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3 years ago

WILL GIVE YOU WHAT EVER YOU WANT JUST HELP ME

Physics

1 answer:

Alona [7]3 years ago

5 0

Answer:

Explanation:

Essential Knowledge

It is essential that the student be able to distinguish between specialized structures that allow protists and fungi to obtain energy and explore their environment.

Protists

Protists are organisms that are classified into the Kingdom Protista. Although there is a lot of variety within the protists, they do share some common characteristics.

● Protists are usually single celled organisms.

● Live in moist environments.

● Vary in the ways they move and obtain energy.

Protists obtain their energy in several ways.

● Animal-like protists ingest or absorb food after capturing or trapping it.

● Plant-like protists produce food through photosynthesis.

● Fungus-like protists obtain their food by external digestion either as decomposers or as parasites.

● Some protists have both autotrophic and heterotrophic characteristi

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Explanation:

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3 years ago

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A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri

GREYUIT [131]

Answer:

$|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))$

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

$\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r$

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)$

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

$\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}$

Now, substitute this into 'dF':

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)$

Now we can integrate dF over the rod.

$\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)$

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3 years ago

Can somebody help me with this I need to pass

son4ous [18]

<h2>Answer:Radiation-3,Conduction-1,Convection-2</h2>

Explanation:

Radiation is the transfer of heat through electromagnetic waves.

These waves do not require any medium.This is the way we get heat from sun.Radiation is the quickest mode to transfer of heat.

Conduction is the transfer of heat through collisions of atomic particles.

This phenomenon largely occurs in solids like metals.The neighbour atoms sets the atoms into random motion thereby raising the temperature.

Convection is the transfer of heat through actual movement of medium particle.

This phenomenon occurs in gases an liquids.The medium particles actually traverse through the space transferring the heat.

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3 years ago

A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at

yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

$\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)$

Therefore, the work done by the worker in lifting the bucket is given as:

$W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)$

Now, plug in the values given and solve for 'W'. This gives,

$W=(20\ kg)(9.8\ m/s^2)(25\ m)+\frac{1}{2}(20\ kg)(4^2-0^2)\ m^2/s^2\\\\W=4900\ J +160\ J\\\\W=5060\ J$

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

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3 years ago