D. Electrons are shared between the bromine atoms and carbon atoms
Answer:
The solution and the explanation are in the Explanation section.
Explanation:
According to the diagram that is in the attached image, the EFFORT force at point A and the load is at O point. The torque due to weight is:
TA = W * (a * cosθ)
The torque due to effort at C point is:
TC = E * (b * cosθ)
The net torque is equal to 0, we have:
Tnet = 0
W * (a * cosθ) - E * (b * cosθ) = 0

From the figure, you can observe that a/b < 1, thus E < W
If an electron, a proton, and a deuteron move in a magnetic field with the same momentum perpendicularly, the ratio of the radii of their circular paths will be:
<h3>How is the ratio of the perpendicular parts obtained?</h3>
To obtain the ratio of the perpendicular parts, one begins bdy noting that the mass of the proton = 1m, the mass of deuteron = 2m, and the mass of the alpha particle = 4m.
The ratio of the radii of the parts can be obtained by finding the root of the masses and dividing this by the charge. When the coefficients are substituted into the formula, we will have:
r = √m/e : √2m/e : √4m/2e
When resolved, the resulting ratios will be:
1: √2 : 1
Learn more about the radii of their circular paths here:
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Answer:
31.905 ft/s²
Explanation:
Given that
Mass of the pilot, m = 120 lb
Weight of the pilot, w = 119 lbf
Acceleration due to gravity, g = 32.05 ft/s²
Local acceleration of gravity of found by using the relation
Weight in lbf = Mass in lb * (local acceleration/32.174 lbft/s²)
119 = 120 * a/32. 174
119 * 32.174 = 120a
a = 3828.706 / 120
a = 31.905 ft/s²
Therefore, the local acceleration due to gravity at that elevation is 31.905 ft/s²