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Naddika [18.5K]
3 years ago
5

FAA advisory circulars containing subject matter specifically related to Airspace are issued under which subject number?

Physics
1 answer:
miskamm [114]3 years ago
6 0

Answer:

70

Explanation:

FAA stands for federal aviation administration.Which offer advisory circular which is a type of publication which provides guidance regarding the airworthiness regulation, training standard of pilots, operational standards.It also regulates the rules related to space and airline.

In this publication the subject matter related to airspace is under 70 subject number

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A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7
maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
From which we find
v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
\frac{1}{2} (m_1+m_2)v_i^2
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
From which we can find the value of the coefficient of kinetic friction:
\mu =  \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49
3 0
3 years ago
A spring with k = 500 N/m stores 704 J. How far is it extended from the equilibrium position
kaheart [24]

Given that:

k = 500 n/m,

work (W) = 704 J

spring extension (x) = ?

         we know that,

                      Work = (1/2) k x²

                          704 = (1/2) × 500 × x²

                            x = 1.67 m

A spring stretched for 1.67 m distance.

4 0
3 years ago
An atom undergoes nuclear decay, but its atomic number is not changed.
andreev551 [17]

Answer:

A. Gamma decay

Explanation:

A form of nuclear decay in which the atomic number is unchanged is a gamma decay.

The atom has undergone a gamma decay.

In a gamma decay, no changes occur to the mass and atomic number of the substance.

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3 years ago
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mrs_skeptik [129]

You can write the equation in 3 different ways, depending on which quantity you want to be the dependent variable.  Any one of the three forms can be derived from either of the other two with a simple algebra operation.  They're all the same relationship, described by "Ohm's Law".

==> Current = (potential difference) / (resistance)

==> Potential difference = (current) x (resistance)

==> Resistance = (potential difference) / (resistance)

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