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Marta_Voda [28]

3 years ago

7

Having _________is the most obvious difference between a eukaryotic and a prokaryotic cell, but there are other differences as w

ell.
a
DNA
b
Cytoplasm
c
a pulse
d
a nucleus

Prokaryotic cells don’t ____________ have like eukaryotic cells.

a
DNA
b
membrane-bound organelles
c
cytoplasam
d
brains

Prokaryotic cells are also usually around ____________ than eukaryotic cells

a
ten times larger
b
half the size
c
ten times smaller
d
twice as large

An example of an organism that is eukaryotic would be

a
bacteria cell
b
animal cell
c
archaea

An example of a prokaryotic organism?

a
plant and animal cell
b
fungi
c
bacteria
d
human cell
GIVE ME ANSWERS AND I WILL GIVE U BRAINLIEST AD 50 POINTS

1 answer:

8_murik_8 [283]3 years ago

7 0

Answer:

1. d. a nucleus

2. b. Membrane-bound organelles

3. c. ten time smaller

4. b. animal cell

5. c. Bacteria

Explanation:

Hope this helps... :)

You might be interested in

Which structure is represented by the letter D? Choose 1 answer: (Choice A) A Chloroplast (Choice B) B Vacuole (Choice C) C Mito

Otrada [13]

Answer:

vacuole

Explanation:

Remember vacuoles in plant cells are a lot bigger than there animal cell counterparts.

6 0

3 years ago

You have two resistors with the same cross sectional area and resistivity. Resistor A has length L1 and resistor B has length L2

oee [108]

Answer:

Explanation:

Given

Resistor A has length $L_1$

and Resistor B has Length $L_2$

and Resistance is given by

$R=\frac{\rho L}{A}$

Considering $\rho$ and A to be constant thus

$R_2>R_1$ because $L_2>L_1$

(a)When they are connected in series

As the current in series is same and power is $i^2R$

therefore $P_2>P_1$ as R is greater for second resistor

(b)if they are connected in Parallel

In Parallel connection Voltage is same

$P=\frac{V^2}{R}$

resistance of 2 is greater than 1 thus Power delivered by 1 is greater than 2

8 0

3 years ago

Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w

Tatiana [17]

Answer:

The average velocity is

$266\frac{m}{s},274\frac{m}{s}$ and $117\frac{m}{s}$ respectively.

Explanation:

Let's start writing the vertical position equation :

$L(t)=2t^{3}+t^{2}-5t+1$

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

$V_{avg}=\frac{Displacement}{Time}$ = Δx / Δt = $\frac{x2-x1}{t2-t1}$

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

$t2-t1=8s-5s=3s$

For the position variation we use the vertical position equation :

$x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m$

$x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m$

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

$\frac{798m}{3s}=266\frac{m}{s}$

For the second time interval :

t1 = 4 s → t2 = 9 s

$x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m$

$x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m$

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

$\frac{1370m}{5s}=274\frac{m}{s}$

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

$x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m$

$x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m$

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

$\frac{702m}{6s}=117\frac{m}{s}$

5 0

3 years ago

A rifle bullet with mass ma = 8.00 g strikes and embeds itself in a block with mass mb = 0.992 kg that rests on a frictionless,

Doss [256]

Answer:

the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

Explanation:

a) Kinetic energy of block = potential energy in spring

½ mv² = ½ kx²

Here m stands for combined mass (block + bullet),

which is just 1 kg. Spring constant k is unknown, but you can find it from given data:

k = 0.75 N / 0.25 cm

= 3 N/cm, or 300 N/m.

From the energy equation above, solve for v,

v = v √(k/m)

= 0.15 √(300/1)

= 2.598 m/s.

b) Momentum before impact = momentum after impact.

Since m = 1 kg,

v = 2.598 m/s,

p = 2.598 kg m/s.

This is the same momentum carried by bullet as it strikes the block. Therefore, if u is bullet speed,

u = 2.598 kg m/s / 8 × 10⁻³ kg

= 324.76 m/s.

Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

6 0

3 years ago

1. Which has more kinetic energy, a 40 kg cheetah running at 25 m/s or a 4,000 kg elephant moving at 2 m/s? How much more energy

s2008m [1.1K]

Answer:

1. The elephant has more kinetic energy at this speed and mass. It has 4,500 J more KE.

2. The elephant would have to go at a speed of 2.5 m/s to reach the same KE as the cheetah.

Explanation:

You would use the formula KE=1/2mv^2.

This formula would be filled in and completed twice, once for the elephant and once for the cheetah.

Cheetah:

KE = 1/2 (40) (25) ^2

KE = 12,500 J

Elephant:

KE = 1/2 (4,000) (2) ^2

KE = 8,000 J

This shows that the cheetah has more KE.

Then you would subtract the elephants amount of J from the cheetahs to find the difference.

Difference = 12,500 J - 8,000 J

Difference = 4,500 J

I hoped this helped with the first part :)

For the second part:

To find the speed the elephant would have to run you would fill in and complete the equation once more with different distance results.

KE = 1/2 (4,000) (2.5) ^2

KE= 12,500 J

7 0

3 years ago