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4 years ago

15

A patient comes in and tells you that he has arthritis and needs Cerebyx. What do you do? A. You know that Cerebyx and Celebrex

are similar, but you assume that the patient knows what drug he needs. B. You know that Cerebyx is used for seizures, while Celebrex is used for arthritis. You check it in the drug reference guide and inform the patient. C. You greet the patient politely and give him the drug. D. You feel sorry for the patient because he mixed up Cerebyx and Celebrex.

1 answer:

Flura [38]4 years ago

6 0

B. truthfully it's the only one that makes sense

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Please help<br> How would you explain 2 m/s/s [1<br> (1 Point)<br> Enter your answer

AlexFokin [52]

2 m/s/s means the velocity increases by 2 m/s every second.

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3 years ago

A charge q of 1.3 × 10-16 coulombs moves from point A to a lower potential at point B in an electric field of 3.2 × 102 newtons/

LUCKY_DIMON [66]

Explanation :

It is given that,

Charge, $q=1.3\times 10^{-16}\ C$

Electric field, $E=3.2\times 10^2\ N/C$

Distance, $d=1.1\times 10^{-2}\ m$

The work done is stored in the form of potential energy.

$W=F.d$

$\because F=qE$

So, $W=qE\ d$

$W=1.3\times 10^{-16}\ C\times 3.2\times 10^2\ N/C\times 1.1\times 10^{-2}\ m$

$W=4.576\times 10^{-16}\ J$

Hence, this is the required solution.

5 0

3 years ago

Read 2 more answers

A 7.0 mm -diameter copper ball is charged to 40 nC. What fraction of its electrons have been removed? The density of copper is 8

mylen [45]

Answer:

$f = 2.6 \times 10^{-13}$

Explanation:

Let the mass of copper ball is "m" gram

now the total number of copper atom present in the ball is given as

$N = \frac{m}{29} \times 6.02 \times 10^{23}$

now the total number of electrons in one copper atom is 29

so total number of electrons in given sample of copper ball is

$N_e = m(6.02 \times 10^{29})$

now diameter of the ball is 7.0 mm

density of the ball = $8900 kg/m^3$

now we have

$m = (\frac{4}{3}\pi r^3)(8900)$

$m = (\frac{4}{3}\pi(\frac{0.007}{2})^3)(8900)$

$m = 1.6 gram$

now we have

$N_e = 9.63 \times 10^{23}$

now the charge on the copper ball is 40 nC

so the number of electrons removed

$Q = ne$

$40 \times 10^{-9} = n(1.6 \times 10^{-19}$

$n = 2.5 \times 10^{11}$

so the fraction of number of electrons removed is given as

$f = \frac{n}{N_e}$

$f = \frac{2.5 \times 10^{11}}{9.63 \times 10^{23}}$

$f = 2.6 \times 10^{-13}$

7 0

4 years ago

a paper airplane gliding down towards the ground will experience the force of air resistance pushing up. the weight of the paper

Oduvanchick [21]

The net force acting on the airplane is 25N.

Forces acting on the paper airplane when it is in the air:

The forward force generated by the engine, propeller, or rotor is called thrust. It resists or defeats the drag force. It operates generally perpendicular to the longitudinal axis. However, as will be discussed later, this is not always the case.
Drag is an airflow disruption generated by the wing, rotor, fuselage, and other projecting surfaces that causes a backward, decelerating force. Drag acts backward and perpendicular to the relative wind, opposing thrust.
Weight is the total load carried by airplane, including the weight of the crew, fuel, and any cargo or baggage. Due to the influence of gravity, weight pulls the airplane downward.
Lift—acts perpendicular to the flight path through the center of lift and opposes the weight's downward force. It is produced by the air's dynamic influence on the airfoil.

Given.

Weight of the paper airplane, F1 = 16N

The force of air resistance, F2 = 9N

Net force = F1 + F2

Net force = 25N

Thus, the net force acting on the airplane is 25N.

Learn more about the net force here:

brainly.com/question/18109210

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1 year ago

How do local action make the cell defective

docker41 [41]

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3 years ago