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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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You run a race with your friend. At first you each have the same kinetic energy, but then you find that she is beating you. When

Blizzard [7]

Answer:

52.49 Kg

Explanation:

Let m1 and v1 denote your mass and velocity respectively

Let m2 and v2 denote your friends mass and velocity respectively

Kinetic energy is given by

$KE= 0.5mv^{2}$

Since your kinetic energies are the same hence

$0.5m1(v1)^{2}=0.5m2(v2)^{2}$

$m1(v1)^{2}=m2(v2)^{2}$ and making m2 the subject then

$m2=\frac { m1(v1)^{2}}{(v2)^{2}}$

Since v2 is v1+0.28v1=1.28v1

Substituting m1 for 86 Kg

$m2=\frac { 86 Kg(v1)^{2}}{(1.28v1)^{2}}= 52.49023\approx 52.49 Kg$

3 0

3 years ago

The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This

erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

$R_{CM} =$ Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

$g = \frac{GM}{(d-R_{CM})^2}$

$g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}$

$g = 3.4*10^{-5}m/s^2$

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

$\omega = \frac{2\pi}{T}$

At the same time we have that centripetal acceleration is given as

$a_c = \omega^2 r$

Replacing

$a_c = (\frac{2\pi}{T})^2 r$

$a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)$

$a_c = 3.34*10^{-5}m/s^2$

3 0

2 years ago

The three forces shown act on a particle. what is the direction of the resultant of these three forces?

melisa1 [442]

Missing figure: http://d2vlcm61l7u1fs.cloudfront.net/media/f5d/f5d9d0bc-e05f-4cd8-9277-da7cdda3aebf/phpJK1JgJ.png

Solution:
We need to find the magnitude of the resultant on both x- and y-axis.

x-axis) The resultant on the x-axis is
$F_x = 65 N\cdot cos 30^{\circ} - 30 N - 20 N\cdot sin 20^{\circ} = 19.45 N$
in the positive direction.

y-axis) The resultant on the y-axis is
$F_y = 65 N \cdot sin 30^{\circ} - 20 N \cdot cos 20^{\circ} = 13.70 N$
in the positive direction.

Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
$\tan \alpha = \frac{F_y}{F_x} = \frac{13.70 N}{19.45 N}=0.7$
from which we find
$\alpha=35^{\circ}$

7 0

3 years ago

Why x ray is called an electromagnetic wave<br>

Dahasolnce [82]

Because they behave just like all the electromagnetic waves of the spectrum. Same equations, just shorter wavelengths and more energy.

Hope you get it :)

7 0

2 years ago

The radius of the earth is 6378 km. What is the diameter of the earth in meters?

Nesterboy [21]

To determine the diameter of the earth in metres first multiply the original value by 2.

6378 X 2 = 12 756 km.

Then convert km - m

1 km = 1000 m
12 756 km = ? m

12 756 • 1000 = 12 756 000 = 12 756 000 m or 1.2756 X 10 ^ 7 m

The final solution for the diameter is 1.2756 X 10 ^ 7 m.

7 0

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.