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3 years ago

7

A 0.144-kg baseball is moving toward home plate with a speed of 43 m/s when

1 answer:

algol [13]3 years ago

4 0

I would say 648858. bc yes

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Chewy travels an average of 22 mph, how many hours does it take him to cover a 110 mike course?

anzhelika [568]

That would be 110/22, which is 5 hours

5 0

3 years ago

A uniform disk of mass 3.25 kg has a radius of 0.100 m and spins with a frequency of 0.550 rev/s. What is its angular momentum?

shutvik [7]

ANSWER:

0.0562 J

STEP-BY-STEP EXPLANATION:

Angular momentum is expressed in terms of moment of inertia and angular velocity. This is expressed as follows:

$L=I\cdot\omega$

Here, I is the angular momentum and ω is the angular velocity.

Angular momentum is mass time the square of the radius of the object. Moment of inertia for a uniform disk is given as,

$I=\frac{1}{2}m\cdot r^2$

Here, m is the mass of the disk and r is the radius of the disk.

Replacing:

$L=\frac{1}{2}m\cdot r^2\cdot\omega$

Convert the units of angular velocity into rad/s.

$\begin{gathered} \omega=0.55\text{ rev/s}\cdot\frac{2\pi\text{ rad}}{1\text{ rev}} \\ \omega=3.46\text{ rad/s} \end{gathered}$

We replace each data to calculate the angular momentum:

$\begin{gathered} L=\frac{1}{2}\cdot3.25\cdot0.1^2\cdot3.46 \\ L=0.0562 \end{gathered}$

The angular momentum of the uniform disk is 0.0562 J

7 0

1 year ago

an n-type semiconductor is known to have an electron concentration of 5.63 x 1019 m-3. if the electron drift velocity is 113 m/s

Grace [21]

Answer:

$1.99581248\ /\Omega m$

Explanation:

e = Charge of electron = $1.6\times 10^{-19}\ C$

n = Electron concentration = $5.63\times 10^{19}\ /m^{3}$

$v_d$ = Drift veloctiy = 113 m/s

E = Electric field = 510 V/m

Electron mobility is given by

$\mu=\dfrac{v_d}{E}\\\Rightarrow \mu=\dfrac{113}{510}\\\Rightarrow \mu=0.22156\ m^2/Vs$

Conductivity is given by

$\sigma=ne\mu\\\Rightarrow \sigma=5.63\times 10^{19}\times 1.6\times 10^{-19}\times 0.22156\\\Rightarrow \sigma=1.99581248\ /\Omega m$

The conductivity of this material is $1.99581248\ /\Omega m$

4 0

3 years ago

Veronika [31]

Answer:

english please

Explanation:

5 0

3 years ago

Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note

anastassius [24]

Answer:

The beat frequency = 2.0 Hz

Explanation:

From the given information

The 3rd harmonic frequency of the note is equivalent to thrice note A fundamental frequency.

i.e.

$f_{3A} = 3f_A$

where;

$f_A =$ note A fundamental freq.

replacing 440 Hz for $f_A$

$f_{#a} = 3 (440 \ Hz)$

$f_{3A} = 1320 Hz$

However, the 2nd harmonic of the E is equivalent to two times the fundamental frequency of the note E.

i.e.

$f_{2E} = 2f_E$

$f_E =$ note E fundamental freq.

replacing 659 Hz for $f_E$

$f_{2E} = 2(659)$

$f_{2E} = 1318 \ Hz$

Finally, the beat frequency when the E string is properly tuned is:

$\Delta f = f_{3A}-f_{2E}$

$\Delta f = 1320 \ Hz - 1318 \ Hz$

$\Delta f = 2.0 \ Hz$

Thus, the beat frequency = 2.0 Hz

6 0

3 years ago