Answer:
θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²
Explanation:
This is an angular kinematic exercise the equation for the angular position
the particle A
θ = θ₀ + ω₀ t + ½ α t²
They say for the particle B
w₀B = ½ w₀
αB = 2 α
In addition, the particle begins at a time t_1 after particle A, in order to use the same timer, we must subtract this time from the initial
t´ = t - t_1
l
et's write the equation of particle B
θ = θ₀ + w₀B t´ + ½ αB t´2
replace
θ = θ₀ + ½ w₀ (t -t_1) + ½ 2α (t -t_1)²
θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²
Work needed: 720 J
Explanation:
The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by
where
k is the spring constant
x is the stretching of the spring from the equilibrium position
In this problem, we have
E = 90 J (work done to stretch the spring)
x = 0.2 m (stretching)
Therefore, the spring constant is
Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of
x = 0.2 + 0.4 = 0.6 m
Substituting,
Therefore, the additional work needed is
Learn more about work:
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Answer:
6ms^-1
Explanation:
Given that the frequency difference is
( 563- 544) = 19
So alsoThe wavelength of each wave is = v/f = 344 /544
and there are 19 of this waves
So it is assumed that each motorcycle has moved 0.5 of this distance
in one second thus the speed of the motorcycles will be
=> 19/2 x 344/544 = 6.0 m/s
it is transfred through the basics
The answer is C. The mass of the platinum sample is greater than the mass of the lead sample. As I explained in a previous answer, if they are the same volume, but one is heavier, then it must be more dense. In this particular example, the platinum is more dense than the lead, and therefore has more mass.
