the brightness of stars as they appear from earth is measured by _______ magnitude. a. light b. apparent c. relative d. absolute

A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit

Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

$P = \frac{1}{2} \mu \omega^2 A^2 v$

Here,

$\mu$ = Linear mass density of the string

$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

$\mu = \frac{m}{l}$

And the angular frequency can be written as

$\omega = 2\pi f$

Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

PART B) The new amplitude A' that is half ot the wavelength of the wave is

$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.058W$

8 0

2 years ago

The science of classifying and naming organisms based on their different characteristics is called ______________.

Fofino [41]

Answer:

Biology

Explanation:

Hopethis helps!

8 0

2 years ago

Read 2 more answers

A tire 0.500 m in radius rotates at a constant rate of 200 rev/min. Find the speed and acceleration of a small stone lodged in t

arsen [322]

Answer:

v = 10.47 m/s

a = 219.32 m/s²

Explanation:

The velocity of the stone can be given by the following formula:

v = r ω

where,

v = linear velocity of the stone = ?

r = radius of tire = 0.5 m

ω = angular velocity = (200 rev/min)(2π rad/1 rev)(1 min/60 s) = 20.94 rad/s

Therefore,

v = (0.5 m)(20.94 rad/s)

v = 10.47 m/s

The acceleration of the stone will be equal to the centripetal acceleration. Therefore,

a = v²/r

a = (10.47 m/s)²/0.5 m

a = 219.32 m/s²

8 0

3 years ago

You’re squeezing a springy rubber ball in your hand. If you push inward on it with a force of 1 N, it dents inward 2 mm. How far

creativ13 [48]

Answer:

10mm

Explanation:

According to Hooke's law which states that "the extension of an elastic material is directly proportional to the applied force provided the elastic limit is not exceeded. Direct proportionality there means, increase/decrease in the force leads to increase/decrease in extension.

Mathematically, F = ke where;

F is the applied force

k is the elastic constant

e is the extension

from the formula k = F/e

k = F1/e1 = F2/e2

Given force of 1N indents the spring inwards by 2mm, this means force of 1N generates extension of 2mm

Let F1 = 1N e1 = 2mm

The extension that will be produced If force of 5N is applied to the string is what we are looking for. Therefore F2 = 5N; e2= ?

Substituting this values in the formula above we have

1/2=5/e2

Cross multiplying;

e2 = 10mm

This shows that we must have dent it by 10mm before it pushes outwards by a 5N force

8 0

3 years ago

A car is stopped at a red light. When the light changes, the car begins to accelerate at a rate of 10 m/s^2 and it continues to

yaroslaw [1]

The velocity of the car after it accelerates for 25 seconds would be 820.2 ft/s or 250 m/s

4 0

3 years ago