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Daniel [21]
3 years ago
8

9) If, after viewing a specimen at low power, you switch to high-dry power and, after using fine focus, cannot find the specimen

, what things could you do to help yourself (before asking the instructor to assist you
Physics
1 answer:
xeze [42]3 years ago
4 0

Answer:

See the answer below

Explanation:

<em>The best thing one can do in this case would be to return the microscope's objective to low power and then </em><em>re-center the specimen </em><em>before switching back to high-dry power.</em>

Most of the time, <u>what makes the specimen under the microscope to be out of focus at higher objective powers after being in focus at low power is because they are not properly centered on the stage</u>.  Hence, before calling on the instructor, it would be wise to first return to low power, re-center the specimen and bring it into focus after which the high power objective can be returned to and the fine focus adjusted to bring the image back to focus.

After doing the above and the specimen still does not come into focus, then the instructor can be called upon.

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A spacecraft is moving past the earth at a constant speed of 0.60 times the speed of light. The astronaut measures the time inte
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Answer:

the time interval that an earth observer measures is 4 seconds

Explanation:

Given the data in the question;

speed of the spacecraft as it moves past the is 0.6 times the speed of light

we know that speed of light c = 3 × 10⁸ m/s

so speed of spacecraft v = 0.6 × c = 0.6c

time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds

Now, from time dilation;

t = Δt₀ / √( 1 - ( v² / c² ) )

t = Δt₀ / √( 1 - ( v/c )² )

we substitute

t = 3.2 / √( 1 - ( 0.6c / c )² )

t = 3.2 / √( 1 - ( 0.6 )² )

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t = 3.2 / √0.64

t = 3.2 / 0.8

t = 4 seconds

Therefore, the time interval that an earth observer measures is 4 seconds

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3 years ago
A string under a tension of 50.4 N is used to whirl a rock in a horizontal circle of radius 2.51 m at a speed of 21.1 m/s. The s
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Answer:

619.8 N

Explanation:

The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:

T=m\frac{v^2}{r}

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m is the mass of the rock

v is the speed

r is the radius of the circular path

At the beginning,

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v = 21.1 m/s

r = 2.51 m

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m=\frac{Tr}{v^2}=\frac{(50.4)(2.51)}{21.1^2}=0.284 kg

Later, the radius of the string is decreased to

r' = 1.22 m

While the speed is increased to

v' = 51.6 m/s

Substituting these new data into the equation, we find the tension at which the string breaks:

T'=m\frac{v'^2}{r'}=(0.284)\frac{(51.6)^2}{1.22}=619.8 N

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