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2 years ago

Which discovery did Galileo make to support the theory that the planets, including Earth, orbit the Sun?

Physics

2 answers:

const2013 [10]2 years ago

7 0

The answer to your quesiton is,

A) Venus has phases.

-Mabel <3

vova2212 [387]2 years ago

6 0

The answer is A- Venus has phases.

The Copernicus's heliocentric (Sun-centered) theory stated that the Sun is at rest near the center of the Universe, and that the Earth, spinning on its axis once daily, revolves annually around the Sun. He proved this theory by discovering that Venus went through phases just like the moon by recording his observations through the telescope.He also disproved the theory that the sun and other planets revolve around the earth through his recorded observations seen via the telescope.

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How many ways can you see that heat can transfer from the inside to the outside of your home on a cold day? List at least three

Yanka [14]

Answer: Heat transfers through brick walls and glass through conduction. In conduction heat is transferred by vibration of molecules. most energetic molecules vibrate and pass on the energy to less energetic molecules. Then they vibrate and further pass on the energy. In this way heat is transferred out of the home. Heat also transfers through the leakage of warm air from doors and windows. This occurs through convection. In convection energy is transferred through bulk movement of liquid and air molecules. Heat also transfers from insulation. in insulation there is no material in between the layers. So heat transfer through insulation occurs through radiations that occurs by X-Rays, Ultravoilet rays etc.

4 0

2 years ago

What are Sir Issac Newton's three laws of motion?

Likurg_2 [28]

The first law is that every object stay at rest or stay in uniform motion in a straight line until it is forced to change its state by the action of an external force. This law is called law of inertia.

The second law is that the acceleration of an object is dependent upon two variables. the net force acting upon the object and the mass of the object. F= ma or force is equal to mass times acceleration. This law is known as the law of force and acceleration.

The third law is that for every action there is an equal and opposite reaction. every interaction there is a pair of forces acting on the two interacting objects. the size of forces on the first object equals the size of the force on the second object.

Hope this helps :)

can you please make this the brainliest answer it would really help . Thanks

4 0

2 years ago

Read 2 more answers

A 1.2 x10 3 kilogram automobile in motion strikes a 1.0 x 10 -4 kilogram insect as a result the insect is accelerated at a rate

Mariana [72]

According to Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
$F=ma$
where F is the magnitude of the force, m is the mass of the object and a its acceleration.

In this problem, the object is the insect, with mass $m=1.0 \cdot 10^{-4} kg$ . The acceleration of the insect is $a=1.0 \cdot 10^2 m/s^2$ , therefore we can calculate the force exerted by the car on the insect:
$F=ma=(1.0 \cdot 10^{-4} kg)(1.0 \cdot 10^2 m/s^2)=0.01 N$

How do we find the force exerted by the insect on the car?
According to Newton's third law (known as action-reaction law), when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. Therefore, the force exerted by the insect on the car is equal to the force exerted by the car on the object, so it is 0.01 N.

6 0

2 years ago

Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .

4 0

2 years ago

Please help with these questions as well! I need urgent help! I will give brainliest! God bless!

Radda [10]

6. Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.

F = ma

F = (15.3)(-9.8)

F = -150N

Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object. So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.

7. Same idea as question 2.

First determine the weight of the object:

F = ma

F = (30)(-9.8)

F = -294N

The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.

-294 + 150N + x = 0

x = 144N

So the person is exerting 144 N.

10. First find the force of block B to the right due to its acceleration:

F = ma

F = (24)(0.5)

F = 12N

So block B is moving 12N to the right relative to block A due to block A's movement to the left. However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A. The force that is causing block B to experience the lower relative force to the right is because of the friction. To find the friction:

The sum of the forces in the leftward and rightward direction for block B must equal 12N.

75 - x = 12

x = 63N

So the force of friction of block A on block B is 63N to the left.

5 0

2 years ago