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4 years ago

9

To meet the minimum required instrument flight experience to act as a pilot in command of an aircraft under IFR, you must have l

ogged within the preceding 6 calendar months in the same category aircraft:_______. six instrument approaches,
A) holding procedures, intercepting and tracking courses through the use of navigation systems.
B) three of which must be in the same category and class of aircraft to be flown, and 6 hours of instrument time in any aircraft.
C) and 6 hours of instrument time in any aircraft.

1 answer:

lora16 [44]4 years ago

6 0

I think its a tbh bc it seems to be the best answer out of a b c and d

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What is momentum equal to?

Setler [38]

Answer:

D

Explanation:

mass times velocity

hope this help

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3 years ago

Which of the following is not a physical property of mattera. melting pointb.heat of combustionc. viscosityd. boiling point

NNADVOKAT [17]

Answer:

The Correct Answer is Heat of Combustion

Explanation:

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3 years ago

A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i

bija089 [108]

Answer:

$v_o\approx0.7059\ m.s^{-1}$

Explanation:

Given:

mass of the boat, $m=120\ kg$

uniform speed of the boat, $v=1\ m.s^{-1}$

rate of accumulation of water mass in the boat, $\dot m=100\ kg.hr^{-1}$

time of observation, $t=0.5\ hr$

The mass of the boat after the observed time:

$m_o=m+\dot m\times t$

$m_o=120+100\times 0.5$

$m_o=170\ kg$

<u>Now using the conservation of momentum:</u>

$m.v=m_o.v_o$

$120\times 1=170\times v_o$

$v_o\approx0.7059\ m.s^{-1}$

4 0

3 years ago

Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k

larisa86 [58]

Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

Mass of the larger disk = $M_2\ =\ 1.60\ kg$
Mass of the smaller disk = $M_1\ =\ 0.900\ kg$
Radius of the larger disk = $R_2\ =\ 5.00\ cm\ =\ 0.05\ m$
Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$
Mass of the block = M = 1.60 kg

Both the disks are welded together, therefore total moment of inertia of the both disks are the summation of the individual moment of inertia of the disks.

$\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}M_1R_1^2\ +\ \dfrac{1}{2}M_2R_2^2\\\Rightarrow I\ =\ \dfrac{1}{2} (0.9\times 0.0245^2\ +\ 1.60\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2$

part (a)

Given that a block of mass m which is hanging with the smaller disk,

Let 'T' be 'a' be the tension in the string and acceleration of the block.

From the free body diagram of the smaller block,

$mg\ -\ T\ =\ ma\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,eqn (1)$

From the pulley,

$\sum \tau\ =\ I\alpha\\\Rightarow T\times R_1\ =\ I\alpha\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{I\alpha}{R_1^2}\,\,\,eqn(2)$

From the equation (1) and (2),

$mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.0245^2}}\ +\ 1.60}\\\Rightarow a\ =\ 2.91\ m/s^2$

part (b)

Above expression for the acceleration of the block is only depended on the radius of the pulley.

Radius of the larger pulley = $R_2\ =\ 0.05\ m$

Let $a_2$ be the acceleration of the block while connecting to the larger pulley. $\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_2^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.05^2}\ +\ 1.60}}\\\Rightarow a\ =\ 6.25\ m/s^2$

4 0

3 years ago

Which characteristic best describes how nuclear decay differs from nuclear fission?

kobusy [5.1K]

Nuclear decay is spontaneous.

Nuclear fission does occur spontaneously but at a very low level in fissionable materials. The main difference between fission and nuclear decay is that fission can occur by neutron bombardment. And when the neutron source is also other fissioning atoms, a chain reaction can may happen. <span>
No chain reaction is possible with nuclear decay.
<span>That would be a much better characteristic to look at.</span></span>

6 0

4 years ago

Read 2 more answers