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2 years ago

What should you do if the high power objective lens touches or breaks the cover slip?

Physics

2 answers:

charle [14.2K]2 years ago

6 0

Answer : If this happens avoid using the high power objective lens, instead use other power lenses.

One should always use the compound microscope with great care. If the high power objective lens breaks the cover slip, it should be immediately informed to the teacher, professor, or any one who is authorized to handle this situation.

You can avoid this situation to occur by being extra careful, while using the high power objective lens adjust the lens using the fine knob adjustment.

barxatty [35]2 years ago

5 0

Firstly, always make it a point that the high power objective lens should never touch or break the coverslip but if the high power objective lens touches or breaks the coverslip,

then don’t use the high power magnification and during usage of high power magnification, only the fine knob adjustment should be used.

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A planar loop consisting of seven turns of wire, each of which encloses 200 cm2, is oriented perpendicularly to a magnetic field

PtichkaEL [24]

Answer:

The induced current is 0.084 A

Explanation:

the area given by the exercise is

A = 200 cm^2 = 200x10^-4 m^2

R = 5 Ω

N = 7 turns

The formula of the emf induced according to Faraday's law is equal to:

ε = (-N * dφ)/dt = (N*(b2-b1)*A)/dt

Replacing values:

ε = (7*(38 - 14) * (200x10^-4))/8x10^-3 = 0.42 V

the induced current is equal to:

I = ε /R = 0.42/5 = 0.084 A

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2 years ago

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A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver im

geniusboy [140]

Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

starting point. On the hill when running out of gas

Em₀ = K + U = ½ m v₀² + m g y₁

final point. Arriving at the gas station

Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

Em₀ = Em_f

½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

v_f ² = v₀² + 2g (y₁ -y₂)

we calculate

v_f ² = 20² + 2 9.8 (10 -15)

v_f = √302

v_f = 17.4 m / s

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2 years ago

A rocket exhausts fuel with a velocity of 1500m/s, relative to the rocket. It starts from rest in outer space with fuel comprisi

babunello [35]

Answer:

v= 2413.5 m/s

Explanation:

maximum change of speed of rocket

=(initial exhaust velocity)×ln [(initialmass/finalmass)]

let initial mass= m

final mass = m-m(4/5) = m/5

[since the 80% of mass which is fuel is exhausted]

V-0 = 1500 ln (1/0.2)

V= 1500×1.609 = 2413.5 m/s

therefore, its exhaust speed v= 2413.5 m/s

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2 years ago

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Artyom0805 [142]

Heat engines were developed during industrial revolution.

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The source of a heat engine is present at a higher temperature as compared to the sink. Due to the temperature difference, the heat will flow from source to sink through working substance.

Let us consider $T_{1}\ and\ T_{2}$ are the temperature of source and sink.

As the source is at higher temperature as compared to sink, heat will flow from source to sink.

$Let\ Q_{1}\ and\ Q_{2}$ are the heat provided by source and heat rejected to sink.

Hence, the work done by the working substance will be -

$W\ =\ Q_{1}-Q_{2}$

The efficiency of a heat engine is defined as the ratio of output to the input energy.

Here output = workdone [W]

Hence, the efficiency of a heat engine is calculated as -

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This is the expression for the efficiency of heat engine.

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Aleonysh [2.5K]

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