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3 years ago

Which approach would be the most interested in studying Phineas Gage

Physics

1 answer:

viva [34]3 years ago

7 0

Biological because its studies the function of the brain’s lobes

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SOMEONE HELP ME PLEASE!!!!!

enyata [817]

Answer:

I think it would be the 3rd or 4th one.

8 0

3 years ago

Please select the word from the list that best fits the definition site of sea-floor spreading

serious [3.7K]

<h3><u>Answer;</u></h3>

Mid-ocean ridges

<h3><u>Explanation</u>;</h3>

Sea-floor spreading is the process by which molten material adds new oceanic crust to the ocean floor.
Mid-ocean ridge is an undersea mountain chain where new ocean floor is produced at a divergent plate boundary.
Mid ocean ridge occurs when convection currents rise in the mantle beneath the oceanic crust and create magma where two tectonic plates meet at a divergent boundary.

8 0

3 years ago

What is the purpose of oil used in a car's engine?

Hoochie [10]

B) to reduce friction

5 0

2 years ago

D. What is the net force on the bowling ball rolling lane

3241004551 [841]

Answer:

Friction

Explanation:

3 0

2 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

3 years ago