What is the mass of a liquid having a density of 1.50 g/ml and a volume of 3.5 liters?

Physics

1 answer:

PIT_PIT [208]2 years ago

6 0

We know the formula for density = Mass/ volume

So Mass, M = Volume * Density

Volume = 3.5 L= 0.0035 $m^3$

Density = 1.50 g/ml = 1500 $kg/m^3$

Mass, M = 0.0035*1500 = 5.25 kg

So mass of liquid having density 1.50 g/ml and volume 3.5 liters is 5.25 kg.

You might be interested in

Which is NOT one of the four principals of Dalton’s atomic theory?

Gre4nikov [31]

Answer:

Your answer would be letter B. Electrons orbit the nucleus in energy level.

Explanation:

Hope it helps..

Just correct me if I'm wrong, okay?

But ur welcome!!

(;ŏ﹏ŏ)(◕ᴗ◕✿)

8 0

2 years ago

Read 2 more answers

A person is singing in the shower. They notice that as certain notes become very loud, the shower walls vibrate. Which statement

AysviL [449]

Answer:

the note is one of the harmonic frequencies for the shower.

I just took the quiz, this is correct. Hope this helps:)

Explanation:

8 0

2 years ago

Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi

nadya68 [22]

Answer:

$E = 1.85*10^{12}\frac{N}{C}$

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction, and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

$E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$