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3 years ago

7

What magnification is best for the observation of most tissue slides? Explain why this has proved to be true?

1 answer:

Komok [63]3 years ago

7 0

Answer:

The 10X objective is use for the identification of actual size of histology tissues and 4X magnification is best for observation of most tissues slides

Explanation:

4X magnification is best for observation of most tissues slides because it has an objective lens that have lower power and have great high field overview which make it very easier to locate specimens on the slide. It is use to get the overview of histology slides. It is use to showcase more detailed observations about histology.

The 40X objective is use majorly to identify tissue , to observe the finer details and study tissue organization on the histology slide.

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Which of the following epidermal cells are the sensory receptors for touch?

never [62]

Merkel cells are the sensory receptors for touch.

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3 years ago

A 70.9-kg boy and a 43.2-kg girl, both wearing skates face each other at rest on a skating rink. The boy pushes the girl, sendin

Lelechka [254]

Answer:

Explanation:

Given

mass of boy $m_b=70.9\ kg$

mass of girl $m_g=43.2\ kg$

speed of girl after push $v_g=4.64\ m/s$

Suppose speed of boy after push is $v_b$

initially momentum of system is zero so final momentum is also zero because momentum is conserved

$P_i=P=f$

$0=m_b\cdot v_b+m_g\cdot v_g$

$v_b=-\frac{m_g}{m_b}\times v_g$

$v_b=-\frac{43.2}{70.9}\times 4.64$

$v_b=-2.82\ m/s$

i.e. velocity of boy is 2.82 m/s towards west

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2 years ago

A hammer strikes one end of a thick iron rail of length 8.80 m. A microphone located at the opposite end of the rail detects two

stepladder [879]

Answer:

ΔT = 0.02412 s

Explanation:

We will simply calculate the time for both the waves to travel through rail distance.

FOR THE TRAVELING THROUGH RAIL:

$T_{rail} = \frac{Distance}{Speed\ of\ Sound\ in\ Rail}\\\\T_{rail} = \frac{8.8\ m}{5950\ m/s}\\\\T_{rail} = 0.00148\ s$

FOR THE WAVE TRAVELING THROUGH AIR:

$T_{air} = \frac{Distance}{Speed\ of\ Sound\ in\ Air}\\\\T_{air} = \frac{8.8\ m}{343\ m/s}\\\\T_{air} = 0.0256\ s$

The separation in time between two pulses can now be given as follows:

$\Delta T = T_{air}-T_{rail} \\\Delta T = 0.0256\ s - 0.00148\ s\\$

<u>ΔT = 0.02412 s</u>

3 0

2 years ago

a basketball player can leap upward .65m how long does the basketball player remain in the air use 9.81m/s²

tigry1 [53]

At the player's maximum height, their velocity is 0. Recall that

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$0^2-{v_i}^2=-2g(0.65\,\mathrm m)\implies v_i=3.6\dfrac{\rm m}{\rm s}$

The player's height at time $t$ is given by

$y=v_it-\dfrac g2t^2$

so we find their airtime to be

$0.65\,\mathrm m=\left(3.6\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2\implies t=0.36\,\mathrm s$

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Anna [14]

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