Data given:
V=45m/s
t=11s
Δx=?
Formula needed:
V=Δx/t
Solution:
Δx=v×t
Δx=45m/s×11s
Δx=495m
According to my solution B) is the most accurate
Answer:
At the highest point the velocity is zero, the acceleration is directed downward.
Explanation:
This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.
I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.
At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.
Answer:
RMS voltage is 113.1370 V
frequency is 780.685 Hz
voltage is −158.66942 V
maximum current is 2.9739 A
Explanation:
Given data
∆V = 160.0 sin(495t) Volts
so Vmax = 160
and angular frequency = 495
time t = 1/106 s
resistor R = 53.8 Ω
to find out
RMS voltage and frequency of the source and voltage and maximum current
solution
we know voltage equation = Vmax sin ωt
here Vmax is 160 as given equation in question
so RMS will be Vmax / √2
RMS voltage = 160/ √2
RMS voltage is 113.1370 V
and frequency = angular frequency / 2π
so frequency = 497 / 2π
frequency is 780.685 Hz
voltage at time (1/106) s
V(t) = 160.0 sin(495/ 108)
voltage = −158.66942 V
so current from ohm law at resistor R 53.8 Ω
maximum current = voltage max / resistor
maximum current = 160 / 53.8
maximum current = 2.9739 A
The root-mean-square voltage of the AC source is 113.14 V, its frequency is 78.75 Hz, and the voltage at time t = 1/106 s is approximately 150.4 V. The current at this peak voltage, when connected to a resistor of 53.8 Ω, is approximately 2.97 A.
The output of an AC voltage source can be represented by the equation V = V₀ sin ωt, where V₀ is the peak voltage, ω is the angular frequency, and t is the time. In this case, V₀ = 160 V and ω = 495 (1/s). The root-mean-square voltage (Vrms), which is commonly used to express AC voltage, can be calculated from the peak voltage using the formula Vrms = V₀/√2 which gives approximately 113.14 V.
The frequency of the source is related to the angular frequency by the equation f = ω/2π, which gives a frequency of approximately 78.75 Hz. To find the voltage at a specific time t = 1/106 s, we substitute these values into the initial equation resulting in V = V₀ sin ωt = approximately 150.4 V.
Finally, the resistance R = 53.8 Ω allows us to calculate the maximum current in the circuit given by I = V/R. The maximum current occurs at the peak voltage, so I(max) = V₀/R = approximately 2.97 A.
#SPJ3
(a) What is the greatest wavelength of light that can be absorbed by the material?
(b) In what region of the electromagnetic spectrum is this wavelength located?
Answer:
a)
b) infrared region
Explanation:
Photon energy is the "energy carried by a single photon. This amount of energy is directly proportional to the photon's electromagnetic frequency and is inversely proportional to the wavelength. If we have higher the photon's frequency then we have higher its energy. Equivalently, with longer the photon's wavelength, we have lower energy".
Part a
Is provide that the smallest amount of energy that is needed to dissociate a molecule of a material on this case 0.42eV. We know that the energy of the photon is equal to:
Where h is the Planck's Constant. By the other hand the know that and if we solve for f we have:
If we replace the last equation into the E formula we got:
And if we solve for we got:
Using the value of the constant we have this:
Part b
If we see the figure attached, with the red arrow, the value for the wavelenght obtained from part a) is on the infrared region, since is in the order of
B.) perception.
C.) similarity.
D.) continuity.
Answer:
C
Explanation:
Similarity
What will his angular velocity be (in rpm) when he pulls in his arms until they are at his sides parallel to his trunk?
To find the final angular velocity when the skater pulls in his arms, we use the conservation of angular momentum.
To find the final angular velocity when the skater pulls in his arms, we can make use of the conservation of angular momentum. Initially, the skater's arms are outstretched, and the moment of inertia can be calculated using the parallel axis theorem. After the skater pulls in his arms, we can calculate the new moment of inertia using the same theorem. Equating the initial and final angular momentum values, we can solve for the final angular velocity.
#SPJ12
The problem involves the concept of conservation of angular momentum. The skater's spinning speed will increase when they pull their arms in. For a precise value of the final velocity, a complex calculation taking into account body mass distribution is needed.
This question involves the principle of conservation of angular momentum, which states that the angular momentum of an object remains constant as long as no external torques act on it. The total initial angular momentum of the skater spinning with outstretched arms is equal to his final angular momentum when he pulls his arms in.
Calculating the skater's initial and final angular momentum, you can then solve for his final velocity.
However, note that the calculation needs to take into account the skater's mass distribution. Specifically, we need to consider the percentage distributions for the arms/hands (13%), head (7%) and trunk/legs (80%), and integrate these over the skater's body.
This can result in a significantly complex calculation if done accurately, involving calculus level mathematics. However, using the qualitative knowledge that the skater's spinning speed will increase when they pull their arms in, it's reasonable to estimate, considering the mass distribution, the final velocity will be somewhere near 2 to 3 times the original rpm. But for an exact value, a detailed and complex calculation is needed.
#SPJ2