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3 years ago

Which cekk part is found in a plant cells but not animal cells?

Physics

2 answers:

liraira [26]3 years ago

7 0

The correct answer should be c

Citrus2011 [14]3 years ago

3 0

Answer:

cell wall ...

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¿ cuales son las especies q mas características tienes en común ?

Katena32 [7]

De los reptiles, anfibios; pájaros, peces; mamíferos y artrópodos, los que más se asemejan en características son los pájaros y los mamíferos porque ambos son de sangre caliente y no varia independientemente de en qué ambiente se encuentre

5 0

3 years ago

PLSSSS HELPPPP PLSSSS 25 POINTS PLS I REALLY NEED THIS AND GET THEM RIGHT SO PLS DONT ANSWER IF U DONT KNOW

denis23 [38]

Answer:

1- C 2-B 3-B - these are ur best answers

Explanation:

4 0

3 years ago

Read 2 more answers

A baseball player friend of yours wants to determine his pitching speed. you have him stand on a ledge and throw the ball horizo

zhenek [66]

Answer:

The pitching speed of your friend is 33.20 m/s

Explanation:

Lets explain how to solve the problem

Your friend throw the ball horizontally that means the vertical initial

component of velocity is zero ( $u_{y}=0$ ).

The ball is thrown from a height 4 meters above the ground.

The height $h=u_{y}t+\frac{1}{2}gt^{2}$

Remember: the height is negative value because its below the point of

thrown (initial position)

h = -4 m , $u_{y}=0$ and g = -9.8 m/s²(downward)

Substitute these values in the rule above

⇒ $4=0-\frac{1}{2}(9.8)t^{2}$

⇒ -4 = -4.9t² (multiply both sides by -1)

⇒ 4 = 4.9t² (divide both sides by 4.9)

⇒ 0.81633 = t² (take √ for both sides)

⇒ t = 0.9035

Then the time of the ball to land on the ground is 0.9035 seconds

The range of the ball on the ground is 30 m

The range $R=u_{x}*t$ , where $u_{x}$ is the horizontal

component of the initial velocity

R = 30 meters and t = 0.9035

⇒ $30=u_{x}(0.9035)$ (divide both sides by 0.9035)

⇒ $u_{x}=33.20$ m/s

The pitching speed of your friend is 33.20 m/s 

4 0

3 years ago

Starting from rest, a 68.0 kg woman jumps down to the floor from a height of 0.790 m, and immediately jumps back up into the air

Aleonysh [2.5K]

Answer:

a) I = 0 N s, b) v = -3.935 m / s, c) vf = 3.935 m / s, d) y = 0.790 m

Explanation:

a) Let's start by defining the upward direction (+ y) as positive. For this part of the exercise we must use the momentum relationship

I = ∫F dt

The force of the woman on the floor is given and by action and rection the floor exerts on the woman a force of equal magnitude, but opposite direction

I = ∫ (9200 t - 11500 t2) dt

I = 9200 t² / 2 - 11500 t³ / 3

We evaluate between the lower limit t = 0 and upper limit t = 0.800 s

I = 9200 (0.8² -0) - 11500 (0.8³ -0)

I = 5888 -5888

I = 0 N s

Directed from the floor to the woman

b) For this part we use kinematics

v² = v₀² - 2g y

v = √ (0 - 2 9.8 (-0.79))

v = 3.935 m / s

The speed direction is down

c) for this we use the relationship between momentum and the amount of movement

I = ΔP

I = m vf - m v₀

vf = (I + m v₀) / m

This is the impulse of women on the floor

vf = ( 0 + 68 (3.935)) / 68

vf = 3.935 m / s

d) let's use kinematics

v₂ = v₀² - 2gy

0 = v₀² - 2gy

y = v₀² / 2g

y = 3.935²/2 9.8

y = 0.790 m

8 0

3 years ago

The planet Krypton has a mass of 8.8 × 1023 kg and radius of 2.5 × 106 m. What is the acceleration of an object in free fall nea

djverab [1.8K]

Answer:

Acceleration, $a=9.39\ m/s^2$

Explanation:

Given that,

Mass of the planet Krypton, $m=8.8\times 10^{23}\ kg$

Radius of the planet Krypton, $r=2.5\times 10^{6}\ m$

Value of gravitational constant, $G=6.6726\times 10^{-11}\ Nm^2/kg^2$

To find,

The acceleration of an object in free fall near the surface of Krypton.

Solution,

The relation for the acceleration of the object is given by the below formula as :

$a=\dfrac{Gm}{r^2}$

$a=\dfrac{6.6726\times 10^{-11}\times 8.8\times 10^{23}}{(2.5\times 10^{6})^2}$

$a=9.39\ m/s^2$

So, the value of acceleration of an object in free fall near the surface of Krypton is $9.39\ m/s^2$

5 0

3 years ago