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3 years ago

How is the kintic energy of nolecules changing when the molecules move faster?

Physics

2 answers:

Nonamiya [84]3 years ago

5 0

Answer:

an increase in temperature will increase the average kinetic energy of the molecules. As the particles move faster, they will likely hit the edge of the container more often. ... Increasing the kinetic energy of the particles will increase the pressure of the gas.

Explanation:

sladkih [1.3K]3 years ago

5 0

Answer:

According to Kinetic Molecular Theory, an increase in temperature will increase the average kinetic energy of the molecules. As the particles move faster, they will likely hit the edge of the container more often.

Explanation:

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An object is dropped and is in free fall. Each second, the position of the object is marked. The distance between each mark is m

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C. hope this helps :)

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3 years ago

What is the final velocity?

Reil [10]

The final velocity is a vector quantity that measures the speed and direction of a moving body after it has reached its maximum acceleration

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3 years ago

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A textbook of mass 2.05 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose d

blsea [12.9K]

Answer:

a. 7.38 N b. 40.87 N c. 0.113 kg-m²

Explanation:

a. Let T be the tension in the cord. For the textbook, T = ma since no other force acts on it and it is an horizontal force, and m = mass = 2.05 kg and a = acceleration. We find the acceleration from s = ut + 1/2at² where u = initial speed = 0 (since it starts from rest), s = distance moved = 1.30 m and t = time = 0.850 s.

Substituting these values into s,

1.30 m = 0 × 0.850 + 1/2a × 0.850² = 0 + 0.36125a

1.30 = 0.36125a

a = 1.30/0.36125 = 3.6 m/s²

Substituting this into T, we have

T = ma = 2.05 kg × 3.6 m/s² = 7.38 N

b. Let T be the tension in the cord attached to the book. The book has the only vertical forces acting on it as the tension, T(acting upwards) and its weight mg (acting downwards). So the net force acting on it is

T - mg = ma

T = m(a + g)

substituting a = 3.6 m/s² and g = 9.8 m/s² and m = 3.05 kg

T = 3.05(3.6 + 9.8) = 3.05 × 13.4 = 40.87 N

c. Since the tangential acceleration of the pulley is also the acceleration of the masses, the a = rα where r = radius of pulley = 0.200 m/2 = 0.100 m and α = angular acceleration of the pulley.

α = a/r = 3.6 m/s² ÷ 0.100 m = 36 rad/s²

Now, the torque on the pulley τ = Tr = Iα where I = moment of inertia of pulley about its rotational axis and T = tension in cord attached to book and r = radius of pulley = 0.200 m/2 = 0.100 m

From the equation above, I = Tr/α

Substituting the variables we have

I = 40.87 N × 0.100 m ÷ 36 rad/s² = 0.113 kg-m²

4 0

3 years ago

Read 2 more answers

A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.

gtnhenbr [62]

Answer:

(a) vmax = 0.34m/s

(b) v = 0.13m/s

(d) x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

$v_{max}=\omega A$ (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

$\omega=\sqrt{\frac{k}{m}}$ (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

$v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}$

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

$v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}$

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

$x=Acos(\omega t)$

You solve the previous equation for t:

$t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s$

With this value of t, you can calculate the speed of the object with the following formula:

$v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}$

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

$\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s$

The position will be:

$x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m$

The position of the object on which its speed is one-half its maximum velocity is 0.039

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3 years ago

An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one fourth of its maximum height above its l

Murrr4er [49]

Answer:

v = 25.45 m/s

Explanation:

In order to calculate the initial speed of the object, you take into account the formula for the maximum height reaches by the object. Such a formula is given by:

$h_{max}=\frac{v_o^2}{g}$ (1)

vo: initial speed of the object = 18 m/s

g: gravitational acceleration = 9.8 m/s²

Furthermore you use the following formula for the final speed of the object:

$v^2=v_o^2-2gh$ (2)

h: height

You know that the speed of the object is 18m/s when it reaches one fourth of the maximum height. You use this information, and you replace the equation (1) in to the equation (2), as follow:

$v^2=v_o^2-2g(\frac{h_{max}}{4})=v_o^2-\frac{1}{2}g(\frac{v_o^2}{g})\\\\v^2=v_o^2-\frac{1}{2}v_o^2=\frac{1}{2}v_o^2$

Then, you solve the previous result for vo:

$v_o=\sqrt{2}v=\sqrt{2}(18m/s)=25.45\frac{m}{s}$

The initial speed of the object was 25.45 m/s

3 0

3 years ago