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2 years ago

Force applied through a rope or string

Physics

1 answer:

neonofarm [45]2 years ago

8 0

The answer is D. Tension force :))

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All objects within a closed system tend to move

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Explanation is in the file

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3 years ago

Identify a mechanical wave that is also a longitudinal wave

nexus9112 [7]

Mechanical waves, it needs a certain medium to travel through. This medium can be a solid, liquid and gas. There are three types of mechanical waves namely, sound wave, water, and waves through the ropes.

A longitudinal wave is a wave that vibrates in a parallel direction with the motion. Also known as a compression wave. Both mechanical and longitudinal waves produce a sound.

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3 years ago

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NeX [460]

Answer:

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2 years ago

You are caring for a person who was involved in an automobile collision. The woman is complaining of pain in her abdomen, nausea

Anna71 [15]

Answer:

I would suspect internal bleeding.

Explanation:

Internal Bleeding. As she has pain in abdomen and nausea means she is loosing blood. As result of low blood in body vessels, she is feeling great thirst as well as sugar level is going down.

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3 years ago

A wooden object (conically shaped) has a diameter of 8cm and height of 14cm. It floats in oil with 6cm of its height above oil l

baherus [9]

Answer:

(a) The density of the object is 316/343 × the density of the oil

(b) The fraction of oil displaced after immersing the object is 0.461 of the oil volume

Explanation:

(a) The volume, V of a cone of height, h and base diameter, D = 2×r is given by the following equation;

$V = \dfrac{\pi r^{2} h}{3}$

The volume of the object is therefore;

$\dfrac{\pi \times 4^{2} \times 14}{3} = 74\tfrac{2}{3}\pi \, cm^3$

Where 6 cm is above the oil level we have;

$\dfrac{\pi \times \left (6 \times \dfrac{4}{14} \right )^{2} \times 6}{3} = 5\tfrac{43}{49}\pi \, cm^3$ above the oil level

Therefore, volume of the oil displaced = $68\tfrac{116}{147}\pi$ cm³ = 216.11 cm³

The density of the object is thus;

$\dfrac{68\tfrac{116}{147}\pi}{ 74\tfrac{2}{3}\pi} \times Density \ of \ the \ oil = \dfrac{316}{343} \right ) \times Density \ of \ the \ oil$

The density of the object = 316/343 × the density of the oil.

(b) The volume of the oil = 2 × Volume of the object = $2 \times 74\tfrac{2}{3}\pi \, cm^3 = 149\tfrac{1}{3}\pi \, cm^3$

The fraction of the volume displaced, x, after immersing the object is given as follows;

$x = \dfrac{68\tfrac{116}{147}\pi}{ 149\tfrac{1}{3}\pi} = \dfrac{158}{343} = 0.461$

The fraction of oil displaced after immersing the object = 0.461 of the volume of the oil

8 0

3 years ago