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Evgen [1.6K]
4 years ago
13

What is the correct sequence for light traveling through a microscope?

Physics
1 answer:
Alexandra [31]4 years ago
3 0
The answer would be B. 

Since the light source comes from the bottom, you just need to find the sequence from the bottom going up. 

It will start with the light source, then go through the slide, which will then go through the objective lens and lastly, the eyepiece lens. In summary, the sequence is:
 
B. Light --- slide ---- objective lens --- eyepiece lens
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7 0
3 years ago
Why is Vy negative and not positive because I solved it and it resulted in a positive number?
11Alexandr11 [23.1K]

Explanation:

You are given the initial velocity, the displacement, and the acceleration.  You're looking for the final velocity.  So you use the equation:

v² = v₀² + 2aΔy

When you solve for v, you take the square root.  Your calculator will return a positive answer, but there are actually two possible answers: positive and negative.

v = ±√(v₀² + 2aΔy)

You must use the physical context of the problem.  If we take up to be the positive direction, then v must be negative, since the projectile is moving down.

8 0
3 years ago
The electric field between two parallel plates is uniform, with magnitude 646 N/C. A proton is held stationary at the positive p
Ahat [919]

Answer:

The distance from the positive plate at which the two pass each other is 0.0023 cm.

Explanation:

We need to find the acceleration of each particle first. Let's use the electric force equation.

F=Eq

ma=Eq

<u>For the proton</u>

m_{p}a_{p}=Eq_{p}

a_{p}=\frac{Eq_{p}}{m_{p}}

a_{p}=\frac{646*1.6*10^{-19}}{1.67*10^{−27}}

a_{p}=6.19*10^{10}\: m/s^{2}

<u>For the electron</u>

m_{e}a_{e}=Eq_{e}

a_{e}=\frac{Eq_{e}}{m_{e}}

a_{e}=\frac{646*1.6*10^{-19}}{9.1*10^{−31}}  

a_{e}=1.14*10^{14}\: m/s^{2}

Now we know that the plate separation is 4.26 cm or 0.0426 m. The travel distance of the proton plus the travel distance of the electron is 0.0426 m.

x_{p}+x_{e}=0.0426

Both of them have an initial speed equal to zero. So we have:

\frac{1}{2}a_{p}t^{2}+\frac{1}{2}a_{e}t^{2}=0.0426

t^{2}(a_{p}+a_{e})=2*0.0426

t^{2}=\frac{2*0.0426}{a_{p}+a_{e}}

t=\sqrt{\frac{2*0.0426}{6.19*10^{10}+1.14*10^{14}}}

t=2.73*10^{-8}\: s    

With this time we can find the distance from the positive plate (x(p)).

x_{p}=\frac{1}{2}a_{p}t^{2}

x_{p}=\frac{1}{2}6.19*10^{10}*(2.73*10^{-8})^{2}

x_{p}=0.0023\: cm

Therefore, the distance from the positive plate at which the two pass each other is 0.0023 cm.

I hope it helps you!

7 0
3 years ago
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