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3 years ago

Why should a glass slide and a coverslip be held by the edges?

1 answer:

8 0

THIS IS NOT THE EXACT ANSWER BUT IT MIGHT HELP
The cover slips serves two purposes: (1) it protects the microscope's objective lens from contacting the specimen, and (2) it creates an even thickness (in wet mounts) for viewing.

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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 $1$ .

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

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 $=0.33\ m$ .

3 0

3 years ago

What occurs when light changes direction after colliding with particles of matter

Slav-nsk [51]

Scattering occurs when light changes direction after colliding with particles of matter.

7 0

3 years ago

A uniform stick 1.5 m long with a total mass of 250 g is pivoted at its center. A 3.3-g bullet is shot through the stick midway

Andru [333]

Answer:

Explanation:

mass of bullet = 3.3 g = 0.0033 kg

initial velocity of bullet $v_{1}$ = 250 m/s

final velocity of bullet $v_{2}$ = 140 m/s

loss of kinetic energy of the bullet = $\frac{1}{2}m(v^{2} _{1} - v^{2} _{2})$

==> $\frac{1}{2}*0.0033*(250^{2} - 140^{2} )$ = 70.785 J

this energy is given to the stick

The stick has mass = 250 g =0.25 kg

its kinetic energy = 70.785 J

from

KE = $\frac{1}{2} mv^{2}$

70.785 = $\frac{1}{2}*0.25*v^{2}$

566.28 = $v^{2}$

$v= \sqrt{566.28}$ = 23.79 m/s

the stick is 1.5 m long

this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m

The angular speed will be

Ω = v/r = 23.79/0.375 =<em> 63.44 rad/s</em>

5 0

2 years ago

Substances that move to the stronger parts of a magnetic field are termed ______ substances; the atomic feature responsible for

Arisa [49]

Substances that move to the stronger parts of a magnetic field are termed paramagnetic substances; the atomic feature responsible for this property is presence of unpaired electrons in atoms.

<h3>What is a paramagnetic substance?</h3>

A paramagnetic substance is the substance that possess unpaired electrons that are heavily attracted in a magnetic field.

A magnetic field is defined as the field that exists around a magnet that produces a field of force.

Examples of paramagnetic substance include the following:

aluminum,
gold,
copper.
Chromium, and
Manganese.

These substances are known as paramagnetic substances because they possess a high number of unpaired electrons.

Other properties of a paramagnetic substance include the following:

They have a permanent dipole moment or permanent magnetic moment.
They are weakly magnetized in the direction of the magnetizing field.
They usually have constant relative permeability (μr) slightly greater than 1.

Therefore, Substances that move to the stronger parts of a magnetic field are termed paramagnetic substances; the atomic feature responsible for this property is presence of unpaired electrons in atoms.

Learn more about magnets here:

brainly.com/question/26171648

#SPJ1

3 0

1 year ago

Plants grow over time by balancing two chemical reactions: photosynthesis and cellular respiration. Plants use the products of p

Andreyy89

The answer to that I think would be D because:

A is wrong because it doesn't make sense the way the order it is in,

B doesn't make sense because it is in the wrong order, and

C says the same exact thing that photosynthesis says which would mean that it is representing the photosynthesis equation not the chemical equation for cellular respiration.

Also it is D because D shows the correct way that the chemical equation representing cellular respiration is supposed to be in.

Hope this helped!

8 0

3 years ago