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4 years ago

The Doppler Effect is used at baseball games to measure ___________. @3.2.P.B5 @1

Physics

2 answers:

pentagon [3]4 years ago

8 0

Answer: The answer is A. Speed <3

Bogdan [553]4 years ago

3 0

Answer:

Explanation:

Those devices the hold up while the pitcher is pitching measures speed. It has nothing to do with weather and temperature.

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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

3 years ago

___________ is the rising of a body of water and its overflowing onto normally dry land.

Ann [662]

Answer:

Flooding

Explanation:

4 0

3 years ago

What's photosynthesis?

lbvjy [14]

Answer:

Photosynthesis is a process used by plants and other organisms to convert light energy into chemical energy that, through cellular respiration, can later be released to fuel the organism's activities.

3 0

3 years ago

Read 2 more answers

By how much would 1500j of heat energy raise the temperature of 0.50kg of aluminum

valentinak56 [21]

pls what is the specific heat capacity of water

7 0

2 years ago

Determine the moment of inertia Ixx of the mallet about the x-axis. The density of the wooden handle is 860 kg/m3 and that of th

Yuki888 [10]

Complete Question

Diagram for this shown on the first uploaded image

Answer:

The moment of inertia Ixx of the mallet about the x-axis is $I{xx}= 0.119 kg \cdot m^2$

Explanation:

From the question we are told that

The density `of wooden handle is $\rho_w = 860 kg/m^3$

The density `of soft-metal head is $\rho_s =8000kg/m^3$

Generally the mass of the wooden can be mathematically obtained with this formula

$m_w = \rho_w A_w l_w$

Where $A_w$ is mass of wooden handle which is mathematically obtain with the formula

$A_w = \frac{\pi}{4} d^2_w$

Where $d_w$ is the diameter of the wooden handle which from the diagram is

$27mm = \frac{27}{1000} = 0.027m$

So $A_w = \frac{\pi}{4} * 0.027^2$

$l_w$ is the length of the the wooden handle which is given in the diagram as $l_w = 315mm = \frac{315}{1000} = 0.315m$

Substituting these value into the formula for mass

$m_w = 860 * (\frac{\pi}{4} * 0.027^2 ) *0.315$

$= 0.155kg$

Generally the mass of the soft-metal head can be mathematically obtained with this formula

$m_s = \rho_s A_s l_s$

Where $A_s$ is mass of soft-metal head which is mathematically obtain with the formula

$A_s = \frac{\pi}{4} d^2_s$

Where $d_s$ is the diameter of the soft-metal head which from the diagram is

$36mm = \frac{36}{1000} = 0.036m$

So $A_s = \frac{\pi}{4} * 0.036^2$

$l_s$ is the length of the the soft-metal head which is given in the diagram

as $l_s = 90mm = \frac{90}{1000} = 0.090m$

Substituting these value into the formula for mass

$m_s = 8000 * (\frac{\pi}{4} * 0.036^2 ) *0.090$

$=0.733kg$

Generally the mass moment of inertia about x-axis for the wooden handle is

$(I_{xx})_w = [\frac{1}{3}m_w + l_w^2 ]$

Substituting values

$(I_{xx})_w = [\frac{1}{3}*0.155 + 0.315^2 ]$

$=5.12*10^{-3}kg \cdot m^2$

Generally the mass moment of inertia about x-axis for the soft-metal head is

$(I_{xx})_s = [\frac{1}{12}m_s l_s ^2 + b^2]$

Where b is the distance from the centroid to the axis of the head which is mathematically given as

$b=l_w +\frac{d_s}{2}$

Substituting values

$b = 0.315 + \frac{0.036}{2}$

$= 0.336m$

Now substituting values into the formula for mass moment of inertia about x-axis for soft-metal head

$(I_{xx})_s = [\frac{1}{12} *0.733* 0.090^2 + 0.336^2]$

$=0.113 kg \cdot m^2$

Generally the mass moment of inertia about x-axis is mathematically represented as

$I_{xx} = (I_{xx})_w + (I_{xx})_s$

$= [\frac{1}{3}m_w + l_w^2 ] + [\frac{1}{12}m_s l_s ^2 + b^2]$

Substituting values

$I_{xx} = 5.12*10^{-3} +0.113$

$I{xx}= 0.119 kg \cdot m^2$

8 0

3 years ago