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3 years ago

Testing Plant Growth with Fertilizer

Physics

1 answer:

kykrilka [37]3 years ago

5 0

b) by the end of the fifth day, group b will be more than one cm taller than group a. eliminate

Explanation:

The correct prediction based on the data shown is that by the end of the fifth day, group B plants will be more than 1cm taller than group A plants.

let us examine the table closely;

Day Group A (Water Only) Plant Height (cm) Group B (Water and Fertilizer)

1 2.0 2.0

2 2.2 2.3

3 2.3 2.8

4 2.5 3.2

5 2.6 3.8

We can see that by the end of the fifth day, the height differences between the two plants, 3.8 - 2.6 = 1.2cm, this is greater than 1cm. This claim from the experiment is very correct.

Other predictions are false.

learn more:

Experiment brainly.com/question/5096428

#learnwithBrainly

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3 years ago

A 15 kg runaway grocery cart runs into a spring with spring constant 230 N/m and compresses it by 56 cm .What was the speed of t

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To solve this problem we will apply the concepts related to the conservation of kinetic energy and elastic potential energy. Thus we will have that the kinetic energy is

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And the potential energy is

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Here,

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Our values are given as,

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Replacing we have that

$\frac{1}{2} (15)v^2 = \frac{1}{2} (230)(0.56)^2$

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4 years ago

A 0.100-kg stone rests on a frictionless, horizontal surface. A bullet of mass 6.00 g, travel- ing horizontally at 350 m>s, s

mariarad [96]

Answer:

Magnitude is 25.8 m/s and direction is $35.5^{o}$

Explanation:

From the law of conservation of linear momentum

$m_{b}v_{ib}+ m_{s}v_{is}= m_{b}v_{fb}+ m_{b}v_{fs}$ where $m_{b}$ and $m_{s}$ are masses of bullet and stones respectively, $v_{ib}$ and $v_{is}$ are the initial velocities of bullet and stone respectively, $v_{fb}$ and $v_{fs}$ are the final velocities of bullet and stone respectively

Substituting 6g=0.006 Kg for mass of bullet, 0.1Kg for mass of stone, 350 m/s for initial velocity of bullet and 250 m/s as final velocity for stone but initial velocity of stone is zero

$0.006 Kg *350 m/s (\hat i)+0.1 Kg*0=0.006 Kg* 250 m/s (\hat j)+0.1*v_{fs}$