Just find the energy of the <span>blueviolet light with a wavelength of 434.0 nm using the formula:
E = hc / lambda
E = energy
c= speed of light = 3 x 10^8 m/s
h = planck's constant = 6.6 x 10^{-34} m^2 kg / s
lambda = 434 nm = 434 x 10^{-9} m
Putting these values (with appropriate units) in the above formula :
we get: Energy, E = 4.5 x 10^{-19} J
E = 0.45 x 10^{-18} J
Now, the </span>minimum energy is 2.18×10^-{18} J but our energy is 0.45 x 10^{-18} J which is less.
<span>Means the electron will not be removed
</span>
The alignments of the planets would be the correct answer.<span />
<h3>
Answer:</h3>
83.33 seconds.
<h3>
Explanation:</h3>
<u>We are given;</u>
- Take off velocity as 300 km/hr
- Acceleration as 1 m/s²
We are required to calculate the take off time of the airplane.
<h3>Step 1: Convert velocity from km/hr to m/s </h3>
We are going to use the conversion factor.
The conversion factor is 3.6 km/hr per m/s
Therefore;
Velocity = 300 km/hr ÷ 3.6 km/hr per m/s
= 83.33 m/s
<h3>Step 2: Calculate the take off time</h3>
We know that;
v = u + at
where, u is the initial velocity, v the final velocity, a the acceleration and t is time.
But, initial velocity is Zero
Therefore;
83.33 m/s = 1 m/s² × t
Thus;
time = 83.33 m/s ÷ 1 m/s²
= 83.33 seconds
Therefore, the take off time is 83.33 seconds.
Answer:
F=Air
Explanation:
Because I took the test hope it helps