All categories

3 years ago

When a spring is compressed, the energy changes from kinetic to potential. Which best describes what is causing this change?

Physics

2 answers:

just olya [345]3 years ago

6 0

Answer:

Well, Your answer will be is Work. Because, When a spring is compressed, the energy changes from kinetic to potential. This change is caused by work.

Good Luck!~

$^{By$ $^{Itsbrazts$

Oxana [17]3 years ago

5 0

Answer:

Hey!

Your answer is WORK!

Explanation:

For the energy store to change from Kinetic to Potential, WORK would have to be done to transfer the energy!

HOPE THIS HELPS!!

<h2>ItsMATT15 </h2>

You might be interested in

How does a generator produce an electric current

love history [14]

Electric generator is a device that converts mechanical energy obtained from an external source into electrical energy as an output. It was discovered that the above flow of electric charges could be induced by moving an electrical conductor such as a wire that contains electric charges in a magnetic field

4 0

2 years ago

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla

Rama09 [41]

Please elaborate more on your question so I can help you

7 0

2 years ago

Where are warm air vents usually located in a forced air heating system

crimeas [40]

The system is located in the HairyFunyon aka near the floor

8 0

3 years ago

Read 2 more answers

You are on a train that is traveling at 3.0 m/s along a level straight track. Very near and parallel to the track is a wall that

loris [4]

Questions Diagram is attached below

Answer:

$T=2.08s$

Explanation:

From the question we are told that:

Speed of Train $V=3.0m.s$

Angle $\theta=12\textdegree$

Height of window $h_w=0.90m$

Width of window $w_w=2.0m$

The Horizontal distance between B and A from Trigonometric Laws is mathematically given by

$b=\frac{0.9}{tan12}$

$b=4.23$

Therefore

Distance from A-A

$d_a=2.0+4.23$

$d_a=6.23$

Therefore

Time Required to travel trough d is mathematically given as

$T=\frac{d_a}{v}$

$T=\frac{6.23}{3}$

$T=2.08s$

5 0

2 years ago

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

8 0

3 years ago