Why is friction a problem in space travellers

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

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What is the energy (in evev) of a photon of visible light that has a wavelength of 500 nmnm?

lisabon 2012 [21]

<h3>Answer:</h3>

E≈2,5 eV

<h3>Explanation:</h3>

_______________

λ=500 nm = 500·10⁻⁹ m

c=3·10⁸ m/s

h=6,63·10⁻³⁴ J·s = 4,14·10⁻¹⁵ eV·s

_______________

E - ?

_______________

$\displaystyle \boldsymbol{E}=h\nu =h \frac{c}{\lambda} =4,14\cdot 10^{-15} \; eV\cdot s\cdot \; \frac{3\cdot 10^8\; m/s}{500\cdot 10^{-9}\; m} =2,484\; eV\approx \boldsymbol{2,5\; eV}$

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Only the smallest particles of soil can be displaced by suspension because

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<span>The choices can be found elsewhere and as follows:

</span><span>a. they are so small that they stay close to the ground due to the attractive properties of charged soil particles.

b. they are easily carried by the wind.

c. they easily dissolve in liquid droplets.

d. it is easier for then to roll along the small crevices in the ground.</span><span>

</span>I think the correct answer from the choices listed above is option B. Only the smallest particles of soil can be displaced by suspension because they are so small that they are easily carried by the wind. Hope this answers the question. Have a nice day. Feel free to ask more questions.

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Why is friction a problem in space travellers​

Why is friction a problem in space travellers