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____ [38]
3 years ago
9

Can someone help me on 2 science question,

Physics
2 answers:
gavmur [86]3 years ago
8 0
1. I think you should compare diagrams of moon phases from the textbook to diagrams of moon phases online. Because if you pick D it will take to long and C will help you out whith 3 different things to look at. 

2. The moon changes in appearances from the perspective of people on earth because it's revolving around the planet and the earth is revolving around the sun, so  A. Hoped this helped.

Amiraneli [1.4K]3 years ago
7 0
For question 1, I would go with option B. This is how science works. You start off with a hypothesis and then you test to see if what you observe matches the hypothesis. For option A, you risk not getting complete data. If the night you pick to start happens to be a full moon, you're done before you start. Option C is only comparing one set of stated facts against another set of stated facts. You have to trust these at face value. Both could be wrong. Option D says "as often as possible" which is too vague. If it was every night, then you would have your evidence six times over. If you can only test every 28th day, then you're going to see the same phase of the moon each time.

Question 2: The answer is B. The moon revolves around the Earth. As the Moon orbits the Earth, it is in a different part of the sky with relation to the Sun. When the Moon is fully behind the Earth, the Sun shines straight on and we see a full moon. When the moon is between the Earth and the Sun the light from the Sun falls on the side farthest away from the Earth and so we don't see it. 
Here is a diagram to illustrate
http://www.moonconnection.com/images/moon_phases_diagram.jpg
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3 years ago
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An object is moving in a straight line along the y axis. As a function of time, its position is given by the equation y=3.0+4.8x
romanna [79]

Answer:

Explanation:

<u>Instant Velocity and Acceleration </u>

Give the position of an object as a function of time y(x), the instant velocity can be obtained by

v(x)=y'(x)

Where y'(x) is the first derivative of y respect to time x. The instant acceleration is given by

a(x)=v'(x)=y''(x)

We are given the function for y

y(x)=3.0+4.8x +6.4x^2

Note we have changed the last term to be quadratic, so the question has more sense.

The velocity is

v(x)=y'(x)=4.8+12.8x

And the acceleration is a(x)=v'(x)=12.8

5 0
3 years ago
A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein
RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of 8m^{3}/s. As you  may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

V_{cone}=\frac{1}{3} \pi r^{2}h

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

\frac {r}{h}=\frac{4}{5}

When solving for r, we get:

r=\frac{4}{5}h

so we can substitute this into our volume of a cone formula:

V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h

which simplifies to:

V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h

V_{cone}=\frac{16}{75} \pi h^{3}

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}

Which simplifies to:

\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}

Now we can substitute the provided values into our equation. So we get:

\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}

so:

\frac{dh}{dt}=0.25m/s

3 0
2 years ago
Two mirrors are touching so they have an angle of 35.4 degrees with one another. A light ray is incident on the first at an angl
alexandr1967 [171]

Answer:

54.6°

Explanation:

From law of reflection i=r.

So, construct the reflected ray at 55.7°degrees from the normal and let it fall on the other mirror.  

Now draw the second normal at the point of incidence and again measure the angle of incidence, and draw the angle of reflection.

If you consider triangle AOB, one angle is ∠AOB=90°

 and ∠OAB is 54.6°

 

From angle sum property third angle ie ∠ABO=180°-90°-54.6°=35.4°

 

So, the second incident angle will be 54.6°

Hence, the second reflected angle will be 54.6 degrees.

8 0
3 years ago
What would occur if the index of refraction of a material was less than n=1?
Bingel [31]

The correct answer is (A). The speed of light would increase to a speed larger than the maximum speed of light in vacuum.

The index of refraction is the ratio of speed of light in vacuum to the speed of light in a medium.

n=C/V

here, n is the index of refraction, c the speed of light in vacuum, v is speed of light in any medium.

Now if the value of index of refraction is less than one, than the value of speed of light would be greater than the speed of light in the vacuum.


3 0
3 years ago
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