Answer:
That is true. They share atoms with each other. I hope this helps. Comment if you have any question
Explanation:
We apply the following equation
T = 2π * sqrt (L/g)
Where g is the gravity = 9.8 m/s^2
L is the longitude of the pendulum (Height of the tower)
T is the period. (T = 18s)
We find L.............> (T /2π)^2 = L/g
L = g*(T /2π)^2...........> L = 80.428 meters
Answer:
1,1 m
Explanation:
Dado que;
coeficiente de fricción = 0,6
sabemos que W = R = mgcos 37 = 3.5Kg * 10m / s ^ 2 * cos37 = 27.95 N
coeficiente de fricción = fuerza / reacción normal (R)
Fuerza = 0.6 * 27.95 N
Fuerza (F) = 16.77 N
Recuerda que F = Ke
dónde;
K = constante de fuerza (15N / m)
e = extensión (lo desconocido)
e = F / K
e = 16,77 N / 15 N / m
e = 1,1 m
Answer:
a) v = 54.7m/s
b) v = (58 - 1.66a) m/s
c) t = 69.9 s
d) v = -58.0 m/s
Explanation:
Given;
The height equation of the arrow;
H = 58t - 0.83t^2
(a) Find the velocity of the arrow after two seconds. m/s;
The velocity of the arrow v can be given as dH/dt, the change in height per unit time.
v = dH/dt = 58 - 2(0.83t) ......1
At t = 2 seconds
v = dH/dt = 58 - 2(0.83×2)
v = 54.7m/s
(b) Find the velocity of the arrow when t = a. m/s
Substituting t = a into equation 1
v = 58 - 2(0.83×a)
v = (58 - 1.66a) m/s
(c) When will the arrow hit the surface? (Round your answer to one decimal place.) t = s
the time when H = 0
Substituting H = 0, we have;
H = 58t - 0.83t^2 = 0
0.83t^2 = 58t
0.83t = 58
t = 58/0.83
t = 69.9 s
(d) With what velocity will the arrow hit the surface? m/s
from equation 1;
v = dH/dt = 58 - 2(0.83t)
Substituting t = 69.9s
v = 58 - 2(0.83×69.9)
v = -58.0 m/s
I dunno, I guess to see if you could get a different answer
