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3 years ago

Differentiate between adhesion and cohesion.

Physics

1 answer:

sergiy2304 [10]3 years ago

8 0

<h2>Answer:</h2>

<h2>Cohesion </h2>

Is the attraction that molecules have for others of their same type.

<h2>Adhesion </h2>

Is the attraction that molecules have for others of different type by intermolecular forces.

A good example of both is water that can stick to itself through hydrogen bonds (cohesion) and can also stick to a glass due to adhesion.

So, while Cohesion is the force of attraction between adjacent particles within the same body, Adhesion is the interaction between the surfaces of different bodies.

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A photon of red light (wavelength = 710 nm) and a ping-pong ball (mass = 2. 40 × 10^-3 kg) have the same momentum. At what speed

Paha777 [63]

The speed of the ball moving is

$v = 3.94 \times 10 {}^{ - 25}m/s$

what is momentum?

The momentum p of a classical object of mass m and velocity v is given by pclassical =mv.

For photons with wavelength λ,this equation does not hold.Instead, the momentum of the Photon is given by p Photon = h/λ

where,h is the planck's constant.

The momentum of the red Photon is

given:

$h = 6.626 \times 10 {}^{ - 34}kgm {}^{2}/s(plancks \: constant)$

$λ = 700 \times 10 {}^{ - 9} m(photons \: wavelength)$

$p \: photons = \frac{6.626 \times 10 {}^{ - 34}kgm {} ^{2}/s }{700 \times 10 {}^{ - 9}m }$

$= 9.47 \times 10 {}^{ - 28}kgm /s$

since,the Photon and the ping-pong ball have the same momentum,we have

$pball = pphotons = \frac{6.626 \times 10 {}^{ - 34}kgm/s }{700 \times 10 {}^{ - 9} }$

$pball = mv,m = 2.40 \times 10 {}^{ - 3}kg$

$v = 3.94 \times 10 {}^{ - 25} m/s$

Therefore, if the red photon and the ping-pong ball have the same momentum, the ping-pong ball must have a speed of approximately

$v = 3.94 \times 10 {}^{ - 25}m/s$

learn more about momentum of photon from here: brainly.com/question/28197406

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3 0

2 years ago

A projectile is launched into the air with the initial speed of vi = 40 m/s at a launch angle of 20 degrees above the horizontal

Sphinxa [80]

The range of the projectile is 188 m

Explanation:

The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

The path of a projectile is the combination of these two motions: see figure in attachment.

In order to find the horizontal range of the projectile, we just need to calculate the horizontal distance travelled.

We have:

t = 5.0 s (time of fligth of the projectile)

and the horizontal velocity is constant, and it is given by

$v_x = v_i cos \theta$

where

$v_i = 40 m/s$ is the initial velocity

$\theta=20^{\circ}$ is the angle of projection

Substituting,

$v_x = (40)(cos 20^{\circ})=37.6 m/s$

And therefore, the range of the projectile is:

$d=v_x t = (37.6)(5.0)=188 m$

Learn more about projectile motion:

brainly.com/question/8751410

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5 0

3 years ago

To win the game, a placekicker must kick a football from a point 44 m (48.1184 yd) from the goal, and the ball must clear the cr

Ganezh [65]

Answer:

The distance by the ball clear the crossbar is 1.15 m

Explanation:

Given that,

Distance = 44 m

Speed = 24 m/s

Angle = 31°

Height = 3.05 m

We need to calculate the horizontal velocity

Using formula of horizontal velocity

$u_{x}=u\cos\theta$

Put the value into the formula

$u_{x}=24\cos(31)$

$u_{x}=20.5\ m/s$

We need to calculate the vertical velocity

Using formula of vertical velocity

$u_{y}=u\sin\theta$

Put the value into the formula

$u_{y}=24\sin(31)$

$u_{y}=12.3\ m/s$

We need to calculate the time

Using formula of time

$t=\dfrac{d}{u_{x}}$

Put the value into the formula

$t=\dfrac{44}{20.5}$

$t=2.1\ sec$

We need to calculate the vertical height

Using equation of motion

$h=u_{y}t+\dfrac{1}{2}at^2$

Put the value into the formula

$h=12.3\times2.1-\dfrac{1}{2}\times9.8\times(2.1)^2$

$h=4.2\ m$

We need to calculate the distance by the ball clear the crossbar

Using formula for vertical distance

$d=h-3.05$

Put the value of h

$d=4.2-3.05$

$d=1.15\ m$

Hence, The distance by the ball clear the crossbar is 1.15 m

7 0

3 years ago

Wagon wheel. While working on your latest novel about settlers crossing the Great Plains in a wagon train, you get into an argum

OverLord2011 [107]

Answer:

I = 16.7kgm²

Explanation:

Since, Torque is given by,

\tau = F*r = I*\alpha

here, I = Moment of inertia = ??

\alpha = angular acceleration of wheel = a/r

F = tangential tension acting on the wheel = T

a = acceleration of bag of sand = 2.95 m/s^2

r = radius of wheel = d/2 = 120/2 = 60 cm = 0.60 m

from force balance on sand bag,

mg - T = m*a

T = m*(g-a)

m = mass of sand bag = 20 kg

So, I = T*r/\alpha = m*(g-a)*r/(a/r)

Using known values:

I = 20*(9.81 - 2.95)*0.60/(2.95/0.60) = 16.74

I = 16.7 kgm² = Moment of inertia of wheel experimentally

also, Moment of inertia of wheel theoretically(I') = M*r²

given, M = mass of wheel = 70 kg

I' = 70*0.60²= 25.2 kgm² = Moment of inertia of wheel theoretically

7 0

3 years ago

All of the halogens are____to humans.

Ymorist [56]

Answer:

b bc they are a dangerous threat to humans health

5 0

1 year ago