Ask question

Ask question

All categories

3 years ago

12

Explain the following behaviour of molecules water rises up in a harrow tubes but mercury which is also a liquid falls in a narr

ow tubes to level below the outside surface

1 answer:

Mandarinka [93]3 years ago

7 0

Answer: find the answer in the explanation

Explanation:

The capillarity of water molecules is different from the mercury molecules.

What is capillarity ?

This is the tendency of a liquid substance to rise in a capillary tube.

Molecules water rises up in a harrow tubes because of the force of adhesion between the water molecules and the tube molecules is greater than the force of cohesion between the water molecules. This helps water to wet the tube and rise. While mercury which is also a liquid falls in a narrow tubes to level below the outside surface because the force of cohesion between the mercury molecules is greater than the force of adhesion between the mercury molecules and the tube molecules. Mercury does not wet.

You might be interested in

How high would a projectile go if it was launched from ground level with an initial speed of 26 m/s at an angle of 30 degrees ab

tigry1 [53]

Answer:

Vy = 26 m/s sin 30 = 13 m/s vertical speed

t = Vy / a = 13 m/s / 9.80 m/s^2 = 1.33 sec time to reach Vy = 0

H = Vy t + 1/2 g t^2

H = 13 m/s * 1.33 sec - 1.33^2 * 9.8 / 2 m = 8.62 m

4 0

1 year ago

A 1.0-kg block moving to the right at speed 3.0 m/s collides with an identical block also moving to the right at a speed 1.0 m/s

____ [38]

Answer:

Speed of both blocks after collision is 2 m/s

Explanation:

It is given that,

Mass of both blocks, m₁ = m₂ = 1 kg

Velocity of first block, u₁ = 3 m/s

Velocity of other block, u₂ = 1 m/s

Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$v=\dfrac{1\ kg\times 3\ m/s+1\ kg\times 1\ m/s}{2\ kg}$

v = 2 m/s

Hence, their speed after collision is 2 m/s.

7 0

3 years ago

A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.010 hour, and the as

Taya2010 [7]

Answer:

$w = - 508.53$ joules

$q = - 3091.47$ joules

Explanation:

Let us convert the time in hours into seconds

$0.010* 3600\\= 36$

Change in internal energy

$\delta E = p * \delta t$

where E is the internal energy in Joules

p is the power in watts

and t is the time in seconds

$\delta E = - 100 * 36\\$

$\delta E = - 3600$ Joules

Amount of work done by the system

$w = - P * \delta V$

where P is the pressure and V is the volume

Substituting the given values in above equation, we get -

$w = - 1 * ( 5.92 -0.90)\\$

$w = -5.02$ liter-atmospheres

Work done in Joules

$- 5.02 * 101.3\\$ $= 508.53$ Joules

$q = \delta E - w\\$

Substituting the given values we get -

$q = - 3600 - (-508.53)\\q = - 3091.47$

Thus

$w = - 508.53$ joules

$q = - 3091.47$ joules

7 0

3 years ago

A crane lifts an air conditioner to the top of a building. If the building is 12 m high, and the air conditioner has a mass of 2

Pepsi [2]

Work needed = 23,520 J

<h3> Further explanation </h3>

Given

height = 12 m

mass = 200 kg

Required

work needed by the crane

Solution

Work is the transfer of energy caused by the force acting on a moving object

Work is the product of force with the displacement of objects.

Can be formulated

W = F x d

W = Work, J, Nm

F = Force, N

d = distance, m

F = m x g

Input the value :

W = mgd

W = 200 kg x 9.8 m/s²x12 m

W = 23520 J

8 0

3 years ago

Two circular coils are concentric and lie in the same plane. The inner coil contains 170 turns of wire, has a radius of 0.0095 m

PIT_PIT [208]

Answer:

Current in outer circle will be 15.826 A

Explanation:

We have given number of turns in inner coil $N_I=170$

Radius of inner circle $r_i=0.0095m$

Current in the inner circle $I_i=8.9A$

Number of turns in outer circle $N_o=150$

Radius of outer circle $r_o=0.015m$

We have to find the current in outer circle so that net magnetic field will zero

For net magnetic field current must be in opposite direction as in inner circle

We know that magnetic field is given due to circular coil is given by

$B=\frac{N\mu _0I}{2r}$

For net magnetic field zero

$\frac{N_I\mu _0I_I}{2r_I}=\frac{N_O\mu _0I_0}{2r_O}$

So $\frac{170\times \mu _0\times 8.9}{2\times 0.0095}=\frac{150\times \mu _0I_O}{2\times 0.015}$

$I_O=15.92A$

4 0

3 years ago