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2 years ago

Answer right I will cashapp $5

Physics

2 answers:

Sholpan [36]2 years ago

6 0

Answer:

The original analog sound waves are perfect but the copy is not because most analog waves have wavelengths that are too long to reproduce in copies

The original digital waves and the copy are perfect because with digital waves no information gets lost in the transfer of data, and there is no additional noise added to the copy.

Analog refers to audio recorded using methods that replicate the original sound waves. Digital audio is recorded by taking samples of the original sound wave at a specified rate.

Explanation:

i dont need $5

i wrote this using information from go ogle and some from brai nly. I hope it helps.

miss Akunina [59]2 years ago

5 0

Analog transfer information quicker, example electromagnetic waves which are used in our modern day phones.

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Find the pressure exerted on the floor by a 100N box whose bottom area is 40cm x 50cm.

Mazyrski [523]

Answer:

Pressure exerted = 500 Pa

Explanation:

The formula for pressure is as follows:

$Pressure = \frac{Force \space\ applied}{Area}$

In this case,

Force applied = 100N

Area = 40cm × 50cm = 2000cm² = 2000 × 10⁻⁴ = 0.2m²

Substituting these values into the formula:

Pressure = $\frac{100}{0.2}$

⇒ Pressure = 500 Pa

3 0

2 years ago

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Which particle is a negatively charged ion?

wel

Answer:

option D is correct

Explanation:

cl has 17 electrons if it has 18 it becomes cl-

6 0

2 years ago

What's an easy way to create an interference pattern of waves?

igor_vitrenko [27]

Answer:

Explanation:

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2 years ago

A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe

Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

$A_{t} = A_{0}\cdot e^{- \lambda t}$ (1)

where $A_{t}$ : is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.

To find A₀ we can use the following equation:

$A_{0} = N_{0} \lambda$ (2)

where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:

$N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}$

where $m_{T}$ : is the tree's carbon mass, $N_{A}$ : is the Avogadro's number and $m_{^{12}C}$ : is the ¹²C mass.

$N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C$

Similarly, from equation (2) λ is:

$\lambda = \frac{Ln(2)}{t_{1/2}}$

where t 1/2: is the half-life of ¹⁴C= 5700 years

$\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}$

So, the initial activity A₀ is:

$A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y$

Finally, we can calculate the time from equation (1):

$t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y$

I hope it helps you!

4 0

2 years ago

A roller coaster car is going over the top of a 18-m-radius circular rise. at the top of the hill, the passengers "feel light,"

Mashutka [201]

The solution for this problem is:

If they feel 50% of their weight that means that the centripetal force is also 50% of their weight 1g - 0.5g = 0.5g

Then 0.5* 9.8m/s² * 18m = 88.2 would be v²

Then get the square root, the answer would be:
and v = 9.391 m/s is the answer.

8 0

3 years ago

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