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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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the frequency of a beam of uv light is 1.0 ×10 ^15hz what is the energy in one quantum of this light express it in ev? ​
kap26 [50]

Answer:

4.14 eV

Explanation:

f = 1.0 ×10^15 Hz

h= 6.63×10^-34 J s (  this is called PLANCK 'S CONSTANT)

ENEGY = E = ?

E = hf  ( THIS IS FORMULA FOR ENERGY OF ONE QUANTA OR ONE PHOTON )

E= 6.63×10^-34×1.0 ×10^15

E = 6.63×10^-19 J

As 1eV = 1.6×10^-19 J so changing energy in eV from joules we will divide energy by 1.6×10^-19

hence E in eV = 6.63×10^-19/(1.6×10^-19)

          E = 4.14 eV

7 0
3 years ago
Helppppppp pleaseeee
SVETLANKA909090 [29]

Answer: (Sorry, but I don't know how to calculate mass)

1. 15 N

2. 0.4921 \frac{ft}{s^2} (feet per second squared)

4. 150 N

5. 8.202 feet per second squared

3 0
2 years ago
Starting at t = 0 s , a horizontal net force F⃗ =( 0.285 N/s )ti^+(-0.460 N/s2 )t2j^ is applied to a box that has an initial mom
Irina-Kira [14]

Answer:

Explanation:

We know that Impulse = force x time

impulse = change in momentum

change in momentum = force x time

Force F = .285 t -.46t²

Since force is variable

change in momentum = ∫ F dt  where F is force

= ∫ .285ti - .46t²j dt

= .285 t² / 2i - .46 t³ / 3 j

When t = 1.9

change in momentum = .285 x 1.9² /2 i  -  .46 x 1.9³ / 3 j

= .514i - 1.05 j

final momentum

= - 3.1 i + 3.9j +.514i - 1.05j

= - 2.586 i + 2.85j

x component = - 2.586

y component = 2.85

7 0
3 years ago
Find the moment of inertia Ihoop of a hoop of radius r and mass m with respect to an axis perpendicular to the hoop and passing
Juliette [100K]

Answer: MR²

is the the moment of inertia  of a hoop of radius R and mass M with respect to an axis perpendicular to the hoop and passing through its center

Explanation:

Since in the hoop , all mass elements  are situated at the same distance from the centre , the following expression for the moment of inertia can be written as follows.

I = ∫ r² dm

= R²∫ dm

MR²

where M is total mass and R is radius of the hoop .

3 0
3 years ago
An 80-cm uniform 10-kg bar is resting on two scales, one at either end. A smaller 4-kg mass (m) is placed at a distance of d = 2
Varvara68 [4.7K]

Answer

given,

length of bar = 80 cm

mass of the bar = 10 kg

smaller mass = 4 kg

distance = 20 cm

s_1 + s_2 = 10 + 4

s_1 + s_2 = 14\ kg

taking moment about B

s_1 \times 0.8 - 10 \times 0.4 - 4 \times 0.6 = 0

s_1 \times 0.8 = 6.4

s_1 = 8\ N

s_2 = 14 - s_1

s_2 = 14 - 8

s_2 = 6 N

difference between two scale = 8 - 6

                                                  = 2 N

7 0
3 years ago
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