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Alex787 [66]
3 years ago
12

Describe the changes in the field of view and the amount of available light when going from low to high power using the compound

microscope
Physics
1 answer:
kogti [31]3 years ago
8 0
A light microscope is a common lab tool used to study microscopic organisms and structures. When increasing the magnification (power) of the microscope, the field of view is decreased, since the microscope is essentially "zooming" in to a smaller area. For a given numerical aperture, the amount of available light decreases when going to a higher magnification (M) because the light collection is proportional to 1 / M^2.
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A power lifter performs a dead lift, raising a barbell with a mass of 305 kg to a height of 0.42 m above the ground, giving the
Ann [662]

Answer:

Explanation:

Before it hits the ground:

The initial potential energy = the final potential energy + the kinetic energy

mgH = mgh + 1/2 mv²

gH = gh + 1/2 v²

v = √(2g (H - h))

v = √(2 * 9.81 m/s² * (0.42 m - 0.21 m))

v ≈ 2.0 m/s

When it hits the ground:

Initial potential energy = final kinetic energy

mgH = 1/2 mv²

v = √(2gH)

v = √(2 * 9.81 m/s² * 0.42 m)

v ≈ 2.9 m/s

Using a kinematic equation to check our answer:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2(9.8 m/s²)(0.42 m)

v ≈ 2.9 m/s

3 0
3 years ago
If you triple the net force on an object, the acceleration will___
gavmur [86]
The acceleration will triple as well
8 0
2 years ago
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Explain whether there can be forces act-
Xelga [282]

Technically friction is acting on the car because it is still rubbing against the street and gravity is pulling the car down preventing it from floating??? lol

5 0
2 years ago
How many oxygen (O) atoms are in a molecule of SiO2?<br> O A. 1<br> OB. 3<br> C. 4<br> D. 2
mafiozo [28]
I believe it is D. 2
3 0
2 years ago
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A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)
kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

W=Fd cos \theta

where

F is the magnitude of the force applied

d is the displacement of the crate

\theta is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied horizontally. Therefore, the work done is

W=(230)(4.0)(cos 0^{\circ})=920 J

b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

W=(1300)(4.0)(cos 0^{\circ})=5200 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
3 years ago
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