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dimulka [17.4K]
3 years ago
8

The average speed of a nitrogen molecule in air is about 6.70 ✕ 102 m/s, and its mass is about 4.68 ✕ 10-26 kg.(a) If it takes 3

.10 ✕ 10-13 s for a nitrogen molecule to hit a wall and rebound with the same speed but moving in an opposite direction (assumed to be the negative direction), what is the average acceleration of the molecule during this time interval?Incorrect: Your answer is incorrect.Your response differs from the correct answer by more than 10%. Double check your calculations. m/s2(b) What average force does the molecule exert on the wall?
Physics
1 answer:
Sveta_85 [38]3 years ago
5 0

Answer:

a = 4.32*10^15 m/s^2

Explanation:

In order to calculate the acceleration of the nitrogen molecule, you take into account that the force experienced by the molecule, is equal to the change in its momentum:

F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}       (1)

m: mass of the nitrogen molecule

Δv: change in the speed of the molecule

Δt: time lapse of the change of the momentum of the molecule = 3.10*10^-13 s

Furthermore, by the Newton second law, you have:

F=ma=m\frac{\Delta v}{\Delta t}\\\\a=\frac{\Delta v}{\Delta t}       (2)

The change in the speed is given by:

\Delta v=v-v_o=6.70*10^2m/s-(-6.70*10^2m/s)=1.34*10^3\frac{m}{s}

where you have taken into account the direction of the initial and final speed.

You replace the values of all parameters in the equation (1) in order to calculate the acceleration:

a=\frac{1.34*10^3m/s}{3.10*10^{-13}s}=4.32*10^{15}\frac{m}{s^2}

The acceleration of the nitrogen molecule is 4.32*10^15m/s^2

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Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.
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a) C.M =(\bar x, \bar y)=(0.767,0.7)m

b) (x_4,y_4)=(-1.917,-1.75)m

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}

\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}

Where M represent the sum of all the masses on the system.

And the center of mass C.M =(\bar x, \bar y)

Part a

m_1= 3 kg, m_2=5kg,m_3=7kg represent the masses.

(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5) represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m

C.M =(\bar x, \bar y)=(0.767,0.7)m

Part b

For this case we have an additional mass m_4=6kg and we know that the resulting new center of mass it at the origin C.M =(\bar x, \bar y)=(0,0)m and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

(x_4,y_4)=(-1.917,-1.75)m

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