All categories

2 years ago

What are the SI units for liquid volume?

1 answer:

5 0

SI units for volume don't care whether you use them to describe
a solid, a liquid, a gas, or a volume of vacuum.

The units are the liter, and the cubic meter (1,000 liters).

You might be interested in

At the surface, atmospheric pressure is 1.013 × 10^5 Pa. People can normally snorkel down to a depth of roughly one meter. What

natulia [17]

Answer:

1.01 × 10⁵ Pa

Explanation:

At the surface, atmospheric pressure is 1.013 × 10⁵ Pa.

We need to find the total pressure on the air in the lungs of a person to a depth of 1 meter.

Pressure at a depth is given by :

$P=\rho gh$

Where

$\rho$ is the density of air, $\rho=1.225\ kg/m^3$

So,

$P=1.225\times 9.8\times 1\\\\=12\ Pa$

Total pressure, P = Atmospheric pressure + 12 Pa

= 1.013 × 10⁵ Pa + 12 Pa

= 1.01 × 10⁵ Pa

Hence, the total pressure is 1.01 × 10⁵ Pa.

5 0

2 years ago

Weight of a man is less at coal mine.Why?

dmitriy555 [2]

If someone is underground, then therefore there is less planet/ground underneath them, so there would be less gravity. Gravity directly affects weight.

5 0

2 years ago

Read 2 more answers

A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect

Stels [109]

Answer:

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

Explanation:

Given that

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field at point P

$E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta$

$sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}$

$r^2=d^2+x^2$

$E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}$

Now by integrating above equation

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

4 0

2 years ago

Which of the following is the best definition of energy?

Montano1993 [528]

Energy is the capacity to do some type of work

8 0

2 years ago

An electron that has a velocity with x component 2.6 × 106 m/s and y component 3.2 × 106 m/s moves through a uniform magnetic fi

elena55 [62]

Answer:

a) $\vec F_{B} = 8.766\times 10^{-14}\,T\,k$ , b) $\vec F_{B} = -8.766\times 10^{-14}\,T\,k$

Explanation:

a) The magnetic force experimented by a particle has the following vectorial form:

$\vec F_{B} = q\cdot \vec v \times \vec B$

The charge of the electron is equal to $-1.602\times 10^{-19}\,C$ . Then, cross product can be solved by using determinants:

$\vec F_{B} = \begin{vmatrix}i&j&k\\-4.165\times 10^{-13}\, C\cdot \frac{m}{s} &-5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}$

The magnetic force is:

$\vec F_{B} = 8.766\times 10^{-14}\,T\,k$

b) The charge of the proton is equal to $1.602\times 10^{-19}\,C$ . Then, cross product has the following determinant:

$\vec F_{B} = \begin{vmatrix}i&j&k\\4.165\times 10^{-13}\, C\cdot \frac{m}{s} &5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}$

The magnetic force is:

$\vec F_{B} = -8.766\times 10^{-14}\,T\,k$

8 0

2 years ago