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qwelly [4]
1 year ago
14

What are the requirements when checking in PSE products?.

Biology
1 answer:
Maurinko [17]1 year ago
3 0

The correct statements are : A ,D ,E , F

A Date received must be documented on each page of the invoice

D Handwritten verification marks (circle, slash, checkmark, etc.) to confirm receipt of each product must be documented on the invoice

E Signature documented on each page of the invoice

F All CIII-V's and PSE products should be checked into the electronic delivery check-in screen via CFRX.

The kind of identification used, the government agency issuing it, the identification number, the name and address of each buyer, the product they purchased that includes ephedrine or pseudoephedrine, including how many grams it contains.

Each page of the invoice must include the date it was received. Documented initials on each page of the invoice If state laws permit, technicians can check in orders for CIII-V and PSE items. Handwritten verification marks (circle, slash, checkmark, etc.) must be recorded on the.

Check-in Requirements: For domestic flights, most airlines advise that you go to the airport at least two hours before departure, and three hours for international flights. Please have your government-issued ID and electronic ticket number available.

Learn more about to PSE products visit here:

brainly.com/question/26603693

#SPJ4

Full Question :

What are the requirements when checking in ClII-Vand PSE products?

A Date received must be documented on each page of the invoice

B Initials documented on each page of the invoice

C Technicians can check in CIII-V and PSE products orders if state regulations allow

D Handwritten verification marks (circle, slash, checkmark, etc.) to confirm receipt of each product must be documented on the invoice

E Signature documented on each page of the invoice

F All CIII-V's and PSE products should be checked into the electronic delivery check-in screen via CFRX.

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Answer:

one that has an unstable nucleus

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An isotope is defined as a chemical variant  present in an atom with different number of neutron but same number of protons.

Radioactive decay is the sudden breakdown of an atomic nucleus allowing matter and radiation to be emitted from the nucleus. A radioisotope with unstable nuclei undergoes radioactive decay because they do not have adequate binding energy to keep the nucleus together. there are four  types of Alpha, Beta, Gamma Decay and Positron Emission.

Hence, the correct option is "one that has an unstable nucleus"

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2.86 A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight
Rashid [163]

Answer:

Distance Below the top = 6,02 m

Explanation:

To get the Final velocity of the first ball (that will be the intial velocity of the second) you need to solve the kinematic equation for Velocity:

(1) V_{1f} =V_{1O}  - gt

As the ball is dropped from rest V_{1O} = 0, so:

(2) V_{1f} = - gt

Note that the velocity is going to be negative as the ball is going down. To get the time it would take the ball to reach de base you can use the kinematic equation for position:

(3) h_{1} = h_{1O} + V_{1O} *t - \frac{1}{2} gt^{2}

We need the answer when h=0, and from the initial conditions V_{1O} = 0, h_{1O} = 24m, you get:

(4) 24m = \frac{1}{2} gt_{1f}^{2}

Solving for t, with 9,8\frac{m}{s^{2}} and :

(5) t_{1f} = \sqrt{\frac{24m*2}{g} } = 2,21 s

Replacing this time in (2), the final velocity is:

(6) V_{1f} = -21,66  \frac{m}{s}

So the initial velocity of ball 2 is equial to this but oppossite in direction so: V_{2O}= -V_{1f} = 21,66  \frac{m}{s}

The general position equation for ball 2 is (considering h_{2O} = 0 :

(7) h_{2} = V_{2O} *t - \frac{1}{2} gt^{2}

They cross paths when h_{1}=h_{2} so:

(8) h_{1O} - \frac{1}{2} gt^{2} = V_{2O} *t - \frac{1}{2} gt^{2}

Rearranging:

(9) t_{cross} =\frac{h_{1O}}{V_{2O} }

Replacing values:

t_{cross} = 1,108 s

To get the absolute position replace t_{cross} on equation (3) or (7):

h_{cross} = 17,98 m

To get it below the top of te cliff:

h_{cross, from above} = 24 m - 17,98 m

h_{cross, from above} = 6,02 m

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