Answer:
A) a = 73.304 rad/s²
B) Δθ = 3665.2 rad
Explanation:
A) From Newton's first equation of motion, we can say that;
a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.
Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s
Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s
We are given; t = 10 s
Thus;
a = 733.04/10
a = 73.304 rad/s²
B) From Newton's third equation of motion, we can say that;
ω² = ω_o² + 2aΔθ
Where Δθ is angular displacement
Making Δθ the subject;
Δθ = (ω² - ω_o²)/2a
At this point, ω = 0 rad/s while ω_o = 733.04 rad/s
Thus;
Δθ = (0² - 733.04²)/(2 × 73.304)
Δθ = -537347.6416/146.608
Δθ = - 3665.2 rad
We will take the absolute value.
Thus, Δθ = 3665.2 rad
Answer:
Explanation:
The work function of the metal corresponds to the minimum energy needed to extract a photoelectron from the metal. In this case, it is:
So, the energy of the incoming photon hitting on the metal must be at least equal to this value.
The energy of a photon is given by
where
h is the Planck's constant
c is the speed of light
is the wavelength of the photon
Using 
and solving for , we find the maximum wavelength of the radiation that will eject electrons from the metal:
And since
1 angstrom = 
The wavelength in angstroms is
Answer:
2.521 (A); 14.0924 (V)
Explanation:
more info in the attachment, the answers are marked with red colour.
What a relief ! That gives her time to step out of the way, before the ball
comes crashing down in the same place where she was standing.
