(Mass does not affect the pendulum's swing. The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period.)
Among the choices above, the one
that is most closely related to an activated complex is the transition state. The
answer is letter D. This formation forms quickly and does not stay in a way
compound is. It usually forms during the enzyme – substrate reaction.
Answer:
you should write about a book you read
Explanation:
because maybe you got really good things in it
or here is an example
Answer:
0.37 m
Explanation:
Let the shoulder be the origin.
The center of mass of the arm bones is 0.60 m/2 = 0.30 m and the center of mass of the hand bones is 0.10 m/2 = 0.05 m since they are modeled as straight rods with uniform density and the center of mass of a rod is x = L/2 where L is the length of the rod.
The center of mass y = (m₁y₁ + m₂y₂)/(m₁ + m₂) where m₁ = mass of arm bones = 4.0 kg, y₁ = distance center of mass of arm bones from shoulder = 0.30 m, m₂ = mass of hand bones = 1.0 kg and y₂ = distance of center of mass hand bones from shoulder = x₁ + distance of center of hand bones from wrist = 0.60 m + 0.05 m = 0.65 m
Substituting these into the equation for the center of mass, we have
y = (m₁y₁ + m₂y₂)/(m₁ + m₂)
y = (4.0 kg × 0.30 m + 1.0 kg × 0.65 m)/(4.0 kg + 1.0 kg)
y = (1.20 kgm + 0.65 kgm)/5.0 kg
y = 1.85 kgm/5.0 kg
y = 0.37 m
The distance of the center of mass from the shoulder is thus y = 0.37 m
1. Vpa = 180m/s. @ 0 deg.
Vag = 40m/s @ 120 deg,CCW.
<span>
Vpg = Vpa + Vag,
Vpg = (180 + 40cos120) + i40sin120,
Vpg = 160 + i34.64,
Vpg=sqrt((160)^2 + (34.64)^2)=163.7m/s.
</span>
<span>2. tanA = Y / X = 34.64 / 160 = 0.2165,
A = 12.2 deg,CCW. = 12.2deg. North of
East. </span>
3. 1 hr = 3600s. <span>d = Vt = 163.7m/s * 3600s = 589,320m.
hope this helps</span>
