Answer:
Technician A
Explanation:
It is seen that a tire pressure will increase or decrease 1 psi for each 
change in temperature.
For Technician B vehicle pressure should not be adjusted after tire has been warmed as the warm air may increase the pressure but it will be auto adjusted as the temperature falls to normal .
The option is Work.
The product of charge and potential is equal to the energy. Adn, as we know work is related to energy as the capacity to do work.
Alos, because, Potential is given as, V = E/q
or E = Vq
Thus, t<span>he product of charge through, and potential across, an electrical device is:work
</span>
Answer:
strong winds that blow for a long time over a great distance
weak winds that blow for short periods of time with a short fetch
Explanation:
When the winds are weak and blow for short periods, we experience the smallest ocean waves but when there are strong winds over a longer duration, the largest ocean waves are seen. Therefore, the conditions to produce the smallest and largest ocean waves are strong winds that blow for a long time over a great distance and weak winds that blow for short periods of time with a short fetch.
Answer: 846°C
Explanation:
The quantity of Heat Energy (Q) required to heat bismuth depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Given that:
Q = 423 joules
Mass of bismuth = 4.06g
C = 0.123 J/(g°C)
Φ = ?
Then, Q = MCΦ
423 J = 4.06g x 0.123 J/(g°C) x Φ
423 J = 0.5J/°C x Φ
Φ = (423J/ 0.5g°C)
Φ = 846°C
Thus, the change in temperature of the sample is 846°C
Answer:
A λ = 97.23 nm
, B) λ = 486.2 nm
, C) λ = 53326 nm
Explanation:
With that problem let's use the Bohr model equation for the hydrogen atom
= -k e² /2a₀ 1/n²
For a transition between two states we have
- 
= -k e² /2a₀ (1/ 
² - 1 / n₀²)
Now this energy is given by the Planck equation
E = h f
And the speed of light is
c = λ f
Let's replace
h c / λ = - k e² /2a₀ (1 / ² - 1 / no₀²)
1 / λ = - k e² /2a₀ hc (1 / ² -1 / n₀²)
Where the constants are the Rydberg constant 
= 1.097 10⁷ m⁻¹
1 / λ = (1 / n₀² - 1 / nf²)
Now we can substitute the given values
Part A
Initial state n₀ = 1 to the final state = 4
1 / λ = 1.097 10⁷ (1/1 - 1/4²)
1 / λ = 1.0284 10⁷ m⁻¹
λ = 9.723 10⁻⁸ m
We reduce to nm
λ = 9.723 10⁻⁸ m (10⁹ nm / 1m)
λ = 97.23 nm
Part B
Initial state n₀ = 2 final state = 4
1 / λ = 1.097 10⁷ (1/2² - 1/4²)
1 / λ = 0.2056 10⁻⁷ m
λ = 486.2 nm
Part C
Initial state n₀ = 3
1 / λ = 1,097 10⁷ (1/3² - 1/4²)
1 / λ = 5.3326 10⁵ m⁻¹
λ = 5.3326 10-5 m
λ = 53326 nm
