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2 years ago

What confirms that the process of seafloor spreading is occurring?

1 answer:

3 0

Mid ocean ridges
The mountain ranges in the middle of the oceans
Sea floor spreading
The process by which new, oceanic crust forms along a mid ocean ridge and older crust moves away from the ridge. :)

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⦁ Consider an atom. Which contributes most to the mass of the atom?

fomenos

Answer:

protons

Explanation:

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3 years ago

A fluid is flowing through a circulat tube at 0.4 kg/s. Tube inner surface is smooth with a diameter 0.014 m. Fluid density is 9

ElenaW [278]

Answer:

The convection coefficient is $15456.48\ W/m^{2}K$

Solution:

Mass flow rate, $\dot{m} = 0.4\ kg$

Inner diameter of the tube, d = 0.014 m

Fluid density, $\rho_{f} = 990\ kg/m^{3}$

Specific Heat, C = 3845 J/K

Thermal Conductivity, K = 0.74

Prandtl Number, $P_{r} = 8.6$

Heat flux, $\dot{q} = 71,297\ W/m^{2}$

Viscosity, $\mu = 0.00079\ Ns/m^{2}$

Now,

To calculate the convection heat coefficient, h:

Determine the cross sectional area of the circular tube:

$A_{c} = \frac{\pi}{4}d^{2} = \frac{\pi}{4}\times (0.014)^{2} = 1.54\time 10^{- 4}\ m^{2}$

Determine the velocity of the fluid inside the tube by mass flow rate:

$\dot{m} = \rho_{f}A_{c}v$

$0.4 = 990\times 1.54\time 10^{- 4}v$

v = 2.624 m/s

Determine the Reynold's Number, $R_{e}$ :

$R_{e} = \frac{\rho_{f}dv}{\mu}$

$R_{e} = \frac{990\times 0.014\times 2.624}{0.00079} = 46036.253$

Thus it is clear that $R_{e}$ > 10,000 hence flow is turbulent.

Now,

Determine the Nusselt Number:

$N_{u} = 0.023R_{e}^{0.8}P_{r}^{0.4}$

$N_{u} = 0.023\times 46036.253^{0.8}\times 8.6^{0.4} = 292.42$

Also,

$N_{u} = \frac{dh}{K}$

where

h = convection coefficient

Now,

$292.42 = \frac{0.014\times h}{0.74}$

$h = 15456.48\ W/m^{2}K$

7 0

3 years ago

Energy that is stored due to types of bonds and arrangement of atoms refers to ___________ energy

Bingel [31]

Potential energy is in short, stored energy

5 0

3 years ago

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_ acceleration occurs when an object speeds up.<br> Answer<br> Positive

Nina [5.8K]

_ acceleration occurs when an object speeds up.

Answer

Positive

6 0

3 years ago

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You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

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1 year ago

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