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kobusy [5.1K]
1 year ago
9

How do characteristics help you to classify seeds?

Physics
1 answer:
Artyom0805 [142]1 year ago
5 0

Characteristics help us to classify seeds because different plants have different features.

<h3>How are characteristics used to identify and classify plants?</h3>

The divisions classify plants that are based on whether they reproduce by spores or seeds. Spore-bearing plants include ferns, club mosses, and horsetail while on the other hand, Seed-bearing plants are divided into gymnosperms and angiosperms. Different plants have different characteristics and features so on the basis of these characteristics we can easily classify seeds whether they belong from angiosperm and gymnosperm.

So we can conclude that characteristics help us to classify seeds because different plants have different features.

Learn more about seeds here: brainly.com/question/18799172

#SPJ1

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A particle of mass m collides with a second particle of mass m. Before the collision, the first particle is moving in the x-dire
oee [108]

Answer:

a) v, v

b) 2mv^2

c) Elastic collion

Explanation:

(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v).  From momentum conservation in x-direction

Here x, y represent direction.They are not variable. 1 and 2 represent before and after.

2vm=v1xm+v2xm, we find v1x=v.

From momentum conservation in y-direction

0 =v1ym+v2ym, we findv1y=v.

(b) By energy conservation principle

Before: K=1/2m(2v)^2=2mv^2.

After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2

(c) The collision is elastic

6 0
3 years ago
How much force is required to accelerate a 12 kg mass at 5 m/s 2
Savatey [412]

Answer:

60 N

Explanation:

This is just Newton's Second Law

F = m*a

F = ?

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a = 5 m/^2

F = 5*12 = 60 Newtons

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g100num [7]

Answer:

u = 7m {s}^{ - 1}

Explanation:

We know that when we don't have air friction on a free fall the mechanical energy (I will symbololize it with ME) is equal everywhere. So we have:

me(1) = me(2)

where me(1) is mechanical energy while on h=10m

and me(2) is mechanical energy while on the ground

Ek(1) + DynamicE(1) = Ek(2) + DynamicE(2)

Ek(1) is equal to zero since an object that has reached its max height has a speed equal to zero.

DynamicE(2) is equal to zero since it's touching the ground

Using that info we have

m \times g \times h   =   \frac{1}{2}  \times m \times u {}^{2} \\

we divide both sides of the equation with mass to make the math easier.

9.8 \times 10 =  \frac{1}{2}  \times u {}^{2}  \\  \frac{98}{2}  = u {}^{2}  \\ u { }^{2} = 49 \\ u = 7

7 0
3 years ago
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