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3 years ago

10

A periodic wave has a fixed pattern that is repeated. What is one completion of the wave pattern called?

1 answer:

Kaylis [27]3 years ago

7 0

It is called cycle. Letter D

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A ball bearing of radius of 1.5 mm made of iron of density

Serjik [45]

Answer:

$\boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise}$

Given:

Radius of ball bearing (r) = 1.5 mm = 0.15 cm

Density of iron (ρ) = 7.85 g/cm³

Density of glycerine (σ) = 1.25 g/cm³

Terminal velocity (v) = 2.25 cm/s

Acceleration due to gravity (g) = 980.6 cm/s²

To Find:

Viscosity of glycerine ( $\sf \eta$ )

Explanation:

$\boxed{ \bold{v = \frac{2}{9} \frac{( {r}^{2} ( \rho - \sigma)g)}{ \eta} }}$

$\sf \implies \eta = \frac{2}{9} \frac{( {r}^{2}( \rho - \sigma)g )}{v}$

Substituting values of r, ρ, σ, v & g in the equation:

$\sf \implies \eta = \frac{2}{9} \frac{( {(0.15)}^{2} \times (7.85 - 1.25) \times 980.6)}{2.25}$

$\sf \implies \eta = \frac{2}{9} \frac{(0.0225 \times 6.6 \times 980.6)}{2.25}$

$\sf \implies \eta = \frac{2}{9} \times \frac{145.6191}{2.25}$

$\sf \implies \eta = \frac{2}{9} \times 64.7196$

$\sf \implies \eta = 2 \times 7.191$

$\sf \implies \eta = 14.382 \: poise$

6 0

3 years ago

A swimming pool measures 5.0 m long x 4.0 m wide x 3.0 mdeep.

QveST [7]

Answer:

a) 588,000 N

b) 294000 N

Explanation:

Given that

Density of water = 1000kg/m3

(g) = 9.8m/s2

volume is given as (V)= 5m*4m*3m

a) force will be equal to weight of water

$W = mg = \rho g V = 1000\times 9.8 \times (5*4*3) = 588,000 kg m/s^2$

b) at either end

$P = \frac{F}{A} = P_o + \rho gh$

$dF = PdA$

$dF = \rho g w \int h dh$

$F = \rho g w \frac{h^2}{2}$

$F = \rho g A \frac{h}{2}$ [A = wh]

$F = 1000\times 9.8 \times 5\times 4 \times\frac{3}{2}$

F = 294000 N

3 0

4 years ago

If the road becomes wet or crowded, you should ____. slow down and increase your following distance All choices are incorrect. m

Vaselesa [24]

Answer:

The first one.

7 0

3 years ago

How does unequal solar heating lead to the Gulf Stream?

adoni [48]

Http://earthguide.ucsd.edu/virtualmuseum/virtualmuseum/OriginofGulfStream.shtml this website might help u find ur answer

8 0

3 years ago

A child is riding a merry-go-round that is turning at 7.18 rpm. if the child is standing 4.65 m from the center of the merry-go-

Alisiya [41]

First we need to convert the angular speed from rpm to rad/s. Keeping in mind that
$1 rev= 2 \pi rad$
$1 min = 60 s$
the angular speed is
$\omega = 7.18 \frac{rev}{min} \cdot \frac{2 \pi}{60} = 0.75 rad/s$

And so now we can calculate the tangential speed of the child, which is the angular speed times the distance of the child from the center of the motion:
$v= \omega r = (0.75 rad/s)(4.65 m)=3.50 m/s$

3 0

4 years ago