Answer:
Explanation:
a)Magnitude = 
84=
x= +50.67 or -50.67 units
b) We are given that the resultant is entirely in the -ve x direction which means that the y-component of the resultant is 0; It means that the y-component of the next vector = -ve of the y component of the initial vector i.e 67.
To make the magnitude 80 units in the negative x direction where the y component is 0, the x component must be -130.67(-50.67 - 80) as the x component is + 50.67units.
Magnitude = 
= 146.85 units
c) The direction vector = 67/146.85 i - 130.67/146.85 j where i corresponds to the vector in y direction and j corresponds to the vector in x direction. Or this vector is at an angle of 180 - 
i.e 152.85 degrees from the +ve x-axis.
It's on your exam boards specification or just google it. the foundation paper will have the same equations as higher, but higher just has more.
A ) v = v o - a t
0 = 22 - a · t
a · t = 22
d = v o · t - a t²/2
0.04 = 22 t - 22 t / 2
0.04 = 11 t
t = 0.04 : 11 = 0.003636 s
a = 22 / t
a = 6050 m/s²
F = m · a = 0.09 kg · 6050 m/s²
F ( target→arrow) = - 544.5 N
b ) F ( arrow→target ) = 544.5 N
c ) If the speed was doubled: v = 44 m/s;
F = a m
a = 6050 m/s²
a · t = 44
t = 6050 : 0.04
t = 0.007272 s
d = 44 t - 44 t/2 = 22 t
d = 22 · 0.007272
d = 0.16 m = 16 cm
Answer:
I think maybe C. They require a large amount of land disturbance and potential displacement of people
Explanation:
Hope this helps.
Answer:
Option b
Solution:
As per the question:
Speed of the mouse, v = 1.3 m/s
Speed of the cat, v' = 2.5 m/s
Angle, 
Now,
To calculate the distance between the mouse and the cat:
The distance that the cat moved is given by:
The position of the cat and the mouse can be given by:
x' = 0.67 t (1)
The initial speed of the cat ahead of the mouse:
u = 
When the time is 0.5t, the speed of the cat is 0, thus:
Substituting the value of t in eqn (1):
x' = 0.67(0.314) = 0.210 m
Thus the distance comes out to be 0.210 m
