Answer:
Number of revolutions = Δθ = 60.63 revs
Explanation:
The angular acceleration of the washer is given by
α = (ωf - ωi)/Δt
Where ωf is the initial angular speed, ωi is the final angular speed and Δt is the interval of time during this acceleration.
α = (4.7 - 0)/8.3
α = 0.56 rev/s²
As we know from the equation of kinematics,
2αΔθ = ωf² - ωi²
Where Δθ is the change in angular displacement of the washer.
Δθ = (ωf² - ωi²)/2α
Δθ = (4.7² - 0²)/2*0.56
Δθ = 19.72 revs
Now the tub decelerates and comes to a stop in 17.5 s
α = (ωf - ωi)/Δt
α = (-4.7 - 0)/17.5
α = -0.27 rev/s²
the corresponding change in angular displacement of the washer is
Δθ = (ωf² - ωi²)/2α
Δθ = (0² - 4.7²)/2*-0.27
Δθ = 40.91 revs
Therefore, the total number of revolutions undergone by the tub during this entire time interval is
Δθ = 19.72 + 40.91
Δθ = 60.63 revs
These tools enable the specific execution of a task or a group of tasks allowing the fulfillment of specific objectives within different stages of product development
The pressure is 5000 Pa
Explanation:
The pressure exerted by a force on a surface is given by the equation:
where
p is the pressure (measured in Pascal, Pa)
F is the magnitude of the force
A is the area of the surface on which the force is exerted
In this problem, we have:
F = 2 N
Substituting into the equation, we find the pressure:
Learn more about pressure:
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Answer: 7.175*10^18
Explanation:
Energy of a wave:
E = nhc/λ, where
E = energy of the wave
n = no of photons
h = Planck's Constant
c = speed of light
λ = wavelength of the wave
4000 = (n * 6.63*10^-34 * 3*10^8) / (510*10^-9)
4000 = 1.989*10^-25 * n / 510*10^-9
4000 = 3.9*10^-19 * n
n = 1.025*10^22 photons per second per meter²
7.0 cm² = 7.0 / 10,000 m²
= 7 x 10⁻⁴
Photons per second = 1.025*10^22 * 7 x 10⁻⁴
= 7.175*10^18 photons per second
This question is incomplete. The complete question is given below:
Question 3 Both the angle and the magnitude of the force have a certain uncertainty: εF = 28 N and εθ = 0.8°. Using the propagation methods described in the video you watched at the beginning of this prelab, calculate the corresponding propagated uncertainty for Fx, in N. For this question, round up your final answer to two significant figures. Do not include the ± sign in your answer. Example: If the x component of F is 200±14 N, you should enter “14”.
Both the force and the angle are measured, and the results are quoted as a central value plus/minus an uncertainty:
F = F0 ± εF
θ = θ0 ± εθ
We would like to evaluate the component of the force in the x direction.
Question 2
Let us first concentrate on the central value. Take F0 = 325 N and θ0 = 57°.
The answer & explanation for this question is given in the attachment below.