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yaroslaw [1]
3 years ago
14

Is the part of the microscope that allows you to adjust the light?

Physics
1 answer:
madreJ [45]3 years ago
7 0
According to funtriva.com, the piece that allows you to adjust the amount of light that's coming through the microscope is called the adjustable diaphragm. It is located under to stage (where what you are observing is placed on) and can be rotated to make the light<span> intensity change</span>
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Based on the Punnett square, what percentage of offspring would have genotype YY?
mel-nik [20]
50 c should be right
6 0
3 years ago
Wavelength of blue photons 495 nm, what is the frequency? and what is the energy?
inessss [21]

Answer:

1.F: About 6*10^14 Hz

2.E: About 4*10^ -19 J

Explanation:

Frequency: We knew that the speed of a wave is its wavelength(λ)* frequency(f, in Hz).  By the wave-particle duality we know we can calculate the frequency of light in the same way. So, c=495nm *f, f=c/495nm=> (299,792,458 m/s) / (4.95*10^-7 m)

=6.05*10^14 /s

Energy: The energy photon contains can be calculate by this formula-- E=hf

f is the frequency and h is Planck's constant which is about 6.62 ×10^-34 *m^2*kg/s (after dimensional analysis ) =6.62*10^ -34 J*s.

So, the energy of a blue photon is (6.05*10^14)*(6.62*10^-34)=40.051*10^-20=  4.051*10^-19 J

3 0
3 years ago
When a 2.50 kg object is hung vertically on a certain light spring described by hookes law the spring stretches 2.76 cm.
Readme [11.4K]
F=K*X,
F=M*a 

M*a=K*X

2.5*9.81=K*0.0276

24.525=K*0.0276

24.525/0.0276=K

K= 888.6 N/m ---- force constant 

assuming 2.5 refers to the new extension, just divide F/ 0.025
to get

981N/m 


8 0
3 years ago
A weight lifter lifts a dumbbell a certain height in 2.0 s, while a competitor does the same workin 1.0 s. Compared to the power
Aloiza [94]

Answer:

a. one-half as great

Explanation:

The power developed by the first lifter is one-half as great as that of the second person.

  Power is defined as the rate at which work is done;

          Power  = \frac{workdone}{time}

Since the two lifters do the same work at different time, let us estimate their power;

       P₁ = \frac{workdone}{2}                     P₂ = \frac{workdone }{1}

   We see that for P₁, power is half of the work done whereas in P₂ power is the same as the work done.

Therefore,

           The power of the first weight lifter is one-half the second lifter.

4 0
3 years ago
A 1.5kw vacuum cleaner is used for half an hour. Calculate its energy usage.​
juin [17]

Answer:

its energy usage is 0.05

Explanation:

1.5kw/30=0.05

8 0
3 years ago
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