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algol13
4 years ago
11

What are properties of a liquid?

Physics
1 answer:
Oksi-84 [34.3K]4 years ago
6 0

Answer:

Pressure

Surface

Flow

Fluency

Explanation:

*Surface

The surface tension is the force with which the surface molecules of a liquid are attracted to bring them inside and thus decrease the surface area.

*Flow

It is defined as resistance to flow. The viscosity of a liquid depends on the intermolecular forces:

-The higher the intermolecular forces of a liquid, its molecules have a greater difficulty moving between them, therefore the substance is more viscous.

-Liquids that are made up of long, flexible molecules that can bend and tangle with each other are more viscous.

* Pressure

if the molecules of the liquid have a greater intensity of intermolecular force, then they will be trapped in the liquid and will have less facility to pass into the gas phase.

    on the contrary to lower intensity of intermolecular force, then the molecules can escape more easily to the gaseous state.

*Fluency

This property allows liquids to easily pass through a hole regardless of size, as long as this hole is at a lower or the same level of the container where the liquid is stored.

This property indicates the deformability of a liquid which is very wide without requiring mechanical stress.

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7. A Cyclist speeds up from 6.00 m/s to 9.70 m/s in a time of 1.40 seconds.
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5 0
3 years ago
A long solenoid with 8.22 turns/cm and a radius of 7.00 cm carries a current of 19.4 mA. A current of 3.59 A exists in a straigh
daser333 [38]

Answer:

a. 3.039cm

b.magnetic field is B=2.958\times10^{-5}T

Explanation:

Direction of the solenoid magnetic field is along the axis of the solenoid. and magnetic field due to the wire perpendicular to that due to the solenoid.. Magnetic field at r is given by:

\overrightarrow B = \overrightarrow B_s+ \overrightarrow B_w,\ \ \ \ \  \overrightarrow B_s\perp \overrightarrow B_w

Angle of net magnetic field from axial direction is given by:

tan\  \theta=\frac{B_w}{B_s},

Field due to solenoid:

B_s=\mu_onI_s,  \ \ \ \ n=(8.22 t/cm)(100cm/m)=822turn/m

Field due to wire:

B_w=\frac{\mu_oI_w}{2\pi r}

Therefore, r:

tan\  \theta=\frac{B_w}{B_s}\\\\=\frac{\mu_oI_w}{2\pi r(\mu_o nI_s)}\\\\r=\frac{I_w}{2\pi  nI_stan \ \theta}\\\\r=\frac{3.59A}{2\pi\times822\times19.4\times10^{-3}A \ tan 49.7\textdegree}\\\\r=3.039cm

Hence, the radial distance is 3.039cm

b.The magnetic field strength is given by:

B=\sqrt{B_w^2+B_s^2}\\\\tan 49.7\textdegree=\frac{B_w}{B_s}\\\\1.179=\frac{B_w}{B_s}\\\\B_w=1.179B_s\\\\B=\sqrt{(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{3}A)+1.179(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{-3}A)}\\\\B=2.958\times10^{-5}T

Hence, the magnetic field is B=2.958\times10^{-5}T

7 0
4 years ago
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