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2 years ago

A(n) ________ microscope keeps an object in focus when the objective lens is changed.

Physics

1 answer:

Anvisha [2.4K]2 years ago

6 0

Parfocal is the term used to describe a microscope that maintains focus when the objective lenses are replaced.

<h3>What is the name of the objective lens ?</h3>

For observing minute features within a specimen sample, a high-powered objective lens, often known as a "high dry" lens, is perfect. You can see a very detailed image of the specimen on your slide thanks to the 400x total magnification that a high-power objective lens and a 10x eyepiece provide.

The four objective lenses on your microscope are for scanning (4x), low (10x), high (40x), and oil immersion (100x).

The first-stage lens used to create a picture from electrons leaving the specimen is referred to as the "objective lens." The objective lens is the most crucial component of the imaging system since the quality of the images is determined by how well it performs (resolution, contrast, etc.,).

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A grocery cart of mass 16 kg is being pushed at a constant speed up a 12-degree ramp by a force of [{MathJax fullWidth='false' f

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Answer:

If the cart is being pushed at a constant speed, then the acceleration in the direction of motion is zero. Hence, the force in the direction of the motion is zero, according to Newton's Second Law.

$\Sigma F_x = ma_x$

For simplicity, I will denote the direction along the inclined ramp as x-direction.

In the question the value of the force is not clearly given, so I will denote it as $F_P$

$\Sigma F_{net_x} = ma_x\\\Sigma F = F_{P}\cos(29^\circ) - mg\sin(12^\circ) = ma_x = 0\\F_{P}\cos(29^\circ) = mgsin(12^\circ)\\F_{P}\times 0.8746 = 16\times 9.8\times 0.2079\\F_{P} = 37.2740$

Here the angle between the applied force and the x-direction is 12° + 17° = 29°

The x-component of the weight of the cart is equal to sine component of the weight.

Since the cart is rolling on tires the kinetic friction does no work.

Work done by the applied force:

$W_{F_P} = F_P_x \cos(29^\circ)\times 7.5 = 244.5 ~J$

Work done by the weight of the cart:

$W_{mg} = -mg\sin(12^\circ)\times 7.5 = -16\times 9.8 \times 0.2079 \times 7.5 = - 244.5~J$

Since the x-component of the weight is in the -x-direction, its work is negative.

Conveniently, the total work done on the particle is zero, since its velocity is constant.

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