Answer:
F = - K x
a) K = 1.3 kg * 9.8 m/s^2 / .096 m = 133 kg/sec^2
b) ω = (K/m)^1/2 angular frequency of SHM
ω = (133 / 1.3)^1/2 = 10.1 / sec
f = 2 π ω = 6.28 * 10.1 / sec = 63.5 / sec
P = 1/f = .0157 sec
Answer: V = 504m/a
F = 4N
Explanation: please find the attached file for the solution
Answer:
First one, third one, and fourth one
Let's ask this question step by step:
Part A)
a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
ab = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
Part (c)
(a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
(a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
(a + b) b = 10 + 36
(a + b) b = 46
Part (d)
comp (ba) = (a.b) / lbl
a.b = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
lbl = root ((2.0) ^ 2 + (4.0) ^ 2) = root (20)
comp (ba) = 26 / root (20)
answer
2k
26
46
26 / root (20)
Assume the snow is uniform, and horizontal.
Given:
coefficient of kinetic friction = 0.10 = muK
weight of sled = 48 N
weight of rider = 660 N
normal force on of sled with rider = 48+660 N = 708 N = N
Force required to maintain a uniform speed
= coefficient of kinetic friction * normal force
= muK * N
= 0.10 * 708 N
=70.8 N
Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.
