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babunello [35]
3 years ago
6

99 POINTS PLEASE HELP

Physics
2 answers:
Ipatiy [6.2K]3 years ago
6 0

The answer is C.) Work

Force is not a quantity of the transfer of energy. An example of force would be when you push on a wall. You are applying force on a wall but the wall doesn't move. Thus, there is no transfer of energy.

  • The SI unit for force is Newtons (N)

Motion is a representation of force, not a quantity of the transfer of energy. When you see a ball rolling on the ground, this is because something pushed it and applied force, or the force of gravity is pulling down on it. There is also no transfer of energy in motion.

  • The SI unit for motion is Meters Per Second (m/s)

Work is the quantity of the transfer of energy. If you were to push on a door and it opens, energy has been transferred from your arms to the door; thus, there is work present.

  • The SI unit for work is a Watt (W), or Joules per Second (j/s)

Let me know if you need any clarifications, thanks!

~ Padoru

yarga [219]3 years ago
3 0

the answer to your problem is work

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Check the attached file for the answer.

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A client is receiving an IV solution of sodium chloride 0.9% (Normal Saline) 250 ml with amiodarone (Cordarone) 1 gram at 17 ml/
tester [92]

Answer:

1.1mg/min

Explanation:

We are given that

Volume of solution=250 ml

Mass of amiodarone=1 g

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We know that

1 g= 1000 mg

Ratio of 1000g:250 ml=\frac{1000}{25}=4 mg/ml

The concentration of solution=4mg/ml

Amiodarone infusing (mg/min)=\frac{infusion \;rate(ml/hr)\times concentration}{60}

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6 0
3 years ago
A block slides to a stop as it goes 47 m across a level floor in a time of 6.35 s. a) What was the initial velocity? b) What is
UNO [17]

Answer :

(a) The initial velocity is, 14.8 m/s

(b) The acceleration is, -2.33m/s^2

Explanation :

By the 1st equation of motion,

v=u+at     ...........(1)

where,

v = final velocity = 0 s

u = initial velocity

t = time = 6.35 s

a = acceleration

The equation 1 will be:

0=u+a(6.35}

u=-6.35a       ..........(2)

By the 2nd equation of motion,

s=ut+\frac{1}{2}at^2     ...........(3)

where,

s = distance = 47 m

Now substitute equation 2 in 3, we get:

47=(-6.35a)\times (6.35)+\frac{1}{2}\times a\times (6.35)^2

By solving the term, we get:

a=-2.33m/s^2

The acceleration is, -2.33m/s^2

Now we have to calculate the initial velocity.

Using equation 2, we gte:

u=-6.35a

u=-6.35s\times (-2.33m/s^2)

u=14.8m/s

The initial velocity is, 14.8 m/s

5 0
3 years ago
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ANTONII [103]

Answer:

The velocity of a particle relative to S is equal to its velocity relative to S′ plus the velocity of S′ relative to S. We can extend Equation 4.35 to any number of reference frames. For particle P with velocities →vPA, →vPB, and →vPC in frames A, B, and C, →vPC=→vPA+→vAB+→vBC.

Explanation:

Resultant Velocity. Multiply the acceleration by the time the object is being accelerated. For example, if an object falls for 3 seconds, multiply 3 by 9.8 meters per second squared, which is the acceleration from gravity. The resultant velocity in this case is 29.4 meters per second.

6 0
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Which is one piece of information that astronomers use to calculate the age of the universe?
Valentin [98]

Answer:

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8 0
4 years ago
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