I am pretty sure that when you travel the first mile in 10 minutes. The last mile takes you 15 minutes. This is an example of negative acceleration. I consider this to be correct because <span>the second mile was slower. Hope you will agree with me. Regards!</span>
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:
9.47 rad/s^2
Explanation:
Diameter = 15 cm, radius, r = diameter / 2 = 7.5 cm = 0.075 m, u = 0, v = 7.1 m/s,
s = 35.4 m
let a be the linear acceleration.
Use III equation of motion.
v^2 = u^2 + 2 a s
7.1 x 7.1 = 0 + 2 x a x 35.4
a = 0.71 m/s^2
Now the relation between linear acceleration and angular acceleration is
a = r x α
where, α is angular acceleration
α = 0.71 / 0.075 = 9.47 rad/s^2
Answer:
B. 1500 kg*m/s
Explanation:
Momentum p = m* v
In any type of collision, the total momentum is preserved!
The total momentum before and the total momentum after the collision is the same. We know the mass and speed after the collision so we can calculate the total momentum.
p1 + p2 =
m1*v1 + m2*v2
m1 = me = 300 kg
v1 = 3 m/s
v2 = 2 m/s
Substitute the given numbers:
300*3 + 300+2
900 + 600
1500 kg*m/s, which is answer B.
