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4 years ago

Which cell part distinguishes a eukaryotic cell?

Physics

2 answers:

Mashcka [7]4 years ago

5 0

Answer:

B. Nucleus

Explanation:

The main difference between eukaryotic and prokaryotic cells is that eukaryotic cells have a nucleus.

Hope this helps!

Marta_Voda [28]4 years ago

4 0

Answer:

The answer is B

Explanation:

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Determine the end (final) value of n in a hydrogen atom transition, if the electron starts in and the atom emits a photon of lig

PSYCHO15rus [73]

Give me some answer choices and i will be happy to help

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4 years ago

What phenomenon causes the fuzzy image? What kind of lens is used to correct this, and how is it corrected?

Amiraneli [1.4K]

Sample Response: This phenomenon is called chromatic aberration. This happens when light of different wavelengths focuses at different points. A converging lens is used to help light of different wavelengths focuses at a common point.

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3 years ago

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The terminals of a 0.70 Vwatch battery are connected by a 80.0-m-long gold wire with a diameter of 0.200 mm What is the current

Komok [63]

Answer:

$I=0.047A$

Explanation:

Let's use Ohm's law:

$V=IR$

$I=\frac{V}{R}$ (1)

Where:

$V=Voltage\\I=Current\\R=Electrical\hspace{2 mm}Resistance$

We know the value of the voltage V, so we need to find the value of R in order to find I. Fortunately there is a relation between the resistivity of a conductor and its electrical resistance given by:

$R=\rho*\frac{l}{A}$ (2)

Where:

$R=Electrical\hspace{2 mm}Resistance\\l=Length\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=80m\\A=Cross\hspace{2 mm}sectional\hspace{2 mm}area\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=1.256637061*10^{-7} \\\rho=Electrical\hspace{2 mm}resistivity\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}material=2.35*10^{-8}$

Keep in mind that the electrical resistivity of the gold is a known constant which is $\rho_g_o_l_d=2.35*10^{-8}$ and the cross sectional area of the conductor is calculated as:

$A=\pi *(r^{2})=\pi *(0.0002m)^{2} =1.256637061*10^{-7} m^{2}$

Because we have a wire in this case, so we assume a cylindrical geometry.

Now replacing our data in (2)

$R=(2.35*10^{-8})*\frac{80}{1.256637061*10^{-7} } =14.96056465\Omega$

Finally, we know R and V, so replacing these values in (1) we will be able to find the current:

$I=\frac{0.7}{14.96056465}\approx0.047A$

7 0

3 years ago

A terrorist throws a grenade with a 6.00 second fuse off a building 150.0 m high at a speed of 10.0 m/s. If the angle at which t

igomit [66]

Here we have a projectile motion. It is type of motion that is made of a vertical shot and a horizontal shot. This is how we will solve it.

Firste step is to find horizontal and vertical component of a speed.
$v_{0x} =v_{0} * cos \alpha$ \\ v_{0y} = v_{0} * sin \alpha [/tex]

We are given this information:
$v_{0} = 10 m/s \\ h=150m \\ \alpha =-30°$
Angle is negative because it is below the horizontal.

VERTICAL SHOT
Time needed for a grenade to fall to the bottom of a building is given by a formula:
$t= \frac{ v_{0y} }{g} \\ t= \frac{v_{0} * sin \alpha}{g} \\ t= \frac{-10*sin(-30)}{9.81} \\ t=0.51s$
We used negative value for a speed because it is considered that upwards shot has positive value and downwards shot has negative value.

The grenade will not explode before it hits the ground.

HORIZONTAL SHOT
<span>The horizontal distance from the building at which the grenade will land is called range. The formula for a range is given by:
</span> $R= v_{0x} * \sqrt{ \frac{2h}{g} } \\ R=v_{0} * cos \alpha* \sqrt{ \frac{2h}{g} } \\ R=10*cos(-30)* \sqrt{ \frac{2*150}{9.81} } \\ R=47.89m$

The grenade will hit the ground at distance of 47.89m.

7 0

3 years ago

Will give

Nataly [62]

Answer:

True

Explanation:

6 0

3 years ago

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