All categories

2 years ago

Air only has a density of 1.3kg/m². What is the mass of3m^3 of air?

Physics

1 answer:

Katarina [22]2 years ago

8 0

Answer:

m = 3.9 [m³]

Explanation:

We must remember that the density of the substance is defined as the relationship between mass over volume. That is expressed in the following equation:

$Ro = m/V$

where:

Ro = density = 1.3 [kg/m³]

m = mass [kg]

V = volume = 3 [m³]

Now replacing the values.

$m =Ro*V\\m = 1.3*3\\\\m=3.9[m^{3}]$

You might be interested in

Knowing the causes of respiratory system diseases, what are

Kamila [148]

Cover your cough
Wash your hands
Keep a clean space
Avoid smoking
Stay active

Hope this helps :)

8 0

3 years ago

The volume electric charge density of a solid sphere is given by the following equation: The variable r denotes the distance fro

qwelly [4]

Answer:

62.8 μC

Explanation:

Here is the complete question

The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?

Solution

The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ

So, Q = ∫∫∫ρdV

Q = ∫∫∫ρr²sinθdθdrdΦ

Q = ∫∫∫(0.2r²)r²sinθdθdrdΦ

Q = ∫∫∫0.2r⁴sinθdθdrdΦ

We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π

So, Q = ∫∫∫0.2r⁴sinθdθdrdΦ

Q = ∫∫∫0.2r⁴[∫sinθdθ]drdΦ

Q = ∫∫0.2r⁴[-cosθ]drdΦ

Q = ∫∫0.2r⁴-[cosπ - cos0]drdΦ

Q = ∫∫∫0.2r⁴-[-1 - 1]drdΦ

Q = ∫∫0.2r⁴-[- 2]drdΦ

Q = ∫∫0.2r⁴(2)drdΦ

Q = ∫∫0.4r⁴drdΦ

Q = ∫0.4r⁴dr∫dΦ

Q = ∫0.4r⁴dr[Φ]

Q = ∫0.4r⁴dr[2π - 0]

Q = ∫0.4r⁴dr[2π]

Q = ∫0.8πr⁴dr

Q = 0.8π∫r⁴dr

Q = 0.8π[r⁵/5]

Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]

Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]

Q = 0.8π[0.025 m⁵ - 0 m⁵]

Q = 0.8π[0.025 m⁵]

Q = (0.02π mC/m⁵) m⁵

Q = 0.0628 mC

Q = 0.0628 × 10⁻³ C

Q = 62.8 × 10⁻³ × 10⁻³ C

Q = 62.8 × 10⁻⁶ C

Q = 62.8 μC

3 0

2 years ago

Pam, wearing a rocket pack, stands on frictionless ice. She has a mass of 49 kg. The rocket supplies a constant force for 22.0 m

ss7ja [257]

Answer:

Magnitude of the force is 4350N

Explanation:

As the woman accelerates at a distance of 22 m to go from rest to 62.5 m / s, we can use the kinematics to find the acceleration

v² = v₀² + 2 a x

v₀ = 0

a = v² / 2x

a = 62.5²/(2 × 22)

a = 88.78m/s²

the time you need to get this speed

v = v₀ + a t

t = v / a

t = 62.5 / 88.78

t = 0.704s

Let's caculate the magnitude of the force

F = ma

= 49 × 88.78

= 4350.22

≅ 4350N

Magnitude of the force is 4350N

t = 1,025 s

a = 55.43 m / s²

4 0

2 years ago

A firewoman dropped a person onto the safety net. Right before the person hit the net he had a

sergejj [24]

Answer:

The Mass of a person is calculated to be 28.6kg .

Explanation:

This is based on Kinetic energy and we know, that Kinetic energy is the energy possessed by body by virtue of its motion .

It can be calculated by expression :

K.E=1/2mv²

Velocity of a person = 11.2m/sec

Kinetic energy = 1800J

Mass of person = ?

We know ,

K.E=1/2mv²

so, putting values we have :

1800=1/2 x m x (11.2)²

That is ,

m=1800 x 2 /11.2 x 11.2

m=3600/125.44

m = 28.6 kg

6 0

2 years ago

A person of mass 55 kg swings on a rope length 4 m from rest (when the rope makes an angle of 30 degrees with the vertical) and

vovangra [49]

Answer:

θ = 19.66°

Explanation:

To determine the angle that the rope makes with the vertical for the two people, you first take into account the potential energy of the first person before he swings on the rope:

$U=mgh$

h: distance to the ground

g: gravitational acceleration = 9.8m/s^2

m: mass of the first person = 55 kg

In the image attache below you can notice that the height h is:

$h=4-4cos(30\°)=0.53m$

Then, the potential energy is:

$U=(55kg)(9.8m/s^2)(0.53m)=285.67J$

When the first person picks up the second person (when the rope is exactly vertical), all the potential energy becomes kinetic energy. Next, when both people reaches the maximum height h' the energy must be equal to the initial potential energy of the first person:

$U'=(m_1+m_2)gh'=285.67\ J$

From the previous equation you can get h':

$h'=\frac{285.67J}{(55kg+70kg)(9.8m/s^2)}=0.2332m$

Finally, you obtain the angle between the rope at the height h,' and the vertical, by calculating the following:

$h'=4-4cos(\theta)\\\\\theta=cos^{-1}(\frac{4-h'}{4})=cos^{-1}(\frac{4-0.2332}{4})=19.66\°$

hence, the angle between the rope and the vertical, when the two people are in the rope is 19.66°

8 0

2 years ago

Air only has a density of 1.3kg/m². What is the mass of3m^3 of air?​

Air only has a density of 1.3kg/m². What is the mass of3m^3 of air?