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2 years ago

On which part of the compound light microscope are specimens placed for viewing?.

1 answer:

7 0

The specimen that will be on display is set up on stage. When working at greater magnifications when careful motions of the specimen slide are needed, a mechanical stage is employed.

<h3>What is a compound light microscope?</h3>

A microscope having several lenses and a separate light source is called a compound light microscope.

Ocular lenses are located in the binocular eyepieces of this type of microscope, while objective lenses are located in a rotating nosepiece that is located closer to the specimen.

Hence, the part of the compound light microscope is specimens placed for viewing is stage.

To learn more about the compound light microscope refer;

brainly.com/question/1600608

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A cosmic ray muon with mass mμ = 1.88 ✕ 10−28 kg impacting the Earth's atmosphere slows down in proportion to the amount of matt

anyanavicka [17]

Answer:

a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b. this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon

Explanation:

F= ma

v²=u² -2aS

(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)

a=1.36×10⁹m/s²

recall

F=ma

F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²

F= 2.55 × 10⁻¹⁹N

the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b. this force compare to the weight of the muon

F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)

= 1.38 × 10⁸

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3 years ago

A boat with a mass of 1000 kg drifts with the current down a straight section of river parallel to the +x+x axis with a speed of

IrinaK [193]

Answer:

Explanation:

Our values are defined by,

$m=1000kg\\v = 2m/s\\a = 2.7 \angle 45\°$

We can express also in a vectorial way, then

$v=2\hat{i} \rightarrow$ no values in $\hat{j}$

Acceleration as follow,

$a= 2.7cos45 \hat{i} + 2.7 sin45 \hat{j}$

We know that velocity is given by,

$V(t) = v_0+at \rightarrow v_0 = 2$

We need to calculate for t=3, then

$v(3) = 2\hat{i} +(2.7cos45 \hat{i} + 2.7 sin45 \hat{j})*3$

$v(3) = (2\hat{i}+3*2.7cos45 \hat{i})+(3*2.7 sin45 \hat{j})$

$v(3) = (2+3*2.7cos45)\hat{i}+(5.7 sin45 )\hat{j}$

$v(3) = 7.7\hat{i}+5.7\hat{j}$

Our mass is 1000Kg, so the momentum is

$P = mv$

$P= 1000(7.7\hat{i}+5.7\hat{j})$

$P=7700\hat{i}+5700\hat{j}$

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4 years ago

A descent vehicle landing on the moon has a vertical velocity toward the surface of the moon of 29.6 m/s. At the same time, it h

inessss [21]

If the vertical component is 29.6 m/s down, and the horizontal component
is 54.8 m/s parallel to the surface, then the magnitude of the slanty vector is

√(29.6² + 54.8²) = √(876.16 + 3003.04) = √3879.2 = 62.28 m/s .

That's 139 mph ! Wow !

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3 years ago

What is the approximate pressure of air at sea level?

Len [333]

My guess would be 101 kPa but I'm not 100% sure, sorry.

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Fed [463]

The correct answer is C. HIPAA

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4 years ago

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