Answer: 1.39 s
Explanation:
We can solve this problem with the following equations:
(1)
(2)
Where:
is the length the steel wire streches (taking into account 1mm=0.001 m)
is the length of the steel wire before being streched
is the force due gravity (the weight) acting on the pendulum with mass 
is the transversal area of the wire
is the Young modulus for steel
is the period of the pendulum
is the acceleration due gravity
Knowing this, let's begin by finding :
(3)
Where 
is the diameter of the wire
(4)
(5)
Knowing this area we can isolate from (1):
(6)
And substitute in (2):
(7)
(8)
Finally:
Answer:
81.6 m
Explanation:
Answer: 81.6 m.
The time it takes gravity to slow 40 m/s to zero when it teaches maximum height is
-v(initial) / -g = t
-40 m/s / -9.8 m/s^2 = 4.08 s
The height reached is the average velocity times this time 4.08 s, with v(avg) = [v(initial) + v(final)] / 2 with v(final) = 0. v(avg) = v(initial) / 2 = 40 m/s / 2 = 20 m/s.
So the distance d of maximum height is
d = v(avg)•t
d = 20 m/s • 4.08 s = 81.6 m.
Rutherford's experiment<span> utilized positively charged alpha particles (He with a +2 charge) which were deflected by the dense inner mass (nucleus). The conclusion that could be formed from this result was that </span>atoms<span> had an inner core which contained most of the mass of an </span>atom<span> and was positively charged.</span>
RDA stands for Recommended Daily Allowance. To determine the amount needed of a certain adult per day, we simply multiply the mass of the adult to the value of RDA. For this case, we do as follows:
Daily requirement = 12 mg/kg ( 78 kg ) = 936 mg of lysine = 936000 g
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
