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4 years ago

Why are some liquids more viscous than others?

1 answer:

7 0

The force that opposes the movement of an object through water is called drag. This is a type of frictional force. This force normally depends on the density and the viscosity of the fluid in question. The liquid which has more density and more viscosity or stickiness will produce a greater amount of drag force on an object than a fluid that is less dense and less viscous in nature. River water normally has less drag than that of sea water.

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In billiards, the 0.165 kg cue ball is hit toward the 0.155 kg eight ball, which is stationary. The cue ball travels at 5.8 m/s

Damm [24]

Answer:

another ball velocity = 3.92 m/s and with 30° clockwise from initial direction

Explanation:

given data

mass m1 = 0.165 kg

mass m2 = 0.155 kg

before collision velocity v1 = 5.8 m/s

before collision velocity v2 = 0

angle = 35.0° from initial direction

after collision 1st ball velocity v3 = 3.2 m/s

to find out

after collision another ball velocity v4

solution

we consider here ball move in x axis and after collision 1st ball move upside of x axis with angle 35 degree and other ball move downside with x axis with angle θ

so from conservation of momentum we say

m1v1 = m1v3cos35 + m2v4cosθ with x axis .............1

m1v3sin35 = m2v4sinθ with y axis .............2

so from 1 equation

0.165 × 5.8 = 0.165(3.2)cos35 + 0.155(v4)cosθ

v4 cosθ = 3.38 .................3

form 2 equation

0.165(3.2)sin35 = 0.155(v4)sinθ

v4 sinθ = 1.95 ......................4

so magnitude of another ball velocity is square and adding equation 3 and 4

another ball velocity = √(3.39²+1.96²)

another ball velocity = 3.92 m/s

and direction is tanθ = 1.96/3.39

θ = 30° clockwise from initial direction

3 0

3 years ago

An electron is accelerqated in the uniform field betwen two parallel charged oplates. The separation of the plates is 1.20 cm

marysya [2.9K]

Answer:

The speed of electron is $8.7\times10^{6}\ m/s$

Explanation:

Given that,

Separation of the plate = 1.20 cm

Suppose the field is $E=1.80\times10^{4}\ N/C$ .

If the electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate.

What the speed does it leave the hole?

We need to calculate the acceleration

Using formula of electric force

$F = qE$

$ma=qE$

$a=\dfrac{qE}{m}$

We need to calculate the speed of electron

Using equation of motion

$v^2=u^2+2as$

$v^2=2as$

Put the value of acceleration in the formula

$v^2=2\times\dfrac{qE}{m}\times s$

Put the value into the formula

$v^2=2\times\dfrac{1.6\times10^{-19}\times1.80\times10^{4}\times1.20\times10^{-2}}{9.11\times10^{-31}}$

$v=\sqrt{2\times\dfrac{1.6\times10^{-19}\times1.80\times10^{4}\times1.20\times10^{-2}}{9.11\times10^{-31}}}$

$v=8.7\times10^{6}\ m/s$

Hence, The speed of electron is $8.7\times10^{6}\ m/s$

7 0

3 years ago

A car drives to the east in a time of 6 hours. Then, immediately (not realistic, but just assume this is the case for this probl

garri49 [273]

Answer:

Average speed of car in the first trip is 10 km/hr

Explanation:

It is given that first the car drives 6 hours to the east

Then travels 12 km to west in 3 hours

Average speed for the entire trip = 8 km/hr

Total time = 3+6 = 9 hour

So distance traveled in 9 hour = 9×8 = 72 km

As the car travel 12 km in west so distance traveled in east = 72-12 = 60 km

Time by which car traveled in east = 6 hour

So speed $=\frac{distance}{time}=\frac{60}{6}=10km/hr$

So average speed of car in the first trip is 10 km/hr

7 0

4 years ago

A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t

lutik1710 [3]

a) E = 0

b) $3.38\cdot 10^6 N/C$

Explanation:

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

where

E is the electric field

q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

$r_1 = 10 cm\\r_2 = 15 cm$

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$

where in this problem:

$Q=15 \mu C = 15\cdot 10^{-6} C$ is the charge on the shell

$r=20 cm = 0.20 m$ is the distance from the centre of the shell

Substituting, we find:

$E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C$

4 0

4 years ago

Which of the following is a physical property of water?

photoshop1234 [79]

Hi there!

First of all, you have to pay attention to the question, and they are asking for something physical, and physical means to be seen on the outside.

The answer is A because you can clearly see that water is a liquid. Being a liquid is a physical property.

It wouldn't be B because I believe it would be a chemical property to react with many substances.

Hope this helps! :D

7 0

3 years ago

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