-19}\ C=-8 \cdot 10^{-19}\ C$

Let's test the possible charges listed in the question:

$-8.0 \times 10 ^{-19 }$ . We have just found it's a possible charge of a particle

$-3.2 \times 10 ^{-19 }$ . Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets

$-1.2 \times 10 ^{-19 }$ this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge

$-5.6 \times 10 ^{-19 },\ -9.4 \times 10 ^{-19 }$ cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6

Finally, the charge $-4.8 \times 10 ^{-19 }\ C$ is four times the charge of the electron, so it is a possible value for the charge of an oil droplet

Summarizing, the following are the possible values for the charge of an oil droplet:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

5 0

3 years ago

Calculate the mass of air in a room of floor dimensions =10M×12M and height 4m(Density of air =1.26kg/m cubic​

Calculate the mass of air in a room of floor dimensions =10M×12M and height 4m(Density of air =1.26kg/m cubic