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3 years ago

states that the velocity of the fluid in contact with the surface is equal to the velocity of that surface

Physics

1 answer:

nikklg [1K]3 years ago

7 0

Answer:

The No-slip boundary condition.

Explanation:

The no-slip boundary condition states that the velocity of the fluid in contact with the surface is equal to the velocity of that surface, both the normal and tangential components of the fluid velocity field are equal to zero.

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An arrow in flight has an initial velocity of 65 meters per second, and 10 seconds later, it has a velocity of 35 meters per sec

-BARSIC- [3]

Acceleration = (final velocity - initial velocity) / time
= (35-65)/10
= -3 m/s2

7 0

4 years ago

In billiards, the 0.165 kg cue ball is hit toward the 0.155 kg eight ball, which is stationary. The cue ball travels at 5.8 m/s

Damm [24]

Answer:

another ball velocity = 3.92 m/s and with 30° clockwise from initial direction

Explanation:

given data

mass m1 = 0.165 kg

mass m2 = 0.155 kg

before collision velocity v1 = 5.8 m/s

before collision velocity v2 = 0

angle = 35.0° from initial direction

after collision 1st ball velocity v3 = 3.2 m/s

to find out

after collision another ball velocity v4

solution

we consider here ball move in x axis and after collision 1st ball move upside of x axis with angle 35 degree and other ball move downside with x axis with angle θ

so from conservation of momentum we say

m1v1 = m1v3cos35 + m2v4cosθ with x axis .............1

m1v3sin35 = m2v4sinθ with y axis .............2

so from 1 equation

0.165 × 5.8 = 0.165(3.2)cos35 + 0.155(v4)cosθ

v4 cosθ = 3.38 .................3

form 2 equation

0.165(3.2)sin35 = 0.155(v4)sinθ

v4 sinθ = 1.95 ......................4

so magnitude of another ball velocity is square and adding equation 3 and 4

another ball velocity = √(3.39²+1.96²)

another ball velocity = 3.92 m/s

and direction is tanθ = 1.96/3.39

θ = 30° clockwise from initial direction

3 0

3 years ago

Which planet is represented by the letter "h"?

Mandarinka [93]

The planet is represented as Saturn ♄

7 0

3 years ago

What is the density (in kg/m3) of a woman who floats in freshwater with 4.92% of her volume above the surface

kipiarov [429]

Answer:

The density of the woman is 950.8 kg/m³

Explanation:

Given;

fraction of the woman's volume above the surface = 4.92%

then, fraction of the woman's volume below the surface = 100 - 4.92% = 95.08%

the specific gravity of the woman $= \frac{95.08}{100 } = 0.9508$

The density of the woman is calculate as;

$Specific \ gravity \ of \ the \ woman = \frac{Density \ of \ the \ woman }{Density \ of \ fresh \ water }\\\\ Density \ of \ the \ woman = Specific \ gravity \ of \ the \ woman \ \times \ Density \ of \ fresh \ water$

Density of fresh water = 1000 kg/m³

Density of the woman = 0.9508 x 1000 kg/m³

Density of the woman = 950.8 kg/m³

Therefore, the density of the woman is 950.8 kg/m³

4 0

3 years ago

The velocity of the transverse waves produced by an earthquake is 8.9 km/s, and that of the longitudinal waves is 5.1 km/s. A se

Brrunno [24]

Answer: The distance is 723.4km

Explanation:

The velocity of the transverse waves is 8.9km/s

The velocity of the longitudinal wave is 5.1 km/s

The transverse one reaches 68 seconds before the longitudinal.

if the distance is X, we know that:

X/(9.8km/s) = T1

X/(5.1km/s) = T2

T2 = T1 + 68s

Where T1 and T2 are the time that each wave needs to reach the sesmograph.

We replace the third equation into the second and get:

X/(9.8km/s) = T1

X/(5.1km/s) = T1 + 68s

Now, we can replace T1 from the first equation into the second one:

X/(5.1km/s) = X/(9.8km/s) + 68s

Now we can solve it for X and find the distance.

X/(5.1km/s) - X/(9.8km/s) = 68s

X(1/(5.1km/s) - 1/(9.8km/s)) = X*0.094s/km= 68s

X = 68s/0.094s/km = 723.4 km

6 0

4 years ago