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3 years ago

How can I create a whirlpool of water?

Physics

1 answer:

RideAnS [48]3 years ago

5 0

Answer:

To make a small scale whirlpool of water, take a cup and fill it with water and stir. It will create a vortex that pulls particles down.

Or you could see this when draining a tub of water.

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Explain 5 factors that act as a stimulant to change.

Tcecarenko [31]

Answer: Stimulant is a factor which brings change in the environment.

Explanation:

The external factors and internal factors both can develop changes in the environment of a living being. The external factors include sunlight, weather, touch, vibration, predation and others. The internal factors include the physiological and biochemical changes in the body, emotions and mental make up.

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4 years ago

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Part B

dimaraw [331]

At the left side of the reaction

Explanation:

the reaction is spontaneous

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4 years ago

You grab a car door handle in the summer (energy transfer to heat up the handle ____) and it burns you through energy transfer.

Cerrena [4.2K]

Answer: conduction

Explanation:

Because you are physically touching the handle

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4 years ago

Which property can be altered by a physical change?

Zolol [24]

A physical change involves a change in physical properties. Examples of physical properties include melting, transition to a gas, change of strength, change of durability, changes to crystal form, textural change, shape, size, color, volume and density.

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4 years ago

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Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax

kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its x-component is

$F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its x-component is

$F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N$

So, the x-component of the resultant force is

$F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N$

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its y-component is

$F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its y-component is

$F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N$