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3 years ago

Which is the best example that something has kinetic energy?

Physics

1 answer:

KonstantinChe [14]3 years ago

6 0

A person running is an example of something that has kinetic energy

(More in general: any object which is moving is an example of kinetic energy)

Explanation:

Kinetic energy is the energy possessed by a body due to its motion.

Mathematically, the kinetic energy of a body is given by:

$K=\frac{1}{2}mv^2$

where

K is the kinetic energy

m is the mass of the body

v is the speed of the body

As we can see:

- The kinetic energy is proportional to the mass of the body

- The kinetic energy is proportional to the square of the speed of the body

This means that an object has kinetic energy whenever its speed is non-zero: so, every object in motion has kinetic energy.

Therefore, a person which is running is an example of something that has kinetic energy.

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

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15) A 328-kg car moving at 19.1 m/s in the +x direction hits from behind a second car moving at 13 m/s in the same direction. If

kolbaska11 [484]

Answer:

Explanation:

Given that,

Mass of first car

M1= 328kg

The car is moving in positive direction of x axis with velocity

U1 = 19.1m/s

Velocity of second car

U2 = 13m/s, in the same direction as the first car..

Mass of second car

M2 = 790kg

Velocity of second car after collision

V2 = 15.1 m/s

Velocity of first car after collision

V1 =?

This is an elastic collision,

And using the conservation of momentum principle

Momentum before collision is equal to momentum after collision

P(before) = P(after)

M1•U1 + M2•U2 = M1•V1 + M2•V2

328 × 19.1 + 790 × 13 = 328 × V1 + 790 × 15.1

16534.8 = 328•V1 + 11929

328•V1 = 16534.8—11929

328•V1 = 4605.8

V1 = 4605.8/328

V1 = 14.04 m/s

The velocity of the first car after collision is 14.04 m/s

5 0

3 years ago

Derive the following equations of motion

xz_007 [3.2K]

Answer:

___________________________________

<h3>a. Let us assume a body has initial velocity 'u' and it is subjected to a uniform acceleration 'a' so that the final velocity 'v' after a time interval 't'. Now, By the definition of acceleration, we have:</h3>

$a = \frac{v - u}{t} \\ or \: at = v - u \\ v = u + at \:$

It is first equation of motion.

___________________________________

<h3>b. Let us assume a body moving with an initial velocity 'u'. Let it's final body 'v' after a time interval 't' and the distance travelled by the body becomes 's' then we already have,</h3>

$v = u + at...........(i) \\ s = \frac{u + v}{2} \times t.........(ii)$

Putting the value of v from the equation (i) in equation (ii), we have,

$s= \frac{u + (u + at)}{2} \times t \: \: \\ or \: s = \frac{(2u + at)t}{2} \\ or \: s = \frac{2ut + a {t}^{2} }{2} \\ s = ut + \frac{1}{2} a {t}^{2}$

It is third equation of motion.

________________________________

<h3>c. Let us assume a body moving with an initial velocity 'u'. Let it's final velocity be 'v' after a time and the distance travelled by the body be 's'. We already have,</h3>

$v = u + at.....(i) \\ s = \frac{u + v}{2} \times t......(ii) \\$

$v = u + at \\ or \: at = v - u \\ t = \frac{v - u}{a}$

Putting the value of t from (i) in the equation (ii)

$s = \frac{u + v}{2} \times \frac{v - u}{a} \\ or \: s = \frac{ {v}^{2} - {u}^{2} }{2a} \\ or \: 2as = {v}^{2} - {u}^{2} \\ {v}^{2} = {u}^{2} + 2as$

It is forth equation of motion.

________________________________

Hope this helps...

Good luck on your assignment..

3 0

3 years ago

There are 5,280 feet in 1 mile and 12 inches in one foot. How many inches are in a mile?

ohaa [14]

63360
5280 feet*12 inches in every foot=63360

4 0

3 years ago

In an experiment, it took 300s to increase the temperature of 0.1kg of the liquid from 25°C to 50°C. During this period, the ene

klemol [59]

Answer:

<h2>4000 $\textbf{J}\text{ }\textbf{kg}^{\textbf{-1}}\text{ }\textbf{K}^{\textbf{-1}}$ </h2>

Explanation:

The temperature of 0.1 kg of liquid rises from 25°C to 50°C in 300 sec. Energy of 13,600 J was supplied during this time. Appartus was losing energy at the rate of 12 J/sec.

Let us assume the Specific heat capacity as $s$ .

As there is no state change from liquid to gas, only Specific heat capacity is involved. Also, work done is approximately zero because volume does not change much. So,

Energy gained = Energy required to rise the temperature

Energy gained by liquid = $13600\text{ }J-(300\text{ }sec)\times(12\text{ }\frac{J}{sec})=10000\text{ }J$

$10000\text{ }J=m\times s\times\Delta T=(0.1\text{ }kg)\times(s)\times(323K-298K)=2.5s\text{ }kgK\\s=4000\text{ }\frac{J}{kg.K}$

∴ Specific heat capacity of liquid = 4000 $\frac{J}{kg.K}$

4 0

3 years ago

The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde

Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

$S_1 S_2$ =3m

$S_1 O$ =4m

By Pythagoras theorem

$S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m$

Now

The intensity at O when both speakers are on is given by

$I=4I_1 cos^2(\pi \frac{\delta}{\lambda})$

Here

I is the intensity at O when both speakers are on which is given as 6 $W/m^2$

I1 is the intensity of one speaker on which is 6 $W/m^2$

δ is the Path difference which is given as

$\delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m$

λ is wavelength which is given as

$\lambda=\frac{v}{f}$

Here

v is the speed of sound which is 320 m/s.

f is the frequency of the sound which is to be calculated.

$16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )$

where k=0,1,2

for minimum frequency $f_1$ , k=1

${f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz$

So the minimum frequency is 702.22 Hz

5 0

3 years ago