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
They are said to be directly related.
a) directly related.
This is Charles' Law.
Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.
At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .
At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .
At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .
Answer:
Pitcher is accelerating the ball at 30 times of acceleration due to gravity = 294 m/s²
Explanation:
Force applied on baseball = 30 times weight of the ball.
Weight of ball = mg, where m is the mass of ball and g is acceleration due to gravity value.
We have force applied is also equal to product of mass and acceleration.
F = ma = 30 x mg
a = 30g
So, pitcher is accelerating the ball at 30 times of acceleration due to gravity = 294 m/s²
That was Tycho Brahe, and I thought it was actually more years than that.
