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1 year ago

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What is the purpose of the scapula to move during arm elevation?

1 answer:

Inessa [10]1 year ago

3 0

The purpose of the scapula to move during arm elevation is increase the range of elevation of the arm.

<h3>What is the importance of movement of the scapula during arm elevation?</h3>

The scapula is an important bone which is found in the shoulder and back region of the body.

The scapula enables and increases the range of motion of the arm with its motions.

During arm elevation, the scapula undergoes an upward rotational motion.

Therefore, the purpose of the scapula to move during arm elevation is increase the range of elevation of the arm.

Learn more about scapula motion at: brainly.com/question/5133017

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A 3.00-kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it encounters a spring. (a) Determ

aniked [119]

Answer:

a) k= 3594,7 N/m

b) v= 0.55 m/s

Explanation:

a)
As the surface is horizontal, the only change in energy will be the change in kinetic energy, as the box comes to an stop after compressing the spring.
As we know that the surface is frictionless also, this change in kinetic energy must be equal to the change in the elastic potential energy of the spring.
So we can write the following equality:

$\Delta K = \Delta U$

where $\Delta K = \frac{1}{2}*m*v^{2}$

and $\Delta U = \frac{1}{2} * k* \Delta x^{2}$

Simplifying and replacing by the values, we get:

$3.00 kg* (1.8 m/s)^{2} = k* (0.052 m) ^{2}$

Solving for k:

$k = \frac{3.00kg*(1.8m/s)^{2} }{(0.052m)^{2}} = 3594.7 N/m$

k = 3594.7 N/m

b)
For this part, we can just apply the same equality, replacing the value of k by the one we got, and solving for the initial speed v:

$v = \sqrt{\frac{k*\Delta x^{2}}{m} } = \sqrt{\frac{3594.7N/m*(0.016m)^{2} }{3.00kg}} = 0.55 m/s$

v = 0.55 m/s

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3 years ago

What is the wavelength of the waves you create in a swimming pool if you splash your hand at a rate of 2.00 Hz and the waves pro

Valentin [98]

Answer:

The wavelength of the waves created in the swimming pool is 0.4 m

Explanation:

Given;

frequency of the wave, f = 2 Hz

velocity of the wave, v = 0.8 m/s

The wavelength of the wave is given by;

λ = v / f

where;

λ is the wavelength

f is the frequency

v is the wavelength

λ = 0.8 / 2

λ = 0.4 m

Therefore, the wavelength of the waves created in the swimming pool is 0.4 m

3 0

3 years ago

An empty graduated cylinder weighs 55.26 g. When filled with 48.1 mL of an unknown liquid, it weighs 92.39 g. The density of the

katrin [286]

<h2>Density of the unknown liquid is 771.93 kg/m³</h2>

Explanation:

An empty graduated cylinder weighs 55.26 g

Weight of empty cylinder = 55.26 g = 0.05526 kg

Volume of liquid filled = 48.1 mL = 48.1 x 10⁻⁶ m³

Weight of cylinder plus liquid = 92.39 g = 0.09239 kg

Weight of liquid = 0.09239 - 0.05526

Weight of liquid = 0.03713 kg

We have

Mass = Volume x Density

0.03713 = 48.1 x 10⁻⁶ x Density

Density = 771.93 kg/m³

Density of the unknown liquid is 771.93 kg/m³

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3 years ago

Describe an object's velocity when an acceleration-time graph is zero?

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Anything times zero is zero

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3 years ago

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The half-life of caffeine is 5 hours. If you ingested a 30 oz Big Gulp, how many oz of caffeine is left after one half life? * Y

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Answer:

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Explanation:

Half life is the time taken for a radioactive substance to degenerate or decay to half of its original size.

The half life of caffeine is 5 hours. So ingesting a 30 oz, this would be reduced to half of its size after the first 5 hours.

So that:

After one half life of 5 hours, the value of caffeine that would be left is;

$\frac{30}{2}$ = 15 oz

The amount of caffeine left after one half life of 5 hours is 15 oz.

8 0

3 years ago