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4 years ago

If more waves start passing a certain point every second,

Physics

1 answer:

ArbitrLikvidat [17]4 years ago

8 0

It would be B. I think

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If we use 1 millimeter to represent 1 light-year, how large in diameter is the Milky Way Galaxy?

storchak [24]

Answer:

d.100 meters

Explanation:

The diameter of the Milky Way Galaxy is approximately 100,000 light years.

Here we are using 1 millimiter (1 mm) to represent 1 light-year (1 ly). So, we can set the following proportion:

$1 mm : 1 ly = x : 100,000 ly$

and by finding x, we find the diameter of the Milky Way Galaxy in the scale used:

$x=\frac{(1mm )(100,000 ly)}{1 ly}=100,000 mm = 100 m$

so the correct answer is

d. 100 meters

4 0

3 years ago

An object 5.Ocm in the length is placed at a distance of 20cm in front of convex mirror

andrew-mc [135]

Answer:

Position = $\frac{60}{7}\ cm$ behind the mirror

Nature = Virtual and Erect

Size = $\frac{15}{7}\ cm$ : Diminished

Explanation:

Sign convention-Distance measured to the left of pole is negative and to the right of pole is positive.

Object distance = u = -20 cm

Focal length = f = Radius of curvature/2 = 30/2 = 15 cm

We have to use mirror formula to find image distance.

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\\ \frac{1}{-20}+\frac{1}{v}=\frac{1}{15}\\ \frac{1}{v}=\frac{7}{60}\\v=\frac{60}{7}\ cm$

Since the image distance is positive, it is formed behind the mirror or a virtual image is formed.

Magnification = $=\frac{h_{image}}{h_{object}}=$ $-\frac{v}{u}=\frac{60}{7\times20}=\frac{3}{7}$

Height of the object = 5 cm

Height of the image = $5\times\frac{3}{7}=\frac{15}{7}\ cm$

Since the height of the image is positive and less than the size of object,it is erect and diminished.

4 0

3 years ago

An athlete competing in long jump leaves the ground with a speed of 9.14 m/s at an angle of 35.00 above the horizontal. What is

Lemur [1.5K]

Answer:

R = 8.01 m

Explanation:

We can solve this problem using the projectile launch equations. The jump length is the throw range

R = v₀² sin 2θ / g

in the exercise they give us the initial speed of 9.14 m / s and in the launch angle 35º

let's calculate

R = 9.14² sin (2 35) / 9.8

R = 8.01 m

this is the jump length

5 0

3 years ago

The focal length is the distance from the lens (or mirror) to the: image object focal point center of curvature

galben [10]

The answer is focal point.
The focal length is the distance from the lens (or mirror) to the focal point. The focal point is <span>the point at which rays of light converge.</span>

5 0

4 years ago

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When liquid water gets into cracks of rock and freezes, it __ and ___.

hoa [83]

It expands and pushes the crack further aprt

3 0

2 years ago

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