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(a) The magnitude of the acceleration of the electron is 5.62 x 10¹³ m/s².
(b) The speed of the electron after the given time is 4.78 x 10⁵ m/s.
<h3>
Acceleration of the electron</h3>
The acceleration of the electron is calculated as follows;
F = qE
ma = qE
a = qE/m
a = (1.6 x 10⁻¹⁹ x 320)/(9.11 x 10⁻³¹)
a = 5.62 x 10¹³ m/s²
<h3>Speed of the electron</h3>
v = at
v = 5.62 x 10¹³ m/s² x 8.50 x 10⁻⁹ s
v = 4.78 x 10⁵ m/s
Learn more about speed here: brainly.com/question/4931057
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Answer: Equal number of protons and electrons. Example: an atom of oxygen atom has 8 electrons and 8 protons and is neutral.
Explanation:
An atom that has no charge is a neutral atom. It contains electrons equal to protons. For example: A neutral atom of oxygen has 8 protons and 8 electrons.
An atom which has charge is said be ionized. It is either positively charged or negatively charged. It is positively charge when the number of electrons is less than the number of protons. For example:
contains 10 electrons and 11 protons.
And when the number of electrons is greater than the number of protons, the atom is negatively charged. For example,
has 17 protons and 18 electrons. It readily accepts an electron to complete its octet.
Answer: In this lab we wanted to know how motion can be described. So the hypothesis is if the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will increase because the toy car will have a greater acceleration. My prediction is that cars travel faster on higher tracts. So the heighten the track was intentionally manipulated. So it is the independent variable the speed of the car is the dependent variable. The speed at the first quarter checkpoint is 1.09 m/s. The speed at the second quarter checkpoint is 1.95 m/s. The speed at the third quarter checkpoint is 2.373.36 m/s. The speed at the finish line is 2.803.00 m/s. The average speed increases as the height increases.
The cars on the higher track travel farther than the cars on the lower track, in the same time.
This means that the cars on the higher track have a greater average speed than those on the lower track. This is demonstrated by the
slope of the higher track line being greater than the slope of the lower track line.
Explanation: put it in notes then send it to files to compress it to submit it.
Answer:
a = 1 m/s² and
Explanation:
The first two parts can be seen in attachment
We use Newton's second law on each axis
Y axis
Ty - W = 0
Ty = w
X axis
Tx = m a
With trigonometry we find the components of tension
Sin θ = Ty / T
Ty = T sin θ
Cos θ = Tx / T
Tx = T cos θ
We calculate the acceleration with kinematics
Vf = Vo + a t
a = (Vf -Vo) / t
a = (20 -10) / 10
a = 1 m/s²
We substitute in Newton's equations
T Sin θ = mg
T cos θ = ma
We divide the two equations
Tan θ = g / a
θ = tan⁻¹ (g / a)
θ = tan⁻¹ (9.8 / 1)
θ = 84º
We see that in the expression of the angle the mass does not appear therefore you should not change the angle