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Troyanec [42]
3 years ago
12

Which of the following is accurate in describing converging and diverging lenses?

Physics
2 answers:
kenny6666 [7]3 years ago
3 0

B. Both of these types of lenses have the ability to produce real images.

koban [17]3 years ago
3 0

Answer:

A. Both of these types of lenses have the ability to produce upright images.

Explanation:

Let's analyze each statement proposed:

A. Both of these types of lenses have the ability to produce upright images.  --> TRUE. In fact, depending on the position of the object relative to the lens, both converging and diverging lenses can produce an image which is upright.

B. Both of these types of lenses have the ability to produce real images.  --> FALSE. In fact, diverging lenses can produce virtual images only, because the rays of light are not focused into a single point.

C. Parallel rays converge into a single point at the focal point of a diverging lens, whereas parallel rays diverge at the focal point of a converging lens and spread out.  --> FALSE. It is actually the opposite: in a diverging lens, parallel rays diverge and spread out, while in a converging lens, parallel rays converge into a single point (the focal point)

D. Diverging lenses will produce images behind the lens, and converging lenses will produce images in front of the lens.  --> FALSE. It's actually the opposite: the image produced by a diverging lens is in front of the lens, while the image produced by a converging lens is behind the lens.


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The combination of all of the forces acting on an object is called the
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Salmon often jump waterfalls to reach their
natta225 [31]

Answer:

5.0 m/s

Explanation:

The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

v_x = u cos \theta

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d=v_x t = (ucos \theta)t

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

t=\frac{d}{v_x}=\frac{d}{u cos \theta} (1)

Along the vertical direction, the equation of motion is

y=h+u_y t -\frac{1}{2}gt^2

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

u_y = u sin \theta is the vertical component of the initial velocity of the salmon

g=9.81 m/s^2 is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}

and we can solve this formula for u, the initial speed of the salmon:

y=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}\\\\u=\sqrt{\frac{gd^2}{2(dtan \theta -y)cos^2 \theta}}=\sqrt{\frac{(9.81)(1.95)^2}{2((1.95)(tan 37.7^{\circ}) -0.311)cos^2 37.7^{\circ}}}=5.0 m/s

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3 years ago
The value of 1.0004 to the power 1 by 2 using Binomial approximation is​
IgorLugansk [536]

Given:

The given value is (1.0004)^{\frac{1}{2}}.

To find:

The value of the given expression by using the Binomial approximation.

Explanation:

We have,

(1.0004)^{\frac{1}{2}}

It can be written as:

(1.0004)^{\frac{1}{2}}=(1+0.0004)^{\frac{1}{2}}

(1.0004)^{\frac{1}{2}}=1+\dfrac{1}{2}\times 0.0004      [\because (1+x)^n=1+nx]

(1.0004)^{\frac{1}{2}}=1+0.0002

(1.0004)^{\frac{1}{2}}=1.0002

Therefore, the approximate value of the given expression is 1.0002.

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2 years ago
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Answer:

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Additionally, the ultraviolet spectrum is divided into three categories and these are; UVA, UVB and UVC.

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