if a star is 20 light minutes away and it stops creating light, how long will it be before we stop seeing the light?

In which part of the ear is the sound wave converted into electrical impulse

Kryger [21]

The part of the ear where the sound wave converted into electrical impulse would be the cochlea. This part is the auditory portion of the inner ear which produces nerve impulses in response to sound vibrations. Hope this answers the question. Have a nice day.

3 0

3 years ago

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An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.50 m above the ground. The railcar

igomit [66]

<span>An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.5 m above the ground. The railcar has a mass of 38,500 kg and is moving to the right at a constant speed of 8.7 m/s on a frictionless rail...
</span>

7 0

3 years ago

What is velocity change divided by time change?

Luda [366]

Acceleration: is the change in velocity divided by the time it takes for the change to occur. . Acceleration is the change in verity divided by the time it takes to make the change, Acceleration has direction.

5 0

3 years ago

How much energy ( in joules ) is released when 0.06 kilograms of mercury is condensed to a liquid at the same temperature ?

aleksley [76]

Answer:

B. 17,705.1 J

Explanation:

The hear released when the mercury condenses into a liquid is given by:

$Q=m \lambda_v$

where

m = 0.06 kg is the mass of the mercury

$\lambda_v$ is the latent heat of vaporization

For mercury, the latent heat of vaporization is $\lambda_v = 296 kJ/kg$ , so the heat released during the process is:

$Q=(0.06 kg)(296 kJ/kg)=17.76 kJ = 17,760 J$

So, the closest option is

B. 17,705.1 J

5 0

3 years ago

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A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium positio

Andru [333]

To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

$f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}$

$f=2.85 Hz$

PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

$\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2$