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3 years ago

What is the relationship between the kinetic energy of molecules in an object and the object's temperature?

Physics

2 answers:

kykrilka [37]3 years ago

7 0

Answer:

As the temp increases the kinetic mol increses

Explanation:

Ghella [55]3 years ago

5 0

Answer:

Correct answer:

As the temperature increases, the kinetic energy of the molecules increases.

Explanation:

The relation between kinetic energy and temperature in an ideal gas which we will take as an example is given by the following formula:

Ek = 3/2 N k T ,

where N is the total number of molecules a k is a Boltzmann constant.

If we assume that the total number of molecules is a constant, kinetic energy only depends on temperature.

God is with you!!!

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What is the relation between centre of gravity and stability

maw [93]

Explanation:

tilting it will raise the height of its center of gravity.

8 0

2 years ago

A plane flies 1800 miles in 9 hours, with a tailwind all the way. the return trip on the same route, now with a headwind, tak

Fittoniya [83]

Initially its moving with tail wind so here the speed of wind will support the motion of the plane

so we can say

$V_{plane} + v_{wind} = \frac{distance}{time}$

$V_{plane} + v_{wind} = \frac{1800}{9}$

$V_{plane} + v_{wind} = 200 mph$

now when its moving with head wind we can say that wind is opposite to the motion of the plane

$V_{plane} - v_{wind} = \frac{distance}{time}$

$V_{plane} - v_{wind} = \frac{1800}{12}$

$V_{plane} - v_{wind} = 150mph$

now by using above two equations we can find speed of palne as well as speed of wind

$V_{plane} = 175 mph$

$v_{wind} = 25 mph$

5 0

3 years ago

The Special Olympics raises money through "plane pull" events in which teams of 25 people compete to see who can pull a 74,000 k

Murrr4er [49]

Answer:

28716.4740661 N

1.2131147541 m/s

51.2474965841%

Explanation:

m = Mass of plane = 74000 kg

s = Displacement = 3.7 m

f = Frictional force = 14000 N

t = Time taken = 6.1 s

u = Initial velocity = 0

v = Final velocity

$s=ut+\frac{1}{2}at^2\\\Rightarrow 3.7=0\times 6.1+\frac{1}{2}\times a\times 6.1^2\\\Rightarrow a=\frac{3.7\times 2}{6.1^2}\\\Rightarrow a=0.198871271164\ m/s^2$

Force is given by

$F=ma+f\\\Rightarrow F=74000\times 0.198871271164+14000\\\Rightarrow F=28716.4740661\ N$

The force with which the team pulls the plane is 28716.4740661 N

$v=u+at\\\Rightarrow v=0+0.198871271164\times 6.1\\\Rightarrow v=1.2131147541\ m/s$

The speed of the plane is 1.2131147541 m/s

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}\times 74000\times 1.2131147541^2\\\Rightarrow K=54450.9540448\ J$

Work done is given by

$W=Fs\\\Rightarrow W=28716.4740661\times 3.7\\\Rightarrow W=106250.954045\ J$

The fraction is given by

$\dfrac{54450.9540448}{106250.954045}=0.512474965841$

The teams 51.2474965841% of the work goes to kinetic energy of the plane.

3 0

3 years ago

Because the time was the same for each segment you know the speed was the same for each segment

netineya [11]

Answer:

Market segmentation helps you send the right message, every time, by efficiently targeting specific groups of consumers.

Explanation:

3 0

1 year ago

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago