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2 years ago

How temperature, surface tension and the diameter of the tube affect capillary action.

1 answer:

5 0

The surface tension acts to hold the surface intact. Capillary action occurs when the adhesion to the surface material is stronger than the cohesive forces between the water molecules. ... Water wants to stick to the glass and surface tension will push the water up, until the force of gravity prevents further rise.

Narrower tube openings allow capillary action to pull water higher

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HELP ME HOW ARE BLACK HOLES FROMED

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Answer:

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2 years ago

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an ice skater starts with a velocity of 2.25 m/s in a 50.0 direction. after 8.33 m/s in a 120 direction. what is the y-component

Nata [24]

The y-component of the acceleration is $0.28 m/s^2$

Explanation:

The y-component of the ice skater acceleration can be calculated with the equation

$a_y = \frac{v_y-u_y}{t}$

where

$v_y$ is the y-component of the final velocity

$u_y$ is the y-component of the initial velocity

t is the time elapsed

Here we have:

Initial velocity is $u=2.25 m/s$ at $\theta_1=50.0^{\circ}$ , so its y-component is $u_y = u sin \theta_1 = (2.25)(sin 50.0^{\circ})=1.72 m/s$

Final velocity is $v=4.65 m/s$ at $\theta_2=120.0^{\circ}$ , so its y-component is $v_y = v sin \theta_2 = (4.65)(sin 120.0^{\circ})=4.03 m/s$

The time elapsed is

t = 8.33 s

Therefore, the y-component of the acceleration is

$a_y = \frac{4.03-1.72}{8.33}=0.28 m/s^2$

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2 years ago

Suppose that a teacher driving a 1972 LeMans zooms out of a darkened tunnel at 34.5 m/s. He is momentarily blinded by the sunshi

valkas [14]

Answer:

489.19m

Explanation:

To find the minimum distance you first calculate the time in which the teacher stops:

$v=v_o-at\\\\t=\frac{v_o-v}{a}=\frac{34.5m/s-0m/s}{2.5m/s^2}=13.8s$

however, the reaction of the teacher is 0.31s later, then you use

t=13.8-0.31s=13.49s

during this time the camper has traveled a distance of:

$x=vt=(15.1m/s)(13.49s)=203.69m$ (1)

Next you calculate the distance that teacher has traveled for 13.6s:

$x=x_o+v_ot+\frac{1}{2}at^2\\\\x=0m+34.5m/s(13.49s)+\frac{1}{2}(2.5m/s^2)(13.49s)^2=692.88m$ (2)

The minimum distance between the driver and the camper will be the difference between (2) and (1):

$x_{min}=692.88m-203.69m=489.19m$

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3 years ago

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A because read the paragraphs

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2 years ago

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3 years ago

How temperature, surface tension and the diameter of the tube affect capillary action.​

How temperature, surface tension and the diameter of the tube affect capillary action.