What are some words that have changed meaning due to technological advances in the last 20 years?

A rock band playing an outdoor concert produces sound at 120 dB 5.0 m away from their single working loudspeaker. What is the so

joja [24]

ANSWER:100dB

from the sound intensity level,the sound intensity is calculated as:

$\beta$ ₁= $\beta$ =(10dB)㏒₁₀(l₁/l₀)

inserting numbers:

120dB=(10dB)㏒₁₀[l₁/10⁻¹²W/m⁸] or 12=㏒₁₀[l₁/(10⁻¹²)W/m²

Getting the antilog of both sides and obtain 10¹²=l₁(10⁻¹²W/m²)which

can be used to solve for l₁ and get

l₁=(10⁻¹²W/m²)(10¹²)=1 W/m²

since the sound intensity is related to the power and that the power does not change,the sound intensity at any other point can be solved.Plugging-in

ᵃ = 4πr²,into P=l₁ₐ₁=l₂ₐ₂ and get:

l₂=l₁(r₁/r₂)² =(1W/m²)(5/35) =2.04×10⁻²W/m²

since we know the sound intensity at the sound point 2r,the sound intensity level at the point can be solved.We have:

$\beta$ ₂= $\beta$ =(10dB)㏒₁₀(l₂/l₀)=(10dB)㏒₁₀(2.04×10⁻²/1×10⁻²)

$\beta$ ₂=(10dB)㏒₁₀(2.04×10¹⁰)

$\beta$ =(10dB)[㏒₁₀(2.04)+㏒₁₀(10¹⁰)]=10dB[0.32+10]=103dB=100dB

3 0

3 years ago

A 3.00 × 10^−9-coulomb test charge is placed near

viktelen [127]

Answer:

B

Explanation:

Given:-

- The charge of the test particle q = 3.0 * 10^-9 C

- The force exerted by the metal sphere F = 6.0 * 10^-5 N

Find:-

The magnitude and direction of the electric field strength at this location?

Solution:-

- The relationship between the electrostatic force F exerted by the metal sphere on the test-charge and the Electric Field strength E at the position of test charge is given by:

F = E*q

- Using the data given we can determine E:

E = F / q

E = (6.0 * 10^-5) / (3.0 * 10^-9)

E = 20,000 N/C

- The direction of electric field is given by the net charge of the source ( metal sphere). The metal sphere is negative charge hence the direction of Electric Field strength E is directed towards the metal sphere.

8 0

3 years ago

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A potter’s wheel moves from rest to an angular speed of 0.20 rev/s in 23.8 s. Assuming constant angular acceleration, what is it

ser-zykov [4K]

Answer:

$0.053 rad/s^2$

Explanation:

0.2 rev/s = 0.2 rev/s * 2π rad/rev = 0.4π rad/s

Since the angular acceleration is assumed to be constant, and the wheel's angular speed is increasing from rest (0 rad/s) to 0.4π rad/s within 23.8s. Then the angular acceleration must be

$\alpha = \Delta \omega / \Delta t = \frac{0.4 \pi - 0}{23.8} = 0.053 rad/s^2$

8 0

3 years ago

A 500kg car is driven forward with a thrust force of 1500N. Air resistance and friction acts against the motion of the car with

Wittaler [7]

2m/s^2, this is because F=ma, meaning a is also equal to F/m. The car applies 1500N in one direction and outside sources apply a total of -500N, meaning the 500kg car is moving forward with a total of 1000N of force. Taking the total 1000N and dividing it by 500kg gives you and acceleration of 2m/s^2. Hope this helps!

8 0

2 years ago

What is the significance of a standard system of measurement?

dimulka [17.4K]

Because scientists all over the world are working together, looking for answers to the same questions, just as much as if they all worked in the same physical laboratory in the same building. They need a way to share data and experimental results in a form that everyone can understand. ( D )

Let's say I perform an experiment and get very exciting results. I'm a good scientist, so the next thing I want to do is to publish a complete description of how I did my experiment, and include all of my results. That way, scientists around the world can read about what I did, they can find any mistakes that I made, and they can even repeat my experiment for themselves and see if they get the same results.

Now let's say that my results looked like this:

Result #1).

The reaction stabilized when it reached the rate of 1.26 briligs per tove.

Result #2).

After running at that constant rate for 35 toves, a pile of product was produced whose mass was exactly 61.284 wibbles.

Result #3).

When the pile of product was allowed to settle for another 20 toves, it had spread out, and covered an area of 14.907 square filks.

Do YOU understand my results ?

All those other scientists would have a tough time trying to decide whether my results made sense. And if they repeated my experiment, they would have no way to tell whether their results matched mine or not.

Without a standard system of measurement, and units that mean the same thing to everybody, us scientists literally could not communicate.

3 0

3 years ago

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