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Leno4ka [110]
2 years ago
11

PLS HELP which ones would be made of cells? and which ones show cell walls?

Physics
1 answer:
maxonik [38]2 years ago
8 0

The first question's answer depends on what you mean by "sponge". If you're talking about sea sponges, then all but plastic are made up of cells. Some sponges used for cleaning are also made of plant material but also other, non-organic materials like dyes.

Cell walls are only present in plant cells, so they would be found in cork (derived from a certain tree bark), wood, and trees. Synthetic sponges made with plant material might also contain them, but they wouldn't be made entirely of cells with walls.

You might be interested in
Do you think there are other planets outside of our solar system? Support your response with facts
Agata [3.3K]


The habitable zone is the range of distances from a star where a planet’s temperature allows liquid water oceans, critical for life on Earth. The earliest definition of the zone was based on simple thermal equilibrium, but current calculations of the habitable zone include many other factors, including the greenhouse effect of a planet’s atmosphere. This makes the boundaries of a habitable zone "fuzzy."



Astronomers announced in August 2016 that they may have found such a planet orbiting Proxima Centauri. The newfound world, known as Proxima b, is about 1.3 times more massive than Earth, which suggests that the exoplanet is a rocky world, researchers said. The planet is also in the star's habitable zone, just 4.7 million miles (7.5 million kilometers) from its host star. It completes one orbit every 11.2 Earth-days. As a result, it's likely that the exoplanet is tidally locked, meaning it always shows the same face to its host star, just as the moon shows only one face (the near side) to Earth.



The young sun would have had a very strong magnetic field, whose lines of force reached out into the disk of swirling gas from which the planets would form. These field lines connected with the charged particles in the gas, and acted like anchors, slowing down the spin of the forming sun and spinning up the gas that would eventually turn into the planets. Most stars like the sun rotate slowly, so astronomers inferred that the same “magnetic braking” occurred for them, meaning that planet formation must have occurred for them. The implication: Planets must be common around sun-like
A Canadian team discovered a Jupiter-size planet around Gamma Cephei in 1988, but because its orbit was much smaller than Jupiter’s, the scientists did not claim a definitive planet detection. “We weren’t expecting planets like that. It was different enough from a planet in our own solar system that they were cautious," Matthews said.
Most of the first exoplanet discoveries were huge Jupiter-size (or larger) gas giants orbiting close to their parent stars. That's because astronomers were relying on the radial velocity technique, which measures how much a star “wobbles” when a planet or planets orbit it. These large planets close in produce a correspondingly big effect on their parent star, causing an easier-to-detect wobble.
Before the era of exoplanet discoveries, instruments could only measure stellar motions down to a kilometer per second, too imprecise to detect a wobble due to a planet. Now, some instruments can measure velocities as low as a centimeter per second, according to Matthews. “Partly due to better instrumentation, but also because astronomers are now more experienced in teasing subtle signals out of the data.”

Today, there are more than 1,000 confirmed exoplanets discovered by a single telescope: the Kepler space telescope, which reached orbit in 2009 and hunted for habitable planets for four years. Kepler uses a technique called the “transit” method, measuring how much a star's light dims when a planet passes in front of it.

Kepler has revealed a cornucopia of different types of planets. Besides gas giants and terrestrial planets, it has helped define a whole new class known as “super-Earths”: planets that are between the size of Earth and Neptune. Some of these are in the habitable zones of their stars, but astrobiologists are going back to the drawing board to consider how life might develop on such worlds.

In 2014, Kepler astronomers (including Matthews’ former student Jason Rowe) unveiled a “verification by multiplicity” method that should increase the rate at which astronomers promote candidate planets to confirmed planets. The technique is based on orbital stability — many transits of a star occurring with short periods can only be due to planets in small orbits, since multiply eclipsing stars that might mimic would gravitationally eject each other from the system in just a few million years.

While the Kepler (and French CoRoT) planet-hunting satellites have ended their original missions, scientists are still mining the data for discoveries, and there are more to come. MOST is still operating, and the NASA TESS (Transiting Exoplanet Survey Satellite), Swiss CHEOPS (Characterizing ExOPlanets Satellite) and ESA’s PLATO missions will soon pick up the transit search from space. From the ground, the HARPS spectrograph on the European Southern Observatory's La Silla 3.6-meter telescope in Chile is leading the Doppler wobble search, but there are many other telescopes in the hunt.

With almost 2,000 to choose from, it’s hard to narrow down a few. Small solid planets in the habitable zone are automatically standouts, but Matthews singled out five other exoplanets that have expanded our perspective on how planets form and
6 0
3 years ago
Read 2 more answers
A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput
Nikitich [7]
There is a missing data in the text of the problem (found on internet):
"with force constant<span> k=</span>450N/<span>m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
</span>E=U+K=  \frac{1}{2}kx^2 + \frac{1}{2} mv^2
<span>where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
</span>E=K_{max} =  \frac{1}{2}mv_{max}^2
<span>Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
</span>E=U_{max}= \frac{1}{2}k A^2
<span>
Since the mechanical energy must be conserved, we can write
</span>\frac{1}{2}mv_{max}^2 =  \frac{1}{2}kA^2
<span>from which we find the maximum speed
</span>v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }=  1.2 m/s
<span>
b) </span><span> the </span>speed<span> of the </span>glider<span> when it is at x= -0.015</span><span>m

We can still use the conservation of energy to solve this part. 
The total mechanical energy is:
</span>E=K_{max}=  \frac{1}{2}mv_{max}^2= 0.36 J
<span>
At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
</span>U= \frac{1}{2}kx^2 =  \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J
<span>And since 
</span>E=U+K
<span>we find the kinetic energy when the glider is at this position:
</span>K=E-U=0.36 J - 0.05 J = 0.31 J
<span>And then we can find the corresponding velocity:
</span>K= \frac{1}{2}mv^2
v=  \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s
<span>
c) </span><span>the magnitude of the maximum acceleration of the glider;
</span>
For a simple harmonic motion, the magnitude of the maximum acceleration is given by
a_{max} = \omega^2 A
where \omega= \sqrt{ \frac{k}{m} } is the angular frequency, and A is the amplitude.
The angular frequency is:
\omega =  \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s
and so the maximum acceleration is
a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2

d) <span>the </span>acceleration<span> of the </span>glider<span> at x= -0.015</span><span>m

For a simple harmonic motion, the acceleration is given by
</span>a(t)=\omega^2 x(t)
<span>where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find 
</span>a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2
<span>
e) </span><span>the total mechanical energy of the glider at any point in its motion. </span><span>

we have already calculated it at point b), and it is given by
</span>E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J
8 0
3 years ago
Removing an electron from a neutral atom will result in an atom that?
Alex73 [517]

Removing an electron from a neutral atom will result in an atom that is positive.

8 0
2 years ago
Read 2 more answers
A wave transfers:<br> Water<br> particles<br> energy<br> matter
SashulF [63]

Answer:

Particles in a water wave exchange kinetic energy for potential energy. When particles in water become part of a wave, they start to move up or down. This means that kinetic energy (energy of movement) has been transferred to them

Explanation:

hope this helps u ....

<em>pls mark this as the brainliest...</em>

6 0
3 years ago
A coaxial cable consists of a solid inner cylindrical conductor of radius 2 mm and an outer cylindrical shell of inner radius 3
4vir4ik [10]

Answer:

d) 1.2 mT

Explanation:

Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.

First of all, we observe that:

- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is

I = 15 A

- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).

Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0 is the vacuum permeability

I = 15 A is the current in the conductor

r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field

Substituting, we find:

B=\frac{(4\pi\cdot 10^{-7})(15)}{2\pi(0.0025)}=1.2\cdot 10^{-3}T = 1.2 mT

8 0
3 years ago
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