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3 years ago

PLEASE HELP! Need this ASAP (in 10 minutes)

2 answers:

6 0

The answer to your question here is A,

(I am dragging a very helpful piece of information from a website call “The Physics Classroom” for an explain took by the way so go check them out if you are still having trouble.)

From the paragraph I read, I could pull out “The data convincingly show that wave frequency does not affect wave speed.”

Thus concluding my answer of why the speed is unchanged.

9966 [12]3 years ago

4 0

I think it’s B

step-by-step explanation:

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For a converging lens, a ray arriving parallel to the optic axis

mixer [17]

Answer:

b. passes through the principal focal point.

Explanation:

Light wave can be defined as an electromagnetic wave that do not require a medium of propagation for it to travel through a vacuum of space where no particles exist.

A lens can be defined as a transparent optical instrument that refracts rays of light to produce a real image.

Basically, there are two (2) main types of lens and these includes;

I. Diverging (concave) lens.

II. Converging (convex) lens.

A converging lens refers to a type of lens that typically causes parallel rays of light with respect to its principal axis to come to a focus (converge) and form a real image. This type of lens is usually thin at the lower and upper edges and thick across the middle.

For a converging lens, a ray arriving parallel to the optic axis passes through the principal focal point.

8 0

3 years ago

A camera lens focuses on an object 75.0 cm from the lensThe image forms 3.50 cm behind the lens. What is the magnification of th

N76 [4]

Answer:

7/150

Explanation:

The following data were obtained from the question:

Object distance (u) = 75cm

Image distance (v) = 3.5cm

Magnification (M) =..?

Magnification is simply defined as:

Magnification (M) = Image distance (v)/ object distance (u)

M = v /u

With the above formula, we can obtain the magnification of the image as follow:

M = v/u

M = 3.5/75

M = 7/150

Therefore, the magnification of the image is 7/150.

6 0

3 years ago

Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point. F⃗ 1F→1F_1_vec has a magnitude of 9.20 NN and is directed at an a

BARSIC [14]

Answer:

The x component of the resultant force is -7.27N.

Explanation:

To obtain the x component of the resultant force, first we have to know the x components of the other forces. To do this, we just have to do some trigonometry:

$|F_{1x}|=|F_1|\cos\theta_1=9.20N\cos62.0\°=4.31N \\|F_{2x}|=|F_2|\cos\theta_2=5.00N\cos53.6\°=2.96N$

Since both vectors are in the left side of the y-axis, they have a negative x component. So:

$F_{1x}=-4.31N;\\F_{2x}=-2.96N$

Finally, we sum both components to obtain the component of the resultant force:

$F_{Rx}=-4.31N-2.96N=-7.27N$

In words, the x component of the resultant force is -7.27N.

6 0

3 years ago

A4 kg bowling ball begins rolling down a at bowling alloy at 6 m/s . When it strikes the pins, it is estimated to be moving at 5

Paul [167]

Answer:

Energy lost due to friction is 22 J

Explanation:

Mass of the ball m = 4 kg

Initially velocity of ball v = 6 m/sec

So kinetic energy of the ball $KE=\frac{1}{2}mv^2$

$KE=\frac{1}{2}\times 4\times 6^2=72J$

Now due to friction velocity decreases to 5 m/sec

Kinetic energy become

$KE=\frac{1}{2}\times 4\times 5^2=50J$

Therefore energy lost due to friction = 72 -50 = 22 J

8 0

3 years ago

Dans un tube en U contenant du mercure ,on verse de l'autre côté de l'acide sulfurique de densité 1,84 et de l'autre côté de l'a

Iteru [2.4K]

Explanation:

Unclear question. The clear rendering reads;

"Into a U-tube containing mercury, pour on the other side sulfuric acid of density 1.84 and on the other side alcohol of density 0.8 so that the levels are in the same horizontal plane. The height of the acid above the mercury being 24 cm. What is the height of the bar and what variation of the level of the acid, when the mercury density is 13.6?

6 0

3 years ago