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3 years ago

What structure help support plants

2 answers:

4 0

The cell structure that helps support plants is the cell wall. It provides a shell structure around the main cell of the plant and its layers so then it will have the adequate amount of strength that it needs. It's also one of the reasons why even non-woody plants such as herbs have the strength of support itself.

Viktor [21]3 years ago

3 0

The structure that help support plants is the stem

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. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of

Olegator [25]

Answer:

See Explanation

Explanation:

Given

$s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}$

Solving (a): Units and dimension of $s_0$

From the question, we understand that:

$s \to L$ --- length

$t \to T$ --- time

Remove the other terms of the equation, we have:

$s=s_0$

Rewrite as:

$s_0=s$

This implies that $s_0$ has the same unit and dimension as $s$

Hence:

$s_0 \to L$ --- dimension

$s_o \to$ Length (meters, kilometers, etc.)

Solving (b): Units and dimension of $v_0$

Remove the other terms of the equation, we have:

$s=v_0t$

Rewrite as:

$v_0t = s$

Make $v_0$ the subject

$v_0 = \frac{s}{t}$

Replace s and t with their units

$v_0 = \frac{L}{T}$

$v_0 = LT^{-1}$

Hence:

$v_0 \to LT^{-1}$ --- dimension

$v_0 \to$ $m/s$ --- unit

Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

Hence:

$a_0 = LT^{-2$ --- dimension

$a_0 \to$ $m/s^2$ --- acceleration

Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

$s=\frac{j_0t^3}{6}$

Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

$j_0 = \frac{6s}{t^3}$

Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

Hence:

$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

Solving (e): Units and dimension of $s_0$

Remove the other terms of the equation, we have:

$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

Make $S_0$ the subject

$S_0 = \frac{24s}{t^4}$

Replace s and t with their units [ignore all constants]

$S_0 = \frac{L}{T^4}$

$S_0 = LT^{-4$

Hence:

$S_0 = LT^{-4$ --- dimension

$S_0 \to$ $m/s^4$ --- unit

Solving (e): Units and dimension of $c$

Ignore other terms of the equation, we have:

$s=\frac{ct^5}{120}$

Rewrite as:

$\frac{ct^5}{120} = s$

Make $c$ the subject

$c = \frac{120s}{t^5}$

Replace s and t with their units [Ignore all constants]

$c = \frac{L}{T^5}$

$c = LT^{-5}$

Hence:

$c \to LT^{-5}$ --- dimension

$c \to m/s^5$ --- units

4 0

2 years ago

Mice21 [21]

Answer:

The displacement in t = 0,

y (0) = - 0.18 m

Explanation:

Given f = 40 Hz , A = 0.25m , μ = 0.02 kg / m, T = 20.48 N

v = √ T / μ

v = √20.48 N / 0.02 kg /m = 32 m/s

λ = v / f

λ = 32 m/s / 40 Hz = 0.8

K = 2 π / λ

K = 2π / 0.8 = 7.854

φ = X * 360 / λ

φ = 0.5 * 360 / 0.8 = 225 °

Using the model of y' displacement

y (t) = A* sin ( w * t - φ )

When t = 0

y (0) = 0.25 m *sin ( w*(0) - 225 )

y (0) = 0.25 * -0.707

y (0) = - 0.18 m

5 0

3 years ago

URGENT!!! NEED ANSWER ASAP Students are completing a table about a particular subatomic particle that helps make up an atom. The

Contact [7]

Hmm, I will come back to this one just to help. :)

3 0

2 years ago

Read 3 more answers

A gardener mows a lawn with an old-fashioned push mower. The handle of the mower makes an angle of 320 with the surface of the l

Virty [35]

Answer:

N = 337.96 N

Explanation:

∅ = 32º

F = 249 N

m = 21 Kg

N = ?

We can apply:

∑ F = 0 (↑)

- Fy - W + N = 0 ⇒ N = Fy + W

⇒ F*Sin ∅ + m*g = N

⇒ N = (249 N*Sin32º) + (21 Kg*9.81 m/s²)

⇒ N = 337.96 N (↑)

8 0

2 years ago

15. Some people have observed that animal behavior changes just before an earthquake. What might explain this?

Ber [7]

Answer:

15: Animals are able to detect the first of an earthquake's seismic waves—the P-wave, or pressure wave, that arrives in advance of the S-wave, or secondary, shaking wave. This likely explains why animals have been seen snapping to attention, acting confused or running right before the ground starts to shake.

16: No, it is not even remotely true. Microwaves cook food from the outside in, just like a regular oven. In fact, most of the cooking on the inside of the food, depending on its thickness, is done by heat conduction from the outside surfaces inwards, as the microwaves do not actually penetrate that far into the food.

3 0

1 year ago