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3 years ago

Describe the kinetic molecular theory

1 answer:

8 0

The kinetic theory of gases is a simple, historically significant model of the thermodynamic behavior of gases, with which many principal concepts of thermodynamics were established. The model describes a gas as a large number of identical submicroscopic particles, all of which are in constant, rapid, random motion

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Which statement accurately describes the motion of the object in the graph above over 10 seconds? Group of answer choices The ob

timurjin [86]

Answer:

Object appears to move forward at 1 cm/sec, then the velocity drops to zero for 3 sec and then moves forward at 2 cm/sec (11 - 3) / (10 - 6) = 2 cm/sec

7 0

2 years ago

An electron is released from rest at a distance of 0.470 m from a large insulating sheet of charge that has uniform surface char

ArbitrLikvidat [17]

Answer:

Part a)

$W = 1.58 \times 10^{-20} J$

Part b)

$v = 1.86 \times 10^5 m/s$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = 4.00 \times 10^{-12} C/m^2$

now the electric field is given as

$E = \frac{4.00 \times 10^{-12}}{2(8.85 \times 10^{-12})}$

$E = 0.225 N/C$

Now acceleration of an electron due to this electric field is given as

$a = \frac{eE}{m}$

$a = \frac{(1.6 \times 10^{-19})(0.225)}{9.1 \times 10^{-31}}$

$a = 3.97 \times 10^{10}$

Now work done on the electron due to this electric field

$W = F.d$

$d = 0.470 - 0.03$

$d = 0.44 m$

So work done is given as

$W = (ma)(0.44)$

$W = (9.11 \times 10^{-31})(3.97 \times 10^{10})(0.44)$

$W = 1.58 \times 10^{-20} J$

Part b)

Now we know that work done by all forces = change in kinetic energy of the electron

so we will have

$W = \frac{1}{2}mv^2 - 0$

$1.58 \times 10^{-20} = \frac{1}{2}(9.1\times 10^{-31})v^2$

$v = 1.86 \times 10^5 m/s$

7 0

3 years ago

Consider an electron, of charge magnitude e = 1.602 10-19 C and mass me = 9.11 10-31 kg, moving in an electric field with an ele

Tems11 [23]

Answer: v= 7.509 x 10^6 m/s

B) the lower plate because the electron is negatively charged

Explanation:

From the question

Electronic charge (q) =1.602 x 10^-19c

Electric field intensity (E) = 8 x 10² = 800N/C

Mass of electron (m) = 9.11 x 10^-31 kg

Length of plate (L) = 50cm=0.5m

Distance between plates (D) = 20cm = 0.2m

Since the electron is entering a uniform electric field, the resulting motion will be of a constant acceleration and can be defined by the equations of motion with a constant acceleration.

From newton's law of motion.

F= ma

The force (F) is coming from the electric field which is

F=Eq.

Thus F = 800 x 1.602 x 10^-19

F = 1.2816 x 10^-16 N

Acceleration of electron (a) = F/m where m is the mass of electron (given above)

Hence

a = 1.2816 x 10^-16 / 9.11 x 10^-31

a= 1.41 x 10¹⁴ m/s².

Using Newton laws of motion to get velocity, we recall that v²= u² + 2ad

Where v is final velocity, u is initial velocity (zero in this case because the electron starts it motion from rest), a= acceleration, d= distance traveled (which in this case is distance between plates)

v² = 0² + 2(1.41 x 10¹⁴) x 0.2

v² = 5.64 x 10¹³

Thus v = √ 5.64 x 10¹³

v = 7.509 x 10^6 m/s

Since the electric field is upwards it denotes that the positive plate is downward and the negative is upward ( this is because electric flux from a positive charge has an outward flow and that from a negative charge has an inward flow. So from our questions, if the electric field is upward, it means it is starting from the bottom plate which will be positive) the electron (which is negatively charged) will be attracted to the positive plate which is downward for this question of ours.

So therefore, the electron is attracted to the downward plate because it (electron) is negative

Option B

4 0

3 years ago

A truck heading east has an initial velocity of 6 m/s. It accelerates at 2 m/s2 for 12 seconds. What distance does the truck tra

gavmur [86]

I believe it would be 144 m!

7 0

3 years ago

Read 2 more answers

A projectile is fired with an initial speed of 40 m/s at an angle of elevation of 30∘. Find the following: (Assume air resistanc

BabaBlast [244]

Answer:

a. 2.0secs

b. 20.4m

c. 4.0secs

d. 141.2m

e. 40m/s, ∅= -30°

Explanation:

The following Data are giving

Initial speed U=40m/s

angle of elevation,∅=30°

a. the expression for the time to attain the maximum height is expressed as

$t=\frac{usin\alpha }{g}$

where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at

$t=40sin30/9.81\\t=2.0secs$

b. the expression for the maximum height is expressed as

$H=\frac{u^{2}sin^{2}\alpha }{2g} \\H=\frac{40^{2}0.25 }{2*9.81} \\H=20.4m$

c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,

Hence T=2t

T=2*2.0

T=4.0secs

d. The range of the projectile is expressed as

$R=\frac{U^{2}sin2\alpha}{g}\\R=\frac{40^{2}sin60}{9.81}\\R=141.2m$

e. The landing speed is the same as the initial projected speed but in opposite direction

Hence the landing speed is 40m/s at angle of -30°

3 0

3 years ago