The answer is Weathering since its nature breaking down the concrete.
Answer:
The average force ≅ 519.44 N.
Explanation:
Impulse = change in momentum of a body
i.e Ft = m(v - u)
where F is the force, t is the time, m is the mass of the body, v is the final velocity and u is the initial velocity.
m = 55.0 g (0.055 Kg), t = 0.00360 s, v = 34.0 m/s, since the ball was initially at rest; u = 0 m/s
So that,
F x 0.00360 = 0.055(34 - 0)
F x 0.00360 = 0.055 x 34
= 1.87
F = 
= 519.4444
The average force exerted on the ball by the club is approximately 519.44 N.
Answer:
Because of the location, humidity and temperatures.
Explanation:
Coca is grown in humid and very humid subtropical forests, called yungas and
they form the lower floor of the upper Jungle, in the Central Andes, mostly in Peru and Bolivia. The yungas are in contact with the rainforests of the lowlands in Amazonia, where it has been started to expand coca cultivation recently (Dourojeanni, 1988). The optimum altitude is 1000 a 2000 meters (where cocaine content is higher), with optimal annual average precipitation, is 2000 meters mm, but it is grown between 700 and 2000 msnm and with an average annual rainfall of 1000 to 4200 mm.
msnm = meters above sea level
Answer:
Explanation:
The formula for this, the easy one, is
where No is the initial amount of the element, t is the time in years, and H is the half life. Filling in:
and simplifying a bit:
and
N = 48.0(.0625) so
N = 3 mg left after 12.3 years
Answer:
F = 63N
Explanation:
M= 1.5kg , t= 2s, r = (2t + 10)m and
Θ = (1.5t² - 6t).
magnitude of the resultant force acting on 1.5kg = ?
Force acting on the mass =
∑Fr =MAr
Fr = m(∇r² - rθ²) ..........equation (i)
∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)
The horizontal path is defined as
r = (2t + 10)
dr/dt = 2, d²r/dt² = 0
Angle Θ is defined by
θ = (1.5t² - 6t)
dθ/dt = 3t, d²θ/dt² = 3
at t = 2
r = (2t + 10) = (2*(2) +10) = 14
but dr/dt = 2m/s and d²r/dt² = 0m/s
θ = (1.5(2)² - 6(2) ) = -6rads
dθ/dt =3(2) - 6 = 0rads
d²θ/dt = 3rad/s²
substituting equation i into equation ii,
Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)
∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]
∑F = 1.5(14*3+0) = 63N
F = √(Fr² +FΘ²) = √(0² + 63²) = 63N