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3 years ago

Unlike eukaryotes, prokaryotes do not have:

Physics

2 answers:

Mashcka [7]3 years ago

7 0

D is the answer

Unlike eukaryotes, prokaryotes have no membrane-bound organelles. This means that they lack a nucleus, mitochondria, and other advanced cell structures

xxMikexx [17]3 years ago

5 0

D. a membrane bound nucleus , lmk if im right

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On a day the air is still a windmill still posses what type of energy is that ?

Montano1993 [528]

Well, it's up on top of a pole or pedestal of some sort,
so it has some gravitational potential energy relative to
the ground. In other words, if it somehow became detached
from its structure and fell to the ground, it would make quite
an energetic splat when it got there.

Also, the windmill is at the temperature of the air around it,
which is far from Absolute Zero, so the windmill holds a lot of
thermal (heat) energy.

Then I guess there's the matter of the chemical energy in the
molecules of the material that the windmill is made of, and the
nuclear energy in its atoms.

6 0

3 years ago

A current of 0.96 amps flows through a copper wire 0.44 mm in diameter when connected to a potential difference of 15 v. how lon

FinnZ [79.3K]

By using Ohm's law, we can calculate the resistance of the wire. Ohm's law states that:
$V=IR$
where V is the potential difference across the conductor, I is the current and R the resistance. Rearranging the equation, we get
$R= \frac{V}{I}= \frac{15 V}{0.96 A}=15.6 \Omega$

Now we can use the following equation to calculate the length of the wire:
$R= \frac{\rho L}{A}$ (1)
where
$\rho$ is the resistivity of the material
L is the length of the conductor
A is its cross-sectional area
In this problem, we have a wire of copper, with resistivity $\rho=1.68 \cdot 10^{-8} \Omega m$ . The radius of the wire is half the diameter:
$r= \frac{d}{2}= \frac{0.44 mm}{2}=0.22 mm=0.22 \cdot 10^{-3} m$
And the cross-sectional area is
$A=\pi r^2=\pi (0.22 \cdot 10^{-3}m)^2=1.52 \cdot 10^{-7} m^2$

So now we can rearrange eq.(1) to calculate the length of the wire:
$L= \frac{RA}{\rho}= \frac{(15.6 \Omega)(1.52 \cdot 10^{-7} m^2)}{1.68 \cdot 10^{-8} \Omega m}=141.1 m$

8 0

4 years ago

How does 'g' vary from place to place?

r-ruslan [8.4K]

Explanation:

The acceleration g varies by about 1/2 of 1 percent with position on Earth's surface, from about 9.78 metres per second per second at the Equator to approximately 9.83 metres per second per second at the poles.

8 0

3 years ago

When the magnetic flux through a single loop of wire increases by , an average current of 40 A is induced in the wire. Assuming

Zielflug [23.3K]

COMPLETE QUESTION:

When the magnetic flux through a single loop of wire increases by 30 Tm^2 , an average current of 40 A is induced in the wire. Assuming that the wire has a resistance of 2.5 ohms , (a) over what period of time did the flux increase? (b) If the current had been only 20 A, how long would the flux increase have taken?

Answer:

(a). The time period is 0.3s.

(b). The time period is 0.6s.

Explanation:

Faraday's law says that for one loop of wire the emf $\varepsilon$ is

$(1). \: \: \varepsilon = \dfrac{\Delta \Phi_B}{\Delta t }$

and since from Ohm's law

$\varepsilon = IR$ ,

then equation (1) becomes

$(2). \: \:IR= \dfrac{\Delta \Phi_B}{\Delta t }.$

(a).

We are told that the change in magnetic flux is $\Phi_B = 30Tm^2$ , the current induced is $I = 40A$ , and the resistance of the wire is $R = 2.5\Omega$ ; therefore, equation (2) gives

$(40A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

which we solve for $\Delta t$ to get:

$\Delta t = \dfrac{30Tm^2}{(40A)(2.5\Omega)},$

$\boxed{\Delta t = 0.3s}$ ,

which is the period of time over which the magnetic flux increased.

(b).

Now, if the current had been $I =20A$ , then equation (2) would give

$(20A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

$\Delta t = \dfrac{30Tm^2}{(20A)(2.5\Omega)},$

$\boxed{\Delta t = 0.6 s\\}$

which is a longer time interval than what we got in part a, which is understandable because in part a the rate of change of flux $\dfrac{\Delta \Phi_B}{\Delta t}$ is greater than in part b, and therefore , the current in (a) is greater than in (b).

7 0

3 years ago

What is the net force on a car if the force of friction is 15 N and the forward force due to the engine is 20 N?

ICE Princess25 [194]

5 newton forward because the force of the engine is greater than the force of friction so that means the car is going forward.
You can think of this: when friction is greater than the force of an object its slowing down but when the force of the car is greater than the force of friction its speeding up.
Hope this helps

6 0

3 years ago