A track coach records how long it takes a sprinter to run 100 meters the track coach is recording what dimension of behavior
1 answer:
You might be interested in
Sally's average speed is <u>35.3 mi/h.</u>
Average speed of a body is the total distance traveled in the given time interval.
Express the distance d traveled in miles.
Express the time t traveled in hours.
Calculate the average speed v.
Her average speed is 35.3 mi/h, which is less than the speed limit of 65 mi/h.
However, the average speed of an object is different from its instantaneous speed. It could be possible that at the time when the officer apprehended her, Sally could have been travelling at a speed greater than the prescribed speed limit, which would have prompted the officer to issue a speeding ticket to her.
Thus, the average speed of a person cannot be considered as a bench mark for speeding offences, since her instantaneous speed could have been higher than the speeding limit and yet she could have had an average speed less than the speeding limit.
Answer:
a) F_net = 30.47 N , θ = 10.6º
b) Fₓ = 29.95 N
Explanation:
For this exercise we use coulomb's law
F₁₂ = k
the direction of the force is on the line between the two charges and the sense is repulsive if the charges are equal and attractive if the charges are different.
As we have several charges, the easiest way to solve the problem is to add the components of the force in each axis, see attached for a diagram of the forces
X axis
Fₓ =
Y axis
Fy =
let's find the magnitude of each force
= 9 10⁹ 2.76 10⁻⁴ 1.02 10⁻⁴ / 3²
F_{ab} = 2.82 10¹ N
F_{ab} = 28.2 N
= 9 10⁹ 6.54 10⁻⁴ 1.02 10⁻⁴ / 4²
F_{bc} = 3.75 10¹ N
F_{bc} = 37.5 N
let's use trigonometry to decompose this force
tan θ = y / x
θ = tan⁻¹ and x
θ= tan⁻¹ ¾
θ = 37º
let's break down the force
sin 37 = F_{bcy} / F_{bc}
F_{bcy} = F_{bc} sin 37
F_{bcy} = 37.5 sin 37
F_{bcy} = 22.57 N
cos 37 = F_{bcx} /F_{bc}
F_{bcx} = F_{bc} cos 37
F_{bcx} = 37.5 cos 37
F_{bcx} = 29.95 N
let's do the sum to find the net force
X axis
Fₓ = 29.95 N
Axis y
Fy = 28.2 -22.57
Fy = 5.63 N
we can give the result in two ways
a) F_net = Fₓ i ^ + j ^
F_net = 29.95 i ^ + 5.63 j ^
b) in the form of module and angle
let's use the Pythagorean theorem
F_net =
F_net = √(29.95² + 5.63²)
F_net = 30.47 N
we use trigonometry for the direction
tan θ=
θ = tan⁻¹ \frac{ F_{y} }{ F_{x} }
θ = tan⁻¹ (5.63 / 29.95)
θ = 10.6º
