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Alja [10]
3 years ago
15

A 2kg book is moved from a shelf that is 2m off the ground to a shelf that is 1.5m off the ground, what is it’s change in gravit

ational potential energy
Physics
1 answer:
Ket [755]3 years ago
3 0
The gravitational energy is going up subtracting the energy that was on the ground
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murzikaleks [220]

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by using it in a such place or thing which needs it or which can work with it

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PLEASE HELP IT'S DUE IN LIKE 2 MINUTES
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1kg

Explanation:

this box is the smallest and weighs the least. Hope this helps :]

4 0
2 years ago
In the formula used to solve problems related to the first law of thermodynamics, what does Q represent? A.internal energy
Blizzard [7]
B, heat, is the correct answer. Heat is represented by a capital q in thermodynamic equations.
7 0
3 years ago
Read 2 more answers
What is the purpose of the scapula to move during arm elevation?
Inessa [10]

The purpose of the scapula to move during arm elevation is increase the range of elevation of the arm.

<h3>What is the importance of movement of the scapula during arm elevation?</h3>

The scapula is an important bone which is found in the shoulder and back region of the body.

The scapula enables and increases the range of motion of the arm with its motions.

During arm elevation, the scapula undergoes an upward rotational motion.

Therefore, the  purpose of the scapula to move during arm elevation is increase the range of elevation of the arm.

Learn more about scapula motion at: brainly.com/question/5133017

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3 0
1 year ago
An intravenous (IV) system is supplying saline solution to a patient at the rate of 0.06 cm3/s through a needle of radius 0.2 mm
horsena [70]

Answer:

Pressure applied to the needle is 7528 Pa

Explanation:

As we know by poiseuille's law of flow of liquid through a cylindrical pipe

the rate of flow through the pipe is given as

Q = \frac{\Delta P \pi r^4}{8\eta L}

now we know that

Q = 0.06 \times 10^{-6} m^3/s

radius = 0.2 mm

Length = 6.32 cm

\eta = 1\times 10^{-3} Pa s

now we have

6 \times 10^{-8} = \frac{\Delta P \pi (0.2 \times 10^{-3})^4}{8(1 \times 10^{-3})6.32 \times 10^{-2}}

3.03 \times 10^{-11} = \Delta P 5.02 \times 10^{-15}

\Delta P = 6028 Pa

now we have

P - 1500 = 6028 Pa

P = 7528 Pa

8 0
3 years ago
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