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Sergio039 [100]
3 years ago
12

What was Galileo’s contribution to the study of motion?

Physics
2 answers:
babunello [35]3 years ago
8 0

Before giving you the answer straight away, I’d like to make sure you understand each option.


FIRST OPTION


“He developed the three laws of motion."


This option is not only the incorrect answer, but it is an incorrect statement. Galileo did not develop the three laws of motion. Isaac Newton was the one to compiled the three laws of motion.


SECOND OPTION


“He was the first to systematically study force and motion."


This option is a statement is true, and it could also be the correct answer. Galileo Galilei used experiments to search for the cause of motion.


THIRD OPTION


“He was the first to discover gravity."


This option is not only the incorrect answer, but it is an incorrect statement. Galileo was not the person to discover gravity. Sir Isaac Newton was the first to discover gravity.


FINAL OPTION


“He improved on Newton's laws."


This option is not only the incorrect answer, but it is an incorrect statement. Galilei did not improve Sir Isaac Newton’s laws.


Now that we’ve gone through each of the options individually, we can go ahead and determine the correct answer to the question, which is...


B ) He was the first to systematically study force and motion.



- Marlon Nunez

Marta_Voda [28]3 years ago
8 0

Answer: Option B

Explanation: Galileo Galilei (1564 - 1642) was an Italian physicist, mathematician, engineer, and philosopher that is mainly known for the telescope and the heliocentrism.

One of the first things Galileo contributed to science was on forces and motion. Because of his mathematic formation, he had a very systematic and logic approach to the science and laws of nature, so his studies where advanced for the time.

The right answer here would be option B: "He was the first to systematically study force and motion."

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Answer:

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Compare the force of air resistance and the force of gravity on an object falling at its terminal velocity.
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The viscous force on an object moving through air is proportional to its velocity.
The only forces acting on an object when falling are air resistance and its weight itself. The weight acts vertically downwards whereas air resistance acts vertically upward.
Let F be the viscous force due to air molecules, B be buoyant force due to air and W be the weight of falling object. Initially, the velocity of falling object and hence the viscous force F is zero and the object is accelerated due to force
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Thus at terminal velocity, air resistance and force of gravity becomes equal.
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3 years ago
If a car changes its velocity from 32 km/hr to 54 km/hr in 8.0 seconds, what is its acceleration?
Stella [2.4K]
Use the equation for the acceleration
A = final velocity - initial velocity divided by time final - time initial
A= 54 - 32 / 8 - 0
A= 22 / 8 
A= 2.75 m/s^2 
Hope this helps!
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3 years ago
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A basketball player is 4.22 m from
max2010maxim [7]

Answer: The height above the release point is 2.96 meters.

Explanation:

The acceleration of the ball is the gravitational acceleration in the y axis.

A = (0, -9.8m/s^)

For the velocity we can integrate over time and get:

V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))

for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)

P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)

now, the time at wich the horizontal displacement is 4.22 m will be:

4.22m = 9.20*cos(69°)*t

t = (4.22/ 9.20*cos(69°)) = 1.28s

Now we evaluate the y-position in this time:

h =  -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m

The height above the release point is 2.96 meters.

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3 years ago
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The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th
Inga [223]

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

E_i = E_f

0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh

v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}

PART B) Replacing the values given as,

h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\

v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}

v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}

v_f = 1.8486m/s

Therefore the speed of the masses would be 1.8486m/s

6 0
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