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4 years ago

7

This is an imaginary line on Earth's surface equidistant from the North Pole and South Pole that divides the Earth into a northe

rn hemisphere and a southern hemisphere

2 answers:

xxTIMURxx [149]4 years ago

3 0

Longitude lines because latitude lines are the ones that run east to west

tigry1 [53]4 years ago

3 0

Answer:

Equator

Explanation:

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According to the third law of planetary motion, the period of revolution of a planet is related to the planet’s _____.

matrenka [14]

Distance from the sun.

<span>The third law of planetary motion states that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit</span>. The semi-major axis is the distance from the sun to the epicenter of the ellipse (which would be the planet in question). So, the revolutionary period is directly related to the distance of the planet from the sun.

7 0

3 years ago

Read 2 more answers

A water balloon is thrown horizontally at a speed of 2.00 m/s from the roof of a building that is 6.00 m above the ground. At th

Elis [28]

Answer:

<u>The water ballon that was thrown straight down at 2.00 m/s hits the ground first, 0.19 s before the other ballon.</u>

Explanation:

The motions of the two water ballons are ruled by the kinematic equations:

$y=y_0+V_0t-gt^2/2$

We are only interested in the vertical motion, so that equation is all what you need.

<u>1. Water ballon is thrown horizontally at sped 2.00 m/s.</u>

The time the ballon takes to hit the ground is independent of the horizontal speed.

Since 2.00 m/s is a horizontal speed, you take the initial vertical speed equal to 0.

Then:

$y=y_0+V_0t-gt^2/2\\ \\ 0=6.00m-9.8\frac{m}{s^2} t^2/2\\ \\ t=\sqrt{2\times6.00m/9.8\frac{m}{s^2}}\\\\ t=1.11s$

<u>2. Water ballon thrown straight down at 2.00 m/s</u>

Now the initial vertical speed is 2.00 m/s down. So, the equation is:

$0=6.00m-2.00\frac{m}{s}t-9.8\frac{m}{s^2}t^2/2\\ \\ 4.9t^2+2t-6=0\\ \\ t=0.92s$

To solve the equation you can use the quadratic formula.

$t=\frac{-2+/-\sqrt{2^2-4(4.9)(-6)} }{2(4.9)}\\ \\ t=-1.33\\ \\ t=0.92$

You get two times. One of the times is negative, thus it does not have physical meaning.

<u>3. Conclusion:</u>

The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.

5 0

3 years ago

2. A student was sitting on one end of a set of metal bleachers outside of a school building. A drumline was practicing on the o

matrenka [14]

The students can feel the vibrations from the bleachers, but not from the drums because of the resonance.

<h3>What is the resonance?</h3>

The resonance condition occurs when the frequency of a wave matches with the resonance frequency.

The transfer of energy becomes maximum when the natural frequency matches with the frequency generated by the source.

The bleacher and student together having the frequency matching with the resonance frequency of the drum.

The bleacher and student with the bag frequency doesn't match with the resonance value of drums. Secondly, as distance increases sound intensity also decreases.

Thus, the student can feel the vibrations from the bleachers, but not from the drums because of the resonance.

Learn more about resonance.

brainly.com/question/15337338

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6 0

2 years ago

1. Find the principal moments of inertia for a uniform square of mass M and side a. 2. Find the kinetic energy of the uniform cu

Ne4ueva [31]

Solution to Question 1 is attached below.

8 0

4 years ago

A 25 {\rm pF} parallel-plate capacitor with an air gap between the plates is connected to a 100 {\rm V} battery. A Teflon slab i

jonny [76]

Answer:

0.275 nF

Explanation:

First of all, let's calculate the initial charge stored by the capactiro when it is in air. This is given by:

$Q_0 = C_0 V$

where

$C_0 = 25 pF=25\cdot 10^{-12} F$ is the initial capacitance of the capacitor

$V=100 V$ is the voltage of the battery

Substituting, we find

$Q_0 = (25\cdot 10^{-12}F)(100 V)=25\cdot 10^{-10} F=0.25 nF$

When the Teflon slab is inserted between the plates, the capacitance changes according to:

$C'=kC_0$

where

$k=2.1$ is the dielectric constant of Teflon. Therefore, the new charge stored on the capacitor will be:

$Q'=C' V=(kC_0)V=(2.1)(25\cdot 10^{-12} F)(100 V)=0.525 nF$

And so, the change in the charge on the capacitor is:

$\Delta Q=Q'-Q=0.525 nF-0.25 nF=0.275 nF$

7 0

3 years ago