Why does paper stand up when a van de graaf generator is on?

What is the biological impact of minimum catch sizes on a population of fish?

ivann1987 [24]

The population comes to be dominated by smaller, slower-growing individuals

7 0

4 years ago

A uniform rod is 2. 0 m long and has mass 15 kg. What is most nearly the rod's mass moment of inertia?

trapecia [35]

The rod's mass moment of inertia is 5kgm².

<h3>Moment of Inertia:</h3>

The "sum of the product of mass" of each particle with the "square of its distance from the axis of rotation" is the formula for the moment of inertia.

The Parallel axis Theorem can be used to compute the moment of inertia about the end of the rod directly or to derive it from the center of mass expression. I = kg m². We can use the equation for I of a cylinder around its end if the thickness is not insignificant.

If we look at the rod we can assume that it is uniform. Therefore the linear density will remain constant and we have;

or = M / L = dm / dl

dm = (M / L) dl

$I = \int\limits^M_0 {r^2} \, dm$

$I = \int\limits^\frac{L}{2} _\frac{-L}{2} {I^2 (M/L)} \, dl$

Here the variable of the integration is the length (dl). The limits have changed from M to the required fraction of L.

$I = \int\limits^\frac{L}{2} _\frac{-L}{2} {I^2 (M/L)} \, dl$

$I = \frac{M}3L}[(\frac{L^3}{2^3} - \frac{-L^3}{2^3} )]\\\\I = \frac{1}{12}ML^2$

Mass of the rod = 15 kg

Length of the rod = 2.0 m

Moment of Inertia, I = $\frac{1}{12}15 (2)^2$

= 5 kgm²

Therefore, the moment of inertia is 5kgm².

Learn more about moment of inertia here:

brainly.com/question/14119750

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4 0

2 years ago

Which of the following keys moves the insertion point to the beginning of data in a cell? (Points : 2)

Korvikt [17]

Insert moves the insertion point to the beginning of data in a cell so the answer is INSERT :)))
i hope this be helpful

6 0

3 years ago

Is space expanding within clusters of galaxies? why or why not?

GarryVolchara [31]

No, since their gravity is powerful enough to keep them together even while the universe expands as a whole. Space is not expanding within clusters of galaxies.

<h3>What is a galaxy?</h3>

A galaxy is a massive clump of gas, dust, and billions of stars and their solar systems bound together by gravity.

No, since their gravity is powerful enough to keep them together even while the universe expands as an entire.

Hence,space is not expanding within clusters of galaxies.

To learn more about the galaxy, refer to the link;

brainly.com/question/2905713

#SPJ1

6 0

2 years ago

A 30 g particle is undergoing simple harmonic motion with an amplitude of 2.0 ✕ 10-3 m and a maximum acceleration of magnitude 8

KiRa [710]

Answer:

Explanation:

By using the Newton second law and the position equation for a simple harmonic motion we have

$F=ma\\a_{max}=\omega^{2}A\\x=Acos(\omega t+ \phi)\\$

where a is the acceleration, w is the angular frecuency and \phi is the phase constant. We can calculate w from the equation for the maximum acceleration

$\omega=\sqrt{\frac{a_{max}}{A}}=\sqrt{\frac{8*10^{3}m/s^{2}}{2*10^{-3}}}=2000rad/s$

(a).

$F=ma=m\omega^{2}Acos(\omega t + \phi)\\F=(30*10^{-3}kg)(2*10^{-3}m)(2000\frac{rad}{s})^{2}cos(2000\frac{rad}{s} t - \frac{\pi}{2})=240N*cos(2000\frac{rad}{s} t - \frac{\pi}{2})$

(b). $T=\frac{2\pi}{\omega}=\frac{2\pi}{2000}=3.14*10^{-13} s$

(c). $v_{max}=A\omega=(2*10^{-3}m)(2000\frac{rad}{s})=4\frac{m}{s}$

(d). The mecanical energy is the kinetic energy when the velocity is a maximum

$E_{m}=E_{k}(v_{max})=\frac{mv_{max}^{2}}{2}=\frac{30*10^{-3}kg(4\frac{m}{s})^{2}}{2}=0.024J$

3 0

4 years ago

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