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3 years ago

Hurry!!!!!!!!!!!!!!! Match the terms to the correct descriptions.

1 answer:

5 0

This can be either kinetic or potential and has to do with the energy of position and motion of an energy- This is related to mechanical energy.It is so because mechanical energy is the sum total of kinetic and potential energy.

This is radiant energy from the sun;it is the only kind of energy we can see with our eyes-It is related to light energy which consists of white light along with its seven colors and these are also electromagnetic radiations.

This is the energy that is transferred by the movement of electrons through a conductor;the electrons create a current-This is related to electric energy.

This type of energy is often found in things like batteries or food-This is related to chemical energy .It is chemical energy which is converted to electrical energy in battery.

This type of energy travels through vibrations on waves-This is related to sound energy which travels in a medium in the form of compressions and rarefactions.

This type of energy is found in the nucleus of an atom-This is related to nuclear energy.For instance the energy produced during nuclear fusion and fission from unstable nuclei.

This is the term used when one type of energy changes from one form to another-This is related to energy conversions.

This type of energy is stored energy;it can be increased by getting into a higher position and/or stretching an object such as a rubber band-This is related to potential energy .It is the energy gained by a body due to its position or change in configurations.

This type of energy can be transferred in three different ways i.e conduction,convection and radiation-We know that heat is transferred in three different ways mentioned above and heat is nothing else than transferred thermal energy.Hence, it is related to thermal energy.

This has to do with the speed of an object and how much mass it has;basically how the how the object is moving-This is related to kinetic energy which is the energy gained by a body due to its motion.

Mathematically it is written as

$kinetic\ energy =\frac{1}{2} mass*[velocity]^2$

You might be interested in

A lever has a total length of 12 meters with a fulcrum at the exact center of the lever. Which best describes how to

valentina_108 [34]

Answer:

The correct option is the last option.

Explanation:

Generally, when trying to create a mechanical advantage of a lever for an apparatus or a machine, the load is usually moved closer to the fulcrum. Hence, if a lever has a total length of 12 meters and the fulcrum is placed at 6 meters (the center), the best way (based on the previous statement) to double the mechanical advantage of the lever is to move the fulcrum 4 meters toward the side on which the force is applied. The correct option is the last option.

3 0

3 years ago

ustapha Jones is speeding on the interstate in his Ferrari at 231km/hr when he passes a police car at rest . If the cop accelera

Lena [83]

Answer:

461 km/h

Explanation:

In order to solve this problem we must first sketch a drawing of what the situation looks like so we can better visualize it. (See attached picture).

We have two situations there, the first one is Mustapha's car that is traveling at a constant speed of 231km/hr.

The second situation is the police that is accelerating from rest until he reaches Mustapha. (We are going to suppose the acceleration is constant and that he will not stop accelerating until he reaches Mustapha). He has an acceleration of 15m/ $s^{2}$ .

We want to find what the final velocity of the police is at the time he reaches Mustapha. From this we can imply that the displacement x will be the same for both particles.

So let's model the first situation.

The displacement of Mustapha can be found by using the following equation:

$V_{M}=\frac{x}{t}$

when solving the equation for the displacement x we get that it will be:

$x=V_{M}t$

Now let's model the displacement of the cop. Since the cop has a constant acceleration, we can model his displacement with the following formula.

$x=V_{0}t+\frac{1}{2}at^{2}$

Since the initial velocity of the cop is zero, we can get rid of that part of the equation leaving us with:

$x=\frac{1}{2}at^{2}$

We can now set both equations equal to each other so we get:

$\frac{1}{2}at^{2}=V_{M}t$

When solving this for t, we get that:

$t=\frac{2V_{M}}{a}$ (let's call this equation 1)

Now, we know the cop has constant acceleration, so we can model it with the following formula too:

$a=\frac{V_{f}-V_{0}}{t}$

since the initial velocity of the cop is zero, we can get rid of that here too, so we get the following formula:

$a=\frac{V_{f}}{t}$

when solving for the final velocity, we get that:

$V_{f}=at$ (let's call this equation 2)

when substituting equation 1 into equation 2 we get:

$V_{f}=a(\frac{2V_{M}}{a})$

we can now cancel a leaving us with:

$V_{f}=2V_{M}$

This tells us that the final velocity of the cop will not depend on his acceleration. (This is only if the acceleration is constant all the time) So we get that the final velocity of the cop is:

$V_{f}=2(231km/hr)$

$V_{f}=461km/hr$

which is our answer.

8 0

3 years ago

Where might you look on the internet to find good scientific information about illness

user100 [1]

It depends on what illness and what country you are in but Mayo Clinic and Johns Hopkins are good sources.

5 0

3 years ago

Read 2 more answers

A 60.0-kg man stands at one end of a 20.0-kg uniform 10.0-m long board. How far from the man is the center of mass of the man-bo

timama [110]

Answer:

x=1.25m

Explanation:

The Center of mass of the system is defined as the point where whole mass of the body is appeared to be concentrated.

The center of mass of the system is given by

x= $\frac{m1x1+m2x2}{m1+m2}$

where m1 is mass of man =60 kg

m2 mass of board =20 kg

let the man be at the origin x1 =0 , x2 =5m

by substituting in above formula

x = $\frac{(60*0)+(20*5)}{60+20}$ = $\frac{100}{80}$ =1.25 m

x=1.25m

So the center of mass of the system is at 1.25 m from man.

6 0

4 years ago

Read 2 more answers

At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constan

Jet001 [13]

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time $t_1$ , the acceleration vector has direction $\theta_1$ such that

$\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ$

which indicates the particle is situated at a point on the lower left half of the circle, while at time $t_2$ the acceleration has direction $\theta_2$ such that

$\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ$

which indicates the particle lies on the upper left half of the circle.

Notice that $\theta_1-\theta_2=90^\circ$ . That is, the measure of the major arc between the particle's positions at $t_1$ and $t_2$ is 270 degrees, which means that $t_2-t_1$ is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

$\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}$

where $R$ is the radius of the circle and $T$ is the period. We have

$t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s$

and the magnitude of the particle's acceleration toward the center of the circle is

$\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}$

So we find that the path has a radius $R$ of

$7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}$

8 0

3 years ago