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MAVERICK [17]
3 years ago
7

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

Physics
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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g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
Veseljchak [2.6K]

Answer:

The speed of q₂ is 4\sqrt{10}\ m/s

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

E_{i}=E_{f}

\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2

\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})

Put the value into the formula

\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})

0.00075(400-v_{f}^2)=0.18


400-v_{f}^2=\dfrac{0.18}{0.00075}

-v_{f}^2=240-400

v_{f}^2=160

v_{f}=4\sqrt{10}\ m/s

Hence, The speed of q₂ is 4\sqrt{10}\ m/s

7 0
2 years ago
What are some ways houses along the coastlines can protect themselves from storm surges?
BARSIC [14]
Build walls around the coast
8 0
3 years ago
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lina2011 [118]

Answer:

the ans is D... good luck

3 0
2 years ago
The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c
eimsori [14]

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

X_f = X_i +\frac{1}{2}(V_i+V_f)t

Where,

X = Desplazamiento

V = Velocity

t = Time

In this case there is no initial displacement or initial velocity, therefore

X_f = \frac{1}{2} (V_i+V_f)t

Clearing for time,

t = \frac{2X_f}{(V_i+V_f)}

t = \frac{2*1.2}{24+0}

t = 0.1s

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

a= \frac{\Delta V}{t}

Then

F = m\frac{\Delta V}{t}

F = m\frac{V_f-V_i}{t}

F = m\frac{-V_i}{t}

F = \frac{(1600kg)(-24m/s)}{(0.1s)}

F = -384000N

Negative symbol is because the force is opposite of the direction of moton.

PART C) Acceleration through kinematics equation is defined as

V_f^2=V_i^2-2ax

0 = (24m/s)^2-2*a(1.2m)

a = \frac{(24m/s)^2}{1.2m}

a=480m/s^2

The gravity is equal to 0.8, then the acceleration is

a = 480*\frac{g}{9.8}

a = 53.3g

3 0
2 years ago
A policeman in a stationary car measures the speed of approaching cars by means of an ultrasonic device that emits a sound with
beks73 [17]

Answer:

4.6 kHz

Explanation:

The formula for the Doppler effect allows us to find the frequency of the reflected wave:

f'=(\frac{v}{v-v_s})f

where

f is the original frequency of the sound

v is the speed of sound

vs is the speed of the wave source

In this problem, we have

f = 41.2 kHz

v = 330 m/s

vs = 33.0 m/s

Therefore, if we substitute in the equation we find the frequency of the reflected wave:

f'=(\frac{330 m/s}{330 m/s-33.0 m/s})(41.2 kHz)=45.8 kHz

And the frequency of the beats is equal to the difference between the frequency of the reflected wave and the original frequency:

f_B = |f'-f|=|45.8 kHz-41.2 kHz|=4.6 kHz

6 0
3 years ago
Read 2 more answers
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