What is the outermost structure in a plant cell

1 answer:

5 0

Molecular structure of the primary cell wall in plants. Up to three strata or layers may be found in plant cell walls: The middle lamella, a layer rich in pectins. This outermost layer forms the interface between adjacent plant cells and glues them together.

You might be interested in

A 25-kg child sits at the top of a 4-meter slide. After sliding down, the child is traveling at 5 m/s. How much PE does he start

Semmy [17]

Daniddmelo says it right there, don't know why he got reported.

The potential energy (PE) is mass x height x gravity. So it would be 25 kg x 4 m x 9.8 = 980 joules. The child starts out with 980 joules of potential energy. The kinetic energy (KE) is (1/2) x mass x velocity squared. KE = (1/2) x 25 kg x 5 m/s2 = 312.5 joules. So he ends with 312.5 joules of kinetic energy. The Energy lost to friction = PE - KE. 980- 312.5 = 667.5 joules of energy lost to friction.

Please don't just copy and paste, and thank you Dan cause you practically did it I just... elaborated more? I dunno.

4 0

3 years ago

a 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its kinetic energy(KE) when it jumps PLEASE HELP ME

Kay [80]

Answer:

0.91 J

Explanation:

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

For the lemming in this problem, when he jumps, we have:

m = 0.0780 kg is the mass

v = 4.84 m/s is the speed

Substituting into the equation, we find:

$K=\frac{1}{2}(0.0780)(4.84)^2=0.91 J$

8 0

3 years ago

A playground carousel is rotating counterclockwise about its center on frictionless bearings. A person standing still on the gro

Shtirlitz [24]

Answer:

1.94601 rad/s

Explanation:

$I_1$ = Moment of inertia of carousel = 124 kgm²

$\omega_1$ = Angular speed of carousel = 3.5 rad/s

$\omega_2$ = Angular speed of person

r = Radius of carousel = 1.53 m

m = Mass of person = 42.3 kg

Moment of inertia of person

$I_2=mr^2\\\Rightarrow I_2=42.3\times 1.53^2\\\Rightarrow I_2=99.02007\ kgm^2$

As the angular momentum is conserved in the system

$I_1\omega_1=(I_1+I_2)\omega_2\\\Rightarrow \omega=\frac{I_1\omega_1}{(I_1+I_2)}\\\Rightarrow \omega_2=\frac{124\times 3.5}{(124+99.02007)}\\\Rightarrow \omega_2=1.94601\ rad/s$