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3 years ago

Which of the following is not a part of dalton s atomic theory?

Physics

2 answers:

larisa [96]3 years ago

8 0

atoms are always in motion. i took the test

Alex777 [14]3 years ago

6 0

<span>c. atoms are always in motion..............</span>

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If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. Wh

In-s [12.5K]

Answer:

v = -v₀ / 2

Explanation:

For this exercise let's use kinematics relations.

Let's use the initial conditions to find the acceleration of the electron

v² = v₀² - 2a y

when the initial velocity is vo it reaches just the negative plate so v = 0

a = v₀² / 2y

now they tell us that the initial velocity is half

v’² = v₀’² - 2 a y’

v₀ ’= v₀ / 2

at the point where turn v = 0

0 = v₀² /4 - 2 a y '

v₀² /4 = 2 (v₀² / 2y) y’

y = 4 y'

y ’= y / 4

We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.

v² = v₀² -2a y’

v² = 0 - 2 (v₀² / 2y) y / 4

v² = -v₀² / 4

v = -v₀ / 2

We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.

7 0

3 years ago

prepara un escrito en el que resumas las ideas principales sobre la evolución de los modelos atómicos

Dvinal [7]

Mexico Mexico language to is the answer to the question

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3 years ago

A dog of mass 18 kg runs at a speed of 4 m/s. What is the momentum of the

Andrej [43]

Answer:

A, 72 kg•m/s

Explanation:

p=mv

p=18x4

p=72

6 0

3 years ago

The temperature and time t given in hours from 0 to 24 after midnight in downtown mathville is given by t=10-5 sin(pi/12 t) degr

Dvinal [7]

Answer:

$T_A_v_g=9.918192559$^{\circ}C$

Explanation:

The problem tell us that the temperature as function of time in downtown mathville is given by:

$T(t)=10-5*sin(\frac{\pi}{12t})$

The average temperature over a given interval can be calculated as:

$T_a_v_g=\frac{T_o+T_f}{2}$

Where:

$T_o=Initial\hspace{3}temperature\\T_f=Final\hspace{3}temperature$

So, the initial temperature in this case, would be the temperature at noon, and the final temperature would be the temperature at midnight:

Therefore:

$T_o=T(12)=10-5*sin(\frac{\pi}{12*12}) =9.890925575$^{\circ}C$

$T_f=T(24)=10-5*sin(\frac{\pi}{12*24}) =9.945459543$^{\circ}C$

Hence, the average temperature between noon and midnight is:

$T_A_v_g=\frac{9.890925575+9.945459543}{2}=9.918192559$^{\circ}C$

3 0

3 years ago

Railroad car A, with mass 6275 kg, is traveling east at 6.5 m/s along a straight track. It strikes railroad car B, with mass 515

LekaFEV [45]

The speed of A and B immediately after collision is 5.28m/s

<u>Explanation:</u>

Mass of A is 6275kg

Speed of A is 6.5m/s

Mass of B is 5155kg

Speed of B is 3.8m/s

Track is frictionless.

A and B stick together.

speed of attached A and B = ?

mₐsₐ + mᵇsᵇ = (mₐ + mb) s

$6275 X 6.5 + 5155 X 3.8 = ( 6275 + 5155) X s\\\\s = \frac{40787.5 + 19589}{11430}\\ \\s = \frac{60376.5}{11430}\\ \\s = 5.28m/s$

Therefore, The speed of A and B immediately after collision is 5.28m/s

4 0

3 years ago