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Klio2033 [76]
3 years ago
5

What is the main reason for using a data table for collecting data?

Physics
1 answer:
Luba_88 [7]3 years ago
7 0

Answer:

to make calculation more easy to get

Explanation:

if you are using chart or calculate Thermodynamic problems you will not never solve this problem with out using data table for thermodynamic

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As mass increases what happens to the kinetic energy
AnnyKZ [126]
As mass increases kinetic energy also increases; kinetic energy is directly proportional to mass so whatever is done to either affects the other one the same. i hope this helps :)
5 0
3 years ago
1. If we want to increase the strength of an electromagnet, which 2 of
kap26 [50]

Answer:

A) Have more coils of wire

F) Use a more powerful battery

4 0
3 years ago
A book prone to air resistance is released from rest 300 m
yaroslaw [1]

Answer:

Approximately 73\%.

(Assuming that g = \rm 9.81\; m \cdot s^{-2}.)

Explanation:

The mechanical energy of an object is the sum of its potential energy and its kinetic energy. It will be shown that the exact mass of this object doesn't matter. For ease of calculation, let m(\text{book}) represent the mass of the book.

The initial potential energy of the book is  

\begin{aligned}U(300\; \text{m}) &= m(\text{book}) \cdot g \cdot \Delta h + U(0\; \text{m}) \cr &=(9.81 \times 300) \cdot m(\text{book})\cr &= \left(2.943\times 10^3\right) \cdot m(\text{book})\end{aligned}.

The book was initially at rest when it was released. Hence, its initial kinetic energy would be zero. Hence, the initial mechanical energy of the book-Earth system would be (2.943\times 10^3) \cdot m(\text{book}).

When the book was about to hit the ground, its speed is \rm 40\; m \cdot s^{-1}. Its kinetic energy would be:

\begin{aligned} \text{KE} &= \frac{1}{2} \, m(\text{book}) \cdot v^{2} \cr &= \left(\frac{1}{2} \times 40^2\right)\cdot m(\text{book}) \cr &= \left(8.00\times 10^2\right)\cdot m(\text{book})\end{aligned}.

The question implies that the potential energy of the book near the ground is zero. Hence, the mechanical energy of the system would be \left(8.00\times 10^2\right)\cdot m(\text{book}) when the book was about to hit the ground.

The amount of mechanical energy lost in this process would be equal to:

\begin{aligned}&\left(2.943\times 10^3\right) \cdot m(\text{book}) - \left(8.00\times 10^2\right)\cdot m(\text{book}) \cr &=\left(2.143\times 10^3\right)\cdot m(\text{book})\end{aligned}.

Divide that with the initial mechanical energy of the system to find the percentage change. Note how the mass of the book, m(\text{book}), was eliminated in this process.

\begin{aligned}&\frac{\left(2.143\times 10^3\right)\cdot m(\text{book})}{\left(2.943\times 10^3\right) \cdot m(\text{book})}\times 100\% \cr &= \frac{2.143\times 10^3}{2.943\times 10^3}\times 100\% \cr & \approx 73\%\end{aligned}.

5 0
4 years ago
Can anyone please explain this point with an example. I have presentation tomorrow.
r-ruslan [8.4K]

Explanation:

Efficiency is a way of describing the amount of useful ​output​ a process or machine can generate as a percentage of the ​input​ required to make it go. In other words, it compares how much energy is used to do work versus how much is lost or wasted to the environment. The more efficient the machine, the less energy wasted.

For example, if a heat engine is able to turn 75 percent of the fuel it receives into motion, while 25 percent is lost as heat in the process, it would be 75 percent efficient. Out of the original 100 percent of the fuel, 75 percent was output as useful work.  

the equation:

energy efficiency =useful output energy/total input energy

8 0
2 years ago
Can someone help me find the which direction is north and which is south on this solenoid?
arsen [322]

Explanation:

Now, looking down the solenoid tube determine what direction is the winding. If clockwise in relation to the positive wire then is the south pole, if anti-clockwise then is the north pole. So, to summarize the magnetic south pole is always clockwise in relation to the positive wire.

7 0
3 years ago
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