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wlad13 [49]
3 years ago
5

Does anyone know the answer to all of these questions

Physics
1 answer:
Rasek [7]3 years ago
5 0

Answer:

i don't understand the hw

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Where is the potential energy equal to zero?
s2008m [1.1K]

Answer:

im sure your already past this but it's E.

Explanation:

This is because in this case potential energy is linear to height, which means that the higher the more potential energy.

4 0
3 years ago
HELP PLEASE I OWE YOU GUYS
Mekhanik [1.2K]

Answer:

no se ingles nu mames

Explanation:

7 0
3 years ago
What variable is represented on the y-axis?
gladu [14]

Distance/ Time which means Distance is on horizontal and time is on vertical

8 0
3 years ago
Calculate the wave number for the infrared absorption of D35Cl when there is a transition from v=0 to v=1 (Unit: cm^-1, 4-digit
AVprozaik [17]

Answer:

wave number = 0.3348 * 10⁻⁸ cm⁻¹

Explanation:

Given data:

K = 4.808 * 10^2 N/m

<u>Determine the wave number for the infrared absorption</u>

considering vibrational Spectre

k' =  2n / λ ---- ( 1 )

λ = c / v ----- ( 2 )

v = √k / u  --- ( 3 )

where : k' = wave number, λ = wavelength, c = velocity of light, v = frequency,  k = force constant, u = reduced mass

u = 1.90415 for  D35Cl

Input equations 2 and 3 into equation 1 to get the final equation

K' = 2n/c * √k / u

   = ( 2 * 3.14 ) / 2.98 * 10^8  ] * (√ 4.808 * 10^2 / 1.90415 )

   = 33.486 * 10⁻⁸ m⁻¹  ≈ 0.3348 * 10⁻⁸ cm⁻¹

6 0
3 years ago
A communications channel has a bandwidth of 4,000 hz and a signal-to-noise ratio (snr of 30 db. what is the maximum possible dat
levacccp [35]
Through Shannon's Theorem, we can calculate the capacity of the communications channel using the value of its bandwidth and signal-to-noise ratio. The capacity, C, can be expressed as 

C = B × log₂(1 + S/N)

where B is the bandwidth of the channel and S/N is its signal-to-noise ratio.

Since the given SN ratio is in decibels, we must first express it as a ratio with no units as

SN (in decibels) = 10 × log (S/N)
30 = 10log(S/N)
log(S/N) = 3
S/N = 10³ = 1000

Now that we have S/N, we can solve for its capacity (in bits per second) as 

C = 4000 × log₂(1 + 1000) 
C = 39868.91 bps

Thus, the maximum capacity of the channel is 39868 bps or 40 kbps.

Answer: 40 kbps

 
4 0
2 years ago
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