Answer:
q = 8.57 10⁻⁵ mC
Explanation:
For this exercise let's use Newton's second law
F = ma
where force is magnetic force
F = q v x B
the bold are vectors, if we write the module of this expression we have
F = qv B sin θ
as the particle moves perpendicular to the field, the angle is θ= 90º
F = q vB
the acceleration of the particle is centripetal
a = v² / r
we substitute
qvB = m v² / r
qBr = m v
q =
The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use
v = d / t
the distance is ¼ of the circle,
d =
d =
we substitute
v =
r =
let's calculate
r =
2 2.2 10-3 88 /πpi
r = 123.25 m
let's substitute the values
q =
7.2 10-8 88 / 0.6 123.25
q = 8.57 10⁻⁸ C
Let's reduce to mC
q = 8.57 10⁻⁸ C (10³ mC / 1C)
q = 8.57 10⁻⁵ mC
Answer:
a. a=33.34ms⁻², V=164.4m/s
Explanation:
Since the dragster started with zero velocity, de determine the acceleration using of the equations of motion.
Below are the data given
Distance, s=404.5m,
time taken,t=4.922secs
Using the equation
S=ut+1/2at²
where u is the initial velocity and u=0
Making the acceleration the subject of the formula, we arrive at
a=2s/t²
a=(2*404.5)/4.922²
a=33.34ms⁻².
To determine the velocity, we use
V=u+at
V=0+33.34ms⁻² *4.922sec
V=164.4m/s
93.5 if it’s wrong sorry sis I need my homework done too
<span>An atom’s emission of light with a specific amount of energy confirms that </span><span>electrons emit and absorb energy based on their position around the nucleus.
The light emitted from an electron is a result of the electron's quantum jumps/leaps ( atomic electron transitions ) to and from different energy levels.</span>