The problem is missing some parts but nevertheless here is
the answer:
Given:
1.7 m long barbell with 23 kg weight on the left and 34 in
the right end.
Solution:
Calculating from the left side:
Center of gravity = ΣM/ΣW = [23*0 + 9*(1.7/2) + 34*1.7]/[23+9+34]
Center of gravity = .992 m from the left end
Question: Initially, the car travels along a straight road with a speed of 35 m/s. If the brakes are applied and the speed of the car is reduced to 13 m/s in 17 s, determine the constant deceleration of the car.
Answer:
1.29 m/s²
Explanation:
From the question,
a = (v-u)/t............................ Equation 1
Where a = deceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.
Given: v = 13 m/s, u = 35 m/s, t = 17 s.
a = (13-35)/17
a = -22/17
a = -1.29 m/s²
Hence the deceleration of the car is 1.29 m/s²
Answer:
It looks yellow because that is the only (major) color reflected.
Visible spectra is from about 4000-7000 Angstroms (10^-10 m).
Red are longer wavelengths and blue are the shorter wavelengths.
The Sodium doublet (yellow) occurs around 5900 Angstroms.
Answer: Yes
Explanation:
A ruber band can be described by his mas, M, the spring constant K, and his rest radius, wich is the radius of the circular rubber band when there is no force applied to it, so yes, a uniform thin circular rubber band of mass M and spring constant k has an original radius R
