Answer with Step-by -step explanation:
We are given that
b.
below the positive x-axis
Therefore, the angle made by vector A in counter clockwise direction when measure from positive x-axis=
x-component of vector A=
y-Component of vector A=
Magnitude of vector B=86 m
The vector B makes angle with positive x- axis=
x-component of vector B=
y-Component of vector B=
Vector A=
Vector B=
Vector C=A+B
Substitute the values
c.Direction=
The direction of the vector C=21.5 degree
In order to answer this exercise you need to use the formulas
S = Vo*t + (1/2)*a*t^2
Vf = Vo + at
The data will be given as
Vf = final velocity = ?
Vo = initial velocity = 1.4 m/s
a = acceleration = 0.20 m/s^2
s = displacement = 100m
And now you do the following:
100 = 1.4t + (1/2)*0.2*t^2
t = 25.388s
and
Vf = 1.4 + 0.2(25.388)
Vf = 6.5 m/s
So the answer you are looking for is 6.5 m/s
The magnitude of vector b is 8.58 Unit.
Since both the vectors a and b are perpendicular to each other, so we can apply the Pythagoras theorem to calculate the magnitude of the vector b.
Applying the Pythagoras theorem
(a-b)^2=a^2+b^2
15^2=12.3^2-b^2
b=8.58 unit
Therefor the magnitude of the vector b is 8.58 unit.
