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1 year ago

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

1 answer:

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:

Coarse adjustment knob
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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A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

Answer:

The value is $T_R = 11.8 \ days$

Explanation:

From the question we are told that

The semi - major axis of the rocky debris $a_R = 45.0\ AU$

The semi - major axis of Planet D is $a_D = 60 \ AU$

The orbital period of planet D is $T_D = 18.164 \ days$

Generally from Kepler third law

$T \ \ \alpha \ \ a^{\frac{3}{2} }$

Here T is the orbital period while a is the semi major axis

$\frac{T_D}{T_R} = \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}$

=> $T_R = T_D * [\frac{a_R}{a_D} ]^{\frac{3}{2} }$

=> $T_R = 18.164 * [\frac{ 45}{60} ]^{\frac{3}{2} }$

=> $T_R = 11.8 \ days$

7 0

2 years ago

If a runner has a speed of 8.66m/s and runs for 46.2s what distance is covered? tv = d

kati45 [8]

Answer:

$\text{Using the formula: }v=\frac{d}{t}\\\therefore vt=d\\\text{Plug and chug:}46.2(8.66)=400.092\text{ metres}$

6 0

2 years ago

An unstretched spring has a length of 0.30 m. When the spring is stretched to a total length of 0.60 m, it supports traveling wa

IceJOKER [234]

Explanation:

Below is an attachment containing the solution

7 0

2 years ago

A 2 kg blue car is moving 6 m/s to the right and collides with a 3 kg red car that is moving 2 m/s to the left. The cars collide

snow_lady [41]

Answer:

Their velocity after the collision is 1.2 m/s, to the right.

Explanation:

Given;

mass of the blue car, m₁ = 2 kg

initial velocity of the blue car, u₁ = 6 m/s

mass of the red car, m₂ = 3 kg

initial velocity of the red car, u₂ = 2 m/s

let the blue car moving to the right be in positive direction

also, let the red car moving to the left be in negative direction

Apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ - m₂u₂ = v(m₁ + m₂)

where;

v is their velocity after the collision

(2 x 6) - (3 x 2) = v(2 + 3)

12 - 6 = 5v

6 = 5v

v = 6/5

v = 1.2 m/s, to the right

Therefore, their velocity after the collision is 1.2 m/s, to the right.

7 0

2 years ago

A. In one short sentence, explain why we call the force of gravity an attractive force.

kondor19780726 [428]

Answer:

Explanation:

(a) The force of gravity is called an attractive force because it is the force (although weak) in which a planetary body or matter uses to attract an object towards itself.

(b) Yes, it does and the formula for force of gravity between any two object is

F = G $\frac{m1m2}{r}$

where m1 and m2 are masses of the first and second object respectively

r is the distance between the center of the two masses

G is the gravitational constant

3 0

2 years ago