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2 years ago

11

The prevailing winds that form in Hadley cells are _____.

2 answers:

dmitriy555 [2]2 years ago

8 0

Answer = C

It blows towards the poles from the sub-tropical high pressure areas

Olin [163]2 years ago

3 0

I also think it is C

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I need help with this<br>

zlopas [31]

Answer:

jejjdedjd sidjjejdd jsms

Explanation:

jdjdndjdjjdj jsnssjns jsjsjs

3 0

2 years ago

Describe the motion of an object that has an acceleration of 0 mi./s squared

miskamm [114]

The object does not move.

5 0

2 years ago

The 5.5 million vinyl long-playing (LP) records sold in the United States per year pales in comparison with the 1.26 billion dig

miv72 [106K]

Answer:

decline

Explanation:

Based on the scenario being described within the question it can be said that these types of firms are in the decline stage of the product life cycle. This stage refers to when a product has already passed it's peak potential and sales begin to decline until production is ultimately halted and the product dies off. Which is exactly what is happening to the LP's since everyone has moved on to digital downloads.

4 0

2 years ago

Bonus: (It's not that hard, you just have to pay attention to units.) The Saturn V rocket first stage

agasfer [191]

$v = 2.45×10^3\:\text{m/s}$

Explanation:

Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as

$F = \dfrac{d{p}}{d{t}}$ (1)

Assuming that the velocity remains constant then

$F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}$

Solving for $v,$ we get

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)$

Before we plug in the given values, we need to convert them first to their appropriate units:

The thrust <em>F</em><em> </em> is

$F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}$

The exhaust rate dm/dt is

$\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}$

$\;\;\;\;\;= 1.36×10^4\:\text{kg/s}$

Therefore, the velocity at which the exhaust gases exit the engines is

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}$

$\;\;\;= 2.45×10^3\:\text{m/s}$

6 0

1 year ago

As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis

klio [65]

Hi there!

We can begin by finding the acceleration of the block.

Use the kinematic equation:

$d = v_0t + \frac{1}{2}at^2$

The block starts from rest, so:

$d = \frac{1}{2}at^2\\\\12 = \frac{1}{2}a(4^2)\\\\\frac{24}{16} = a = 1.5 m/s^2$

Now, we can do a summation of forces of the block using Newton's Second Law:

$F = ma = m_bg - T$

mb = mass of the block

T = tension of string

Solve for tension:

$T = m_bg - ma = 8.2(9.8) - 8.2(1.5) = 68.06 N$

Now, we can do a summation of torques for the wheel:

$\Sigma \tau = rF\\\\\Sigma\tau = rT$

Rewrite:

$I\alpha = rT$

We solved that the linear acceleration is 1.5 m/s², so we can solve for the angular acceleration using the following:

$\alpha = a/r\\\\\alpha = 1.5/.42= 3.57 rad/sec^2$

Now, plug in the values into the equation:

$I(3.57) = (0.42)(68.06)\\\\I = (0.42)(68.06)/(3.57) = \boxed{8.00 kgm^2}$

8 0

1 year ago