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USPshnik [31]
2 years ago
6

which part of the microscope will be used first to adjust the focus when starting with the lowest power lens?

Physics
1 answer:
noname [10]2 years ago
7 0

The first part of the microscope that should first be used to adjust the focus when starting with the lowest power lens would be the coarse adjustment knob.

There are two knobs in a typical light microscope with which objects on slides can be brought into focus:

  1. Coarse adjustment knob
  2. Fine adjustment knob

The 2 knobs are used to adjust the stage to either bring it up towards the objective lens or down away from them. The coarse adjustment knob, however, moves the stage a considerable distance with each turn. The fine adjustment knob, on the other hand, only moves the stage very little with each turn.

The lowest power lenses are often short. Hence, using the coarse adjustment knob is ideal in order to quickly bring objects on slides into focus.

The fine adjustment knob comes highly recommended at high objectives because high objectives lenses are usually long and using the coarse adjustment knob can lead to a breakage of the slide by the lens.

More on bringing objects into focus on a microscope can be found here: brainly.com/question/24319677

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Answer:

when you are pushing the pedal you are causing the pedal to move done and then you will move 100cm

Explanation:

10 cm= 100 cm moved so when you move you will move because you are timeing  the 10 by 100 to get the spped

3 0
3 years ago
PLEASEE HELPP
topjm [15]

Explanation:

u=166m/s, v=0(at it's highest point final velocity is zero), a=9.8m/s², t=8.6s

by the formula, S=ut+½at².

S=[166×8.6+½.×9.8×(8.6)²]. ...by calculation

S = 1427.6+362.404

S=1790.004m

hope this helps you.

4 0
3 years ago
⦁ Determining the magnetic flux, A rectangular piece of stiff paper measures 10 cm x 5 cm. You hold the piece of paper in a unif
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Explanation:

4 0
3 years ago
Estimate the constant rate of withdrawal (in m3 /s) from a 1375 ha reservoir in a month of 30 days during which the reservoir le
kap26 [50]

Answer:

Explanation:

1 ha = 10⁴ m²

1375 ha = 1375 x 10⁴ m² = 13.75 x 10⁶ m²

In flow in a month = .5 x 10⁶ x 30 m³ = 15 x 10⁶ m³

Net inflow after all loss = 18.5 - 9.5 - 2.5 cm = 6.5 cm = .065 m

Net inflow in volume = 13.75 x 10⁶ x .065 m³= .89375 x 10⁶ m³

Let Q be the withdrawal in m³

Q - 15 x 10⁶ - .89375 x 10⁶ = 13.75 x 10⁶ x .75 = 10.3125 x 10⁶

Q = 26.20 x 10⁶ m³

rate of withdrawal per second

= 26.20 x 10⁶ / 30 x 24 x 60 x 60

= 26.20 x 10⁶ / 2.592 x 10⁶

= 10.11 m³ / s

6 0
3 years ago
H. A truck starts to move from rest. If it gains the acceleration of 3 m/s2 in 5 sec,
SashulF [63]

Answer:

1. The final velocity of the truck is 15 m/s

2. The distance travelled by the truck is 37.5 m

Explanation:

1. Determination of the final velocity

Initial velocity (u) = 0 m/s

Acceleration (a) = 3 m/s²

Time (t) = 5 s

Final velocity (v) =?

The final velocity of the truck can be obtained as follow:

v = u + at

v = 0 + (3 × 5)

v = 0 + 15

v = 15 m/s

Therefore, the final velocity of the truck is 15 m/s

2. Determination of the distance travelled

Initial velocity (u) = 0 m/s

Acceleration (a) = 3 m/s²

Time (t) = 5 s

Distance (s) =?

The distance travelled by the truck can be obtained as follow:

s = ut + ½at²

s = (0 × 5) + (½ × 3 × 5²)

s = 0 + (½ × 3 × 25)

s = 0 + 37.5

s = 37.5 m

Therefore, the distance travelled by the truck is 37.5 m

6 0
2 years ago
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