Average <u>speed</u> = (distance covered) / (time to cover the distance) =
(5m) / (15 sec) =
(5/15) (m/s) = <em>1/3 m/s</em> .
Average <u>velocity</u> =
(displacement) / (time spent traveling) in the direction of the displacement
Average velocity = (5m) / (15 sec) left =
(5/15) / (m/sec) left =
<em>1/3 m/s left</em>.
A number without a direction is a speed, not a velocity.
The solution and answer are well written in the Pic above.
Answer:
mb = 3.75 kg
Explanation:
System of forces in balance
ΣFx =0
ΣFy = 0
Forces acting on the box
T₁ : Tension in string 1 ,at angle of 50° with the horizontal on the left
T₂ = 40 N : Tension in string 2, at angle of 75° with the horizontal on the right.
Wb :Weightt of the box (vertical downward)
x-y T₁ and T₂ components
T₁x= T₁cos50°
T₁y= T₁sin50°
T₂x= 30*cos75° = 7.76 N
T₂y= 30*sin75° = 28.98 N
Calculation of the Wb
ΣFx = 0
T₂x-T₁x = 0
T₂x=T₁x
7.76 = T₁cos50°
T₁ = 7.76 /cos50° = 12.07 N
ΣFy = 0
T₂y+T₁y-Wb = 0
28.98 + 12.07(cos50°) = Wb
Wb = 36.74 N
Calculation of the mb ( mass of the box)
Wb = mb* g
g: acceleration due to gravity = 9.8 m/s²
mb = Wb/g
mb = 36.74 /9.8
mb = 3.75 kg
Answer:
245.45km in a direction 21.45° west of north from city A
Explanation:
Let's place the origin of a coordinate system at city A.
The final position of the airplane is given by:
rf = ra + rb + rc where ra, rb and rc are the vectors of the relative displacements the airplane has made. If we separate this equation into its x and y coordinates:
rfX = raX+ rbX + rcX = 175*cos(30)-150*sin(20)-190 = -89.75km
rfY = raY + rbY + rcT = 175*sin(30)+150*cos(20) = 228.45km
The module of this position is:
And the angle measure from the y-axis is:
So the answer is 245.45km in a direction 21.45° west of north from city A