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2 years 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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The vertical displacement of the wave is measured from the ?

sergey [27]

The whole question is talking about the amplitude of a wave
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Equilibrium to the crest . . . that's the amplitude.

Crest to trough . . . that's double the amplitude.

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3 years ago

Suppose you are on a cart, initially at rest, which rides on a frictionless horizontal track. You throw a ball at a vertical sur

Len [333]

Answer:

$F_c t_ c = -F_b t_b$

And the forces are equal but in the opposite direction. So then we can write by general rule:

$m_c \Delta V_{c} = -m_b \Delta V_b$

Or equivalently:

$m_c \Delta V_{c} +m_b \Delta V_b =0$

Where: $V_c$ represent the speed of the car and $V_b$ the speed of the ball

$m_c$ represent the mass of the car

$m_b$ represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

Explanation:

If we assume that we have the situation in the figure attached.

For this case we assume that the momentum changes are equal in magnitude and opposite in direction, so then we satisfy this:

$F_c t_ c = -F_b t_b$

And the forces are equal but in the opposite direction. So then we can write by general rule:

$m_c \Delta V_{c} = -m_b \Delta V_b$

Or equivalently:

$m_c \Delta V_{c} +m_b \Delta V_b =0$

Where: $V_c$ represent the speed of the car and $V_b$ the speed of the ball

$m_c$ represent the mass of the car

$m_b$ represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.