The three pictures you attached represent three different runs of the toy car.
Here's the data for the first run:
The change in time for the first quarter is <em>1.38</em> seconds.
The change in time for the second quarter is (2.19-1.38) = <em>0.81</em> seconds.
The change in time for the third quarter is (2.80 - 2.19) = <em>0.61</em> seconds.
The change in time for the fourth quarter is (3.31 - 2.80) = <em>0.51</em> seconds.
The quarters are all the same length, but the times are getting shorter.
The car must be getting faster !
Maybe gravity is pulling it down, do ya reckon ?
"Measuring the angle at which a gymnast’s body bends during an aerial flip" is the one data-gathering activity among the choices given in the question that would be useful for a biomechanical analysis. The correct option among all the options that are given in the question is the fourth option or the last option.
Hope this Helps! :)
Answer:
v = 
the speed in the two planes will be the same since it does not depend on the angle of the same
Explanation:
In this exercise we are told that the two inclined planes have no friction force, so we can apply the conservation of energy for each one, we will assume that the initial height in the two planes is the same
starting point. Highest part of each plane
Em₀ = U = m g h
final point. Lowest part of each plane
= K = ½ m v²
as there is no friction, the mechanical energy is preserved
Em₀ = Em_{f}
mg h = ½ m v²
v =
As we can see, the speed in the two planes will be the same since it does not depend on the angle of the same
Answer:
42000N
Explanation:
First you calculate how much it would contract, and secondly you then calculate the force to stretch it by that amount.
1) linear thermal expansion coef brass 19e-6 /K
∆L = αL∆T = (19e-6)(1.85)(110) = 0.00387 meter or 3.87 mm
Second part involves linear elasticity.
for brass, young's modulus is 15e6 psi or 100 GPa
cross-sectional area of rod is π(0.008)² = 0.0002 m²
F = EA∆L/L
F = (100e9)(0.0002)(0.00387) / (1.85)
F = 42000 or 42 kN
