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3 years ago

Explain why surface tension occurs

1 answer:

5 0

Hey there! The cohesive forces between liquid molecules are responsible for the phenomenon known as surface tension. The molecules at the surface do not have other like molecules on all sides of them and consequently they cohere more strongly to those directly associated with them on the surface. Hope this helps! :)

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A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou

Hoochie [10]

Answer:

e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

Explanation:

This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.

The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.

Mathematical expression for the Newton's second law of motion is given as:

$F=\frac{dp}{dt}$ ............................................(1)

where:

dp = change in momentum

dt = time taken to change the momentum

We know, momentum:

$p=m.v$

Now, equation (1) becomes:

$F=\frac{d(m.v)}{dt}$

∵mass is constant at speeds v << c (speed of light)

$\therefore F=m.\frac{dv}{dt}$

and, $\frac{dv}{dt} =a$

where: a = acceleration

$\Rightarrow F=m.a$

also

$F\propto \frac{1}{dt}$

so, more the time, lesser the force.

& Impulse:

$I=F.dt$

$I=m.a.dt$

$I=m.\frac{dv}{dt}.dt$

$I=m.dv=dp$

∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.

So, the impulse in both the cases will be same.

4 0

3 years ago

An expert archer tried to shoot an arrow into a target lying at the bottom of a shallow pool. Even though the water did not chan

frutty [35]

The interaction that caused the archer to miss was probably refraction.

7 0

3 years ago

Read 2 more answers

A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of

Lynna [10]

We have the relation

$\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}$

where $v_{A \mid B}$ denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

$v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}$

$v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}$

Let's assume the river flows South-to-North, so that

$\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath$

and let $-90^\circ < \theta < 90^\circ$ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

$\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)$

Then the velocity of the boat relative to the Earth is

$\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath$

The crossing is 153.0 m wide, so that for some time $t$ we have

$153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s$

which is minimized when $\theta=0^\circ$ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

$\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}$