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2 years ago

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

Physics

2 answers:

babunello [35]2 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]2 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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$U(r)=-\frac{Gm_Emr^2}{2R^3_E}$

Explanation:

We are given that

Gravitational force= $F_g=\frac{Gm_Emr}{R^3_E}$

r=0,U(0)=0

We know that

Gravitational potential energy= $-\int F_gdr$

$U(r)=-\int\frac{Gm_Emr}{R^3_E}dr$

$U(r)=-\frac{Gm_Em}{R^3_E}\times \frac{r^2}{2}+C$

Substitute r=0 ,U(0)=0

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2 years ago

A vector → A has a magnitude of 56.0 m and points in a direction 30.0° below the negative x axis. A second vector, → B , has a m

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Answer:

The magnitude of the vector $\vec{C}$ is 107.76 m

Explanation:

To find the components of the vectors we can use:

$\vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ is the magnitude of the vector, and θ is the angle over the positive x axis.

The negative x axis is displaced 180 ° over the positive x axis, so, we can take:

$\vec{A} = 56.0 \ m \ ( \ cos( 180 \° + 30 \°) \ , \ sin (180 \° + 30 \°) \ )$

$\vec{A} = 56.0 \ m \ ( \ cos( 210 \°) \ , \ sin (210 \°) \ )$

$\vec{A} = ( \ -48.497 \ m \ , \ - 28 \ m \ )$

$\vec{B} = 82.0 \ m \ ( \ cos( 180 \° - 49 \°) \ , \ sin (180 \° - 49 \°) \ )$

$\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )$

$\vec{B} = ( \ -53.797 \ m \ , \ 61.886\ m \ )$

Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

$(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)$

So, for our vectors:

$\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ , ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )$

$\vec{C} = ( \ - 102.294 \ m , \ 33.886 m \ )$

To find the magnitude of this vector, we can use the Pythagorean Theorem

$|\vec{C}| = \sqrt{C_x^2 + C_y^2}$

$|\vec{C}| = \sqrt{(- 102.294 \ m)^2 + (\ 33.886 m \)^2}$

$|\vec{C}| =107.76 m$

And this is the magnitude we are looking for.

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