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2 years ago

5

a star is seen to have two transiting planets. planet 1 transits every 3 months, and planet 2 transits every 15 months. what can

we infer about their orbits?

1 answer:

WINSTONCH [101]2 years ago

4 0

Planet 1 orbits 3x as quick as planet 2

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Two different substances, Substance A and Substance B, are in direct contact with each other and are at different temperatures.

Alenkinab [10]

Answer:

resonance

Explanation:

The particles of substance B will cause the particle of substance A to vibrate at the same frequency

8 0

3 years ago

Read 2 more answers

Explain the process of why the balloon is attracted to the wall, and why electrons are not transferred in this process. Is the w

strojnjashka [21]

Answer:

The process by which the balloon is attracted and possibly sticks to the wall is known as static electricity which is the attraction or repulsion between electric charges which are not free to move.

The wall is an insulator.

Explanation:

When a balloon is blown and tied off, and then the balloon is rubbed on the woolly object once in one direction, and the side that was rubbed against the wool is brought near a wall and then released, it is observed that the balloon is attracted to and sticks to the wall. The above observation is due to static electricity.

Static electricity refers to electric charges that are not free to move or that are static. One of the means of generating such charges is by friction. When the balloon is rubbed on the woollen material, electrons are given away to the balloon's surface. Since the balloon is an insulator (materials which do not allow electricity to pass through them easily), the electrons are not free to move. When the balloon is brought near to a wall, there is a rearrangement of the charges present on the wall. Negative charges on the wall move farther away while the positive charges on the wall are attracted to the electrons on the balloon's surface. Because the wall is also an insulator, the charges are not discharged immediately. Therefore, this attraction between opposite charges as well as the static nature of the charges results in the balloon sticking to the wall.

6 0

3 years ago

What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ is the pressure difference between the two ends

$\mu$ is viscosity of the fluid

L is the length of the pipe

$Q=Av$ is the volumetric flow rate, with $A=\pi r^2$ being the section of the tube and $v$ the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

$\mu=0.001 Pa/s$ is the dynamic water viscosity at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Using these data in the formula, we get:

$\Delta P = 3200 Pa$

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0

3 years ago

How many electrons are in the third energy level?<br><br> 2<br> 8<br> 18<br> 20

Blababa [14]

Answer:

B)8

Explanation:

In the first energy level you can have, at most, 2. Every energy level after that wants to have 8 electrons. Valence electrons I believe.

Hope this helps, have a nice day! (^-^)

6 0

3 years ago

Read 2 more answers

To navigate, a porpoise emits a sound wave that has a wavelength of 3.3 cm. The speed at which the wave travels in seawater is 1

dedylja [7]

Answer:

$2.2\times 10^{-5} s$

Explanation:

We are given that

The wavelength of sound wave= $\lambda=3.3 cm=3.3\times 10^{-2}m/s$

1 cm/s= $10^{-2}m/s$

Speed of sound wave,v= $1522 m/s$

We have to find the period of the wave.

We know that

Frequency= $\nu=\frac{v}{\lambda}$

Using the formula

Frequency = $\frac{1522}{3.3\times 10^{-2}}=4.6\times 10^{4}$ Hz

Time period= $\frac{1}{4.6\times 10^4}=0.22\times 10^{-4}\times \frac{10}{10^1}=2.2\times 10^{-4-1}=2.2\times 10^{-5}s$

Using identity: $\frac{a^x}{a^y}=a^{x-y}$

Hence, the time period of the wave= $2.2\times 10^{-5} s$

4 0

3 years ago