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3 years ago

The attraction of liquid particles for a solid surface is due to ____.

Physics

2 answers:

Kisachek [45]3 years ago

8 0

Answer:

the force of adhesion.

Explanation:

The attractive force between the two types of molecules is called adhesive force.

So, the force between the liquid molecule and the solid molecule is adhesive in nature.

The attractive force acts between the two same type of molecules is called cohesive force.

White raven [17]3 years ago

6 0

This attraction occurs from adhesion, also known as adsorption <span />

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Which of the following statements about the force on a charged particle due to a magnetic field are not valid?

Vinil7 [7]

The correct answer is "None of the above; all of these statements are valid." All the statements namely, it depends on the particle's charge, it depends on the strength of the external magnetic field, it depends on the particle's velocity, and it acts at right angles to the direction of the particle's motion are all valid. Thank you for posting your question. I hope this answer helped you. Let me know if you need more help.

5 0

4 years ago

a high speed train travels with an average speed of 250 km/h. the train travels for 2 hrs. how far does the train travel

steposvetlana [31]

Answer:

<h3>The answer is 500 km </h3>

Explanation:

The distance covered by an object given it's velocity and time taken can be found by using the formula

<h3>distance = average velocity × time</h3>

From the question

average speed = 250 km/h

time = 2 hrs

We have

distance = 250 × 2

We have the final answer as

Hope this helps you

5 0

3 years ago

Read 2 more answers

We can compare these two interactions on the basis of impulse (see above), but sometimes, we are more interested in the forces (

kiruha [24]

Complete Question

The complete question is shown on the first uploaded image

Answer:

$I = 476 \ N \cdot s$

$I_1 = 14.21 \ N\cdot s$

$F = 20300 \ N$

Explanation:

Considering the first question

From the question we are told that

The force produced is $F = 3400 \ N$

The duration of the punch is $t = 0.14 \ s$

Generally the impulse delivered is mathematically represented as

$I = F * t$

=> $I = 3400 * 0.14$

=> $I = 476 \ N \cdot s$

Considering the second question

The approaching velocity of the ball is $v_b = 45 \ m/s$

The leaving velocity of the ball is $v_l = -53 \ m/s$

The mass of the ball is $m_b = 0.145 \ kg$

Generally the magnitude of the impulse delivered is mathematically represented as

$I_1 = m* v_b - m * v_l$

=> $I_1 = [0.145 * 45] - [0.145 * -53]$

=> $I_1 = 14.21 \ N\cdot s$

Considering the third question

The duration of the impact of the bat is $t _1 = 0.7 \ ms = 0.7 *10^{-3} \ s$

Generally the average force exerted by the bat is mathematically represented as

$F = \frac{I_1}{t_1}$

=> $F = \frac{14.21 }{0.7 *10^{-3}}$

=> $F = 20300 \ N$

7 0

4 years ago

The Sun has a mass of 1.99x10^30 kg and a radius of 6.96x10^8 m. Calculate the acceleration due to gravity, in meters per second

just olya [345]

Answer:

$g=274\ m/s^2$

Explanation:

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

The radius of the Sun, $r=6.96\times 10^8\ m$

We need to find the acceleration due to gravity on the surface of the Sun. It is given by the formula as follows :

$g=\dfrac{GM}{r^2}\\\\g=\dfrac{6.67\times 10^{-11}\times 1.99\times 10^{30}}{(6.96\times 10^8)^2}\\\\g=274\ m/s^2$

So, the value of acceleration due to gravity on the Sun is $274\ m/s^2$ .

8 0

3 years ago

Please help!!! I have a physics exam tomorrow and I just can't wrap my head around this one!

Ainat [17]

The total electrostatic force on charge A is $28 \mu N$

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

Here we have three positively charged particles A,B and C, located at the following positions:

$x_A = 0\\x_B = 10 m\\x_C = 20 m$

The magnitudes of the three charges are:

$q_A = q_B = q_C = 0.5 \mu C = 0.5\cdot 10^{-6}C$

The force exerted by B on A is to the left (because the force between two positive charges is repulsive), and the force exerted by C on A is also to the left (also repulsive). Therefore, the net force on A is just the sum of the two forces exerted by charges B and C:

$F_A = F_{BA} + F_{CA} = k\frac{q_B q_A}{(x_B-x_A)^2}+k\frac{q_C q_A}{(x_C-x_A)^2}=\\=(8.99\cdot 10^9) \frac{(0.5\cdot 10^{-6})^2}{(10)^2}+(8.99\cdot 10^9) \frac{(0.5\cdot 10^{-6})^2}{(20)^2}=2.8\cdot 10^{-5} N = 28 \mu N$

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

3 0

4 years ago