In component form, the displacement vectors become
• 350 m [S] ==> (0, -350) m
• 400 m [E 20° N] ==> (400 cos(20°), 400 sin(20°)) m
(which I interpret to mean 20° north of east]
• 550 m [N 10° W] ==> (550 cos(100°), 550 sin(100°)) m
Then the student's total displacement is the sum of these:
(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m
≈ (280.371, 328.452) m
which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].
Answer:
A point on the outside rim will travel 157.2 meters during 30 seconds of rotation.
Explanation:
We can find the distance with the following equation since the acceleration is cero (the disk rotates at a constant rate):
Where:
v: is the tangential speed of the disk
t: is the time = 30 s
The tangential speed can be found as follows:
Where:
ω: is the angular speed = 100 rpm
r: is the radius = 50 cm = 0.50 m
Now, the distance traveled by the disk is:
Therefore, a point on the outside rim will travel 157.2 meters during 30 seconds of rotation.
I hope it helps you!
A) the forces are acting in the same direction.. B) Together, forces are acting in opposite directions
Answer:
A) 80 N
B) 20 N
Explanation:
A) If the forces acting are in the same direction, then the net force will be a sum of both so many faces..
Thus;
ΣF = 50 + 30
ΣF = 80 N
B) If the forces are acting in the in opposite directions with the larger force pointing in the positive y-axis then, the net force is;
ΣF = 50 - 30
ΣF = 20 N
Given a = 10 cm/s²
u = 0 cm/s
v = 50 cm/s
we know that
v²=u²+2aS
2500=2×10×S
2500÷20 = S
S= 125 cm
The ramp is 125 cm
