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3 years ago

Which two structures are not found in animal cells?

Physics

1 answer:

Alinara [238K]3 years ago

7 0

Answer:

Explanation:

So, lets start off by thinking about what we know is in both cells, and rule out from there.

There are just some basic things you need in eukaryotic cells, those include the following:

Nucleus
cell membrane
mitochondria
cytoplasm(dont get confused with chloroplast)
Rough and smooth endoplasmic reticulum
Golgi vescile
Golci apparatus
Nucleolus(Dont get confused with nulceus)

Here are jsut a few examples of what can be found in both plant and animal cells, however we can already rule out all of the following except for:

Because remember, c talks about the chlorplast, which is not the chlorplasm found in both plant anf animal cells, its just the chloroplast found in plant cells.

Although I gave the asnwer to you. Here are some things that are in plant cells that arent in animal cells:

Cell wall
vacioles
central vaciole
chlroplast
plastids

Animal cells on the other hand, have a few things that plant cells do not as well:

Lysome
Centrolsome

Hope this helps! ;)

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Explanation:

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From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 8.10 m/s and angle

GarryVolchara [31]

Answer:

Part a)

$y = 88.5 m$

Part b)

$v_x = 7.7 m/s$

$v_y = 2.5 m/s$

Part c)

equation of position in x direction is given as

$x = v_x t$

equation of position in y direction is given as

$y = v_y t + \frac{1}{2}gt^2$

Part d)

$x = 30.8 m$

Part e)

H = 88.5 m

Part f)

t = 1.2 s

Explanation:

As we know that ball is projected with speed

v = 8.10 m/s at an angle 18 degree below the horizontal

so we will have

$v_x = 8.10 cos18$

$v_x = 7.7 m/s$

$v_y = 8.10 sin18$

$v_y = 2.5 m/s$

Part a)

Since it took t =4 s to reach the ground

so its initial y coordinate is given as

$y = v_y t + \frac{1}{2}a_y t^2$

$y = 2.5(4) + \frac{1}{2}(9.81)(4^2)$

$y = 88.5 m$

Part b)

components of the velocity is given as

$v_x = 8.10 cos18$

$v_x = 7.7 m/s$

$v_y = 8.10 sin18$

$v_y = 2.5 m/s$

Part c)

equation of position in x direction is given as

$x = v_x t$

equation of position in y direction is given as

$y = v_y t + \frac{1}{2}gt^2$

Part d)

distance where it will strike the floor is given as

$x = v_x t$

$x = 7.7 \times 4$

$x = 30.8 m$

Part e)

Height from which it is thrown is same as initial y coordinate of the ball

so it is given as

H = 88.5 m

Part f)

time taken by ball to reach 10 m below is given as

$y = v_y t + \frac{1}{2}gt^2$

$10 = 2.5t + \frac{1}{2}(9.81) t^2$

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3 years ago

A block with mass of 10 kg is on a frictionless surface. One hand on the left side of the block is pushing it to the right. A se

igomit [66]

Answer:

$W_2=-12J$

Explanation:

The work of force 2 will be given by the vectorial equation $W_2=F_2.d$ . We know the value of $F_1$ and have information about its movement, which relates to the net force $F=F_1+F_2$ .

About this movement we can obtain the acceleration using the equation $v_f^2=v_i^2+2ad$ . Since it departs from rest we have $a=\frac{v_f^2}{2d}$ .

And then using Newton's 2dn Law we can obtain the net force F=ma, thus we will have $F_2=F-F_1=ma-F1=\frac{mv_f^2}{2d}-F_1$

And we had the work done by force 2 as:

$W_2=F_2.d=\frac{mv_f^2}{2}-F_1d$

(The sign will be given algebraically since we take positive the direction to the right.)

With our values:

$W_2=\frac{(10kg)(2m/s)^2}{2}-(8N)(4m)=-12J$

Another (shorter but maybe less intuitive way for someone who is learning) way of doing this would have been to say that the work done by both forces would be equal to the variation of kinetic energy:

$W_1+W_2=K_f-K_i=K_f=\frac{mv_f^2}{2}$

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3 years ago

If the mass of block B is 2kg, the gravitational force exerted on block B is most nearly which of the following?

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Answer:

20 N.

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Gravitational force (F) =..?

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Force = mass (m) × acceleration (a)

F = ma

Thus, we can say that the gravitational force on block B will be the product of the mass of block B and acceleration due to gravity i.e

Gravitational force (F) = mass of block B (m) × acceleration due to gravity (g)

F = mg

Mass of block B (m) = 2 Kg

Acceleration due to gravity (g) = 10 m/s²

Gravitational force (F) =..?

F = mg

F = 2 × 10

F = 20 N

The, the gravitational force exerted on block B is 20 N

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3 years ago