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marissa [1.9K]
2 years ago
11

Based on the video, which venus phase would be impossible to see (from earth) of venus orbited earth as described in ptolemy's e

arth-centered model
Physics
2 answers:
sergey [27]2 years ago
7 0

In Ptolemy's Earth-focused model for the nearby planetary group, Venus consistently remains nearby to the Sun in the sky and, in light of the fact that it generally remains among Earth and the Sun, its stages run uniquely among new and bow.

Further Explanation:  

phase of Venus when it is closest which when farthest How do you know:

At the point when Venus is nearest to the Earth it is in its new stage and can't be seen. At the point when Venus is in its full stage it is at its most far off from the Earth yet at the same time can't be viewed for what it's worth behind the Sun. Venus is most splendid when it is 39 degrees from the Sun.

phases of Venus reveal about its orbit:

The periods of Venus result from the planet's circle around the Sun inside the Earth's circle giving the adaptive onlooker a grouping of dynamic lighting comparable in appearance to the Moon's stages. It demonstrates a quarter stage when it is at its greatest lengthening from the Sun.  

phases of Venus proved:

Utilizing his telescope, Galileo found that Venus experienced stages, much the same as our Moon. Be that as it may, the nature of these stages must be clarified by Venus circumventing the Sun, not the Earth. Galileo's perceptions of the periods of Venus practically demonstrated that the Earth was not the focal point of the universe.

Subject:  physics

Level: High School

Keywords: phase of Venus when it is closest which when farthest How do you know, phases of Venus reveal about its orbit, phases of Venus proved.

Related links:  

Learn more about evolution on

brainly.com/question/4596821

brainly.com/question/11220201

kaheart [24]2 years ago
3 0
Ptolemy's model theorizes that Earth is the center of the universe, and that all planets and the Sun revolve around it. If Venus is between the Earth and the Sun, then the Venus must always be crescent or all dark. <em>Thus, it is impossible to see a full or gibbous face of Venus. </em>Later on, this was disproved by Galileo when he saw through the telescope all phases of Venus. Hence, the heliocentric theory was approved instead.
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(a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500 - μC charge and flies due west at a sp
12345 [234]

(a) 2.64\cdot 10^{-8} N north

We can treat the aircraft as a single point charge moving in a magnetic field. In this case, the magnetic force exerted on the plane is

F=qvB sin \theta

where

q=0.500 \mu C = 0.500\cdot 10^{-6} C is the charge on the plane

v = 660 m/s is the velocity

B=8.00\cdot 10^{-5} T is the magnitude of the magnetic field

\theta=90^{\circ} is the angle between the direction of motion of the jet and of the magnetic field

Substituting,

F=(0.5\cdot 10^{-6})(660)(8.0\cdot 10^{-5})=2.64\cdot 10^{-8} N

The direction can be found by using Fleming's left hand rule. We have:

- index finger: magnetic field direction (straight up)

- middle finger: velocity of the plane (due west)

- force: thumb --> north

(b) Not negligible

As we can see from part (a), the magnitude of the force is not really big, so the effects are negligible.

For instance, we can compare this force with the weight of a plane. If we take a Boeing 737, its mass is about 80,000 kg, so its weight is

W=mg=(80000)(9.8)=784,000 N

As we can see, this is several orders of magnitude bigger than the magnetic force calculated at point (a), so the effects of the magnetic force are negligible.

8 0
2 years ago
A 2200 kg truck has put its front bumper against the rear bumper of a 2400 kg SUV to give it a push. With the engine at full pow
leonid [27]

Answer:

a) The maximum possible acceleration the truck can give the SUV is 7.5 meters per second squared

b) The force of the SUV's bumper on the truck's bumper is 18000 newtons

Explanation:

a) By Newton's second law we can find the relation between force and acceleration of the SUV:

F=ma

With F the maximum force the truck applies to the SUV, m the mass of the SUV and a the acceleration of the SUV; solving for a:

a=\frac{F}{m}=\frac{18000}{2400}\approx7.5\,\frac{m}{s^{2}}

b) Because at this acceleration the truck's bumper makes a force of 18000 N on the SUV’s bumper by Third Newton’s law the force of the SUV’s bumper on the truck’s bumper is 18000 N too because they are action-reaction force pairs.

7 0
2 years ago
Which of the following best describes the dependant variable?
weeeeeb [17]
Hello!

The independent variable is the variable deliberately changed.

The dependent variable is the variable that responds to change. So the answer is A.

Hope this helps. Any questions please just ask! Thank you!
5 0
2 years ago
A 3kW oven supplied with 9mJ of energy.How many minutes can it run for?
andreev551 [17]

<span>Hello!
 
We have the following data:
</span>
Time (T) = ? (in minutes)
Power (P) = 3 kW → 3000 W
Energy (E) = 9 MJ → 9000000 J or (W/s)

Formula of the consumption of electric energy:

P =  \frac{E}{T}

Solving:

P = \frac{E}{T}

P = \frac{E}{T} \to T =  \frac{E}{P}

T =  \frac{9000\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W/s}{3\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W}

\boxed{T = 3000\:seconds}

How many minutes can it run for? (<span>Let's convert in minutes)
</span>
1 minute --------- 60 seconds
y minute --------- 3000 seconds

\frac{1}{y} = \frac{60}{3000}

<span>Product of extremes equals product of means
</span>
60*y = 1*3000

60y = 3000

y =  \frac{3000}{60}

\boxed{\boxed{y = 50\:minutes}}\end{array}}\qquad\quad\checkmark


I hope this helps! =)
<span>

</span>
7 0
2 years ago
Read 2 more answers
A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro
yanalaym [24]

Answer:

a) v_{o} =16m/s

b) v=9.8m/s

c) \beta =-35.46º

Explanation:

From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial

Being said that, we can calculate the initial velocity of the ball

a) First we analyze its horizontal motion

x=v_{ox}t

x=v_{o}cos(60)t

v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)} (1)

That would be our first equation

Now, we need to analyze its vertical motion

y=y_{o}+v_{oy}t+\frac{1}{2}gt^2

y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2

Knowing v_{o} in our first equation (1)

8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2

\frac{1}{2}(9.8)t^2=16tan(60)-8

Solving for t

t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s

So, the ball takes to seconds to get to the other building. Now we can calculate its <u>initial velocity</u>

v_{o}=\frac{16m}{(2s)cos(60)}=16m/s

b) To find the <u>magnitude of the ball just before it strikes the building</u> we need to calculate its x and y components

v_{x}=v_{ox}+at=16cos(60)=8m/s

v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s

So, the magnitude of the velocity is:

v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s

c) The <u><em>direction of the ball</em></u> is:

\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º

4 0
2 years ago
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