As the Moon revolves around the Earth, it also rotates on its axis. Why is it that the same side of the Moon is always visible from Earth?

Answers

Answer 1
Answer:

Answer: The speed of the moon's rotation keeps the same side always facing Earth.

Explanation: Please mark me brainiest

Answer 2
Answer:

Answer:

The speed of the Moon's rotation keeps the same side always facing Earth.

Explanation:

got it right on study island :)


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How does the geosphere interact with the hydrosphere

Answers

 Plants (biosphere) draw water (hydrosphere) and nutrients from the soil (geosphere) and release water vapor into the atmosphere.

Question 7 of 10A railroad freight car with a mass of 32,000 kg is moving at 2.0 m/s when it
runs into an at-rest freight car with a mass of 28,000 kg. The cars lock
together. What is their final velocity?
A.1.1 m/s
B. 2.2 m/s
C. 60,000 kg•m/s
D. 0.5 m/s

Answers

Answer:

a

Explanation:

you take 32,000kg ÷2.0m

Complete the calculations for total magnification produced by various combinations of the eyepiece and objective lenses. You may assume that the magnification for the eyepiece is 10X for each question. 1. When the scanning (4X) objective is used the total magnification will be:________
2. When the low power (10X) objective is used the total magnification will be:________
3. When the high power (40X) objective is used the total magnification will be:________
4. When the oil immersion (100X) objective is used the total magnification will be:_________

Answers

Answer:

a) m_ttoal = 40x, b)     m_total = 100X, c) m_total = 400X,

d)     m_total = 1000 X

Explanation:

La magnificación o aumentos es el incremento de del tamaño de la imagen con respecto al tamaño original del objeto, en la mayoria del os sistema optico la magnificacion total es  el producoto de la magnificación del objetivo por la magnificación del ocular

       

    m_total = m_ objetivo  * m=ocular

apliquemos esto a nuestro caso

1)  m_total = 4 x * 10 x

    m_ttoal = 40x

2)  m_total = 10X * 10X

    m_total = 100X

3)mtotal = 40X * 10X

     m_total = 400X

4) m _totla = 100x * 10 X

     m_total = 1000 X

en este ultimo caso para magnificación grandes es decalcificar el objeto

The total magnification produced by different combinations of eyepiece and objective lenses in a microscope.

1. When the scanning (4X) objective is used, the total magnification will be 40X because the eyepiece magnification is 10X and the objective magnification is 4X.

2. When the low power (10X) objective is used, the total magnification will be 100X because the eyepiece magnification is 10X and the objective magnification is 10X.

3. When the high power (40X) objective is used, the total magnification will be 400X because the eyepiece magnification is 10X and the objective magnification is 40X.

4. When the oil immersion (100X) objective is used, the total magnification will be 1000X because the eyepiece magnification is 10X and the objective magnification is 100X.

Learn more about Microscope magnifications here:

brainly.com/question/36268796

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\huge{\gray{\sf Question:} }A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm, (b) 3 cm and (c) 0 cm.​

Answers

Answer:

here \: amplitude = 5cm (convert \: to \: si \: unit) = .05m \n t = .2sec \n omega =  (2\pi)/(t) \n  =  (2\pi)/(.2 )  = 10\pi  (rad)/(s) \n  we \: want \: find \: a \: and \: v \n we \: know \: that \: a =  -  {omega}^(2) x \n v = omega \sqrt{ {r}^(2)  -  {x}^(2) }  \n(1)x = .05m \n a =  -  {10\pi}^(2)  * .05 =  - 5 {\pi}^(2)  \frac{m}{ {s}^(2) }  \n v = 10\pi \sqrt{ {.05}^(2)  -  {.05}^(2) }  = 0 \n 2)x = 3cm = .03m \n a =  {(10\pi)}^(2)  * .03 =  - 3 {\pi}^(2)  \frac{m}{ {s}^(2) }  \n v = 10\pi \sqrt{ {.05}^(2)  -  {.03}^(2) }  = 10\pi * .04 = .4\pi (m)/(s)  \n 3)x = 0 \n a =  -  {(10\pi)}^(2)  * 0 =  0 \n v = 10\pi *  \sqrt{ {.05}^(2)  -  {0}^(2) }  =  - 10\pi * .05 = .5\pi (m)/(s)  \n thank \: you

Answer:

ANSWER

A = 5 cm = 0.05 m

T = 0.2 s

ω=2π/T=2π/0.2=10πrad/s

plz give brainlist

hope this helped

An object essentially at infinity is moved to a distance of 90 cm in front of a thin positive lens. In the process its image distance triples. Determine the focal length of the lens.

Answers

Answer:

67.5 cm

Explanation:

u = - 90 cm, v = 3 x u = 3 x 90 = 270 cm

let f be the focal length

Use lens equation

1 / f = 1 / v - 1 / u

1 / f = 1 / 270 + 1 / 90

1 / f = 4 / 270

f = 67.5 cm

Final answer:

To determine the focal length of the lens, we use the lens formula and set up an equation based on the given information. Solving for the image distance, we find that it is zero, indicating the image is formed at infinity. Therefore, the focal length of the lens is 90 cm.

Explanation:

To determine the focal length of the lens, we can use the lens formula:

1/f = 1/v - 1/u

Where f is the focal length, v is the image distance, and u is the object distance.

Given that the image distance triples when the object is moved from infinity to 90 cm in front of the lens, we can set up the following equation:

1/f = 1/(3v) - 1/(90)

Multiplying through by 90*3v, we get:

90*3v/f = 270v - 90*3v

90*3v/f = 270v - 270v

90*3v/f = 0

Simplifying further, we find that: v = 0

When the image distance is zero, it means the image is formed at infinity, so the lens is focused at the focal point. Therefore, the focal length of the lens is 90 cm.

It's a little hard can u help

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I can't really tell what (I) is, but F) blender def converts electric into motion of spinning blades
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