Which state of matter would be described as a highly energized charge particles with moving extremely fast

A student is investigating the relationship between sunlight and plant growth for her science expieriment. Determine which of th

lions [1.4K]

The question is incomplete as it does not have the options which have been provided in the attachment.

Answer:

Option-D

Explanation:

In the given question, the effect of the sunlight on the growth of the plant has been studied. The values provided in the Option-D can be considered correct as the values are measured in the decimal value up to two decimal value.

The values are measured after the first week, second week, and the initial readings. The difference in the values provided in Option-D does not show much difference as well as are up to two decimal places.

Thus, Option-D is the correct answer.

4 0

3 years ago

Please help! What's the answer to this question, A 5.6 nC electric charge is placed in an Electric Field and experiences a force

zvonat [6]

Answer:

1321 N/C

Explanation:

From the question,

Electric Field (E) = Electric Force(F)/Electric Charge(q)

E = F/q............ Equation 1

Given: F = 7.4 μN = 7.4×10⁻⁶ N, q = 5.6 nC = 5.6×10⁻⁹ C

Substitute these value into equation 1

E = ( 7.4×10⁻⁶)/(5.6×10⁻⁹)

E = 1.321×10³ N/C

E = 1321 N/C

Hence the magnitude of the electric field at that location is 1321 N/C

3 0

3 years ago

A solid, horizontal cylinder of mass 10.6 kg and radius 1.00 m rotates with an angular speed of 8.00 rad/s about a fixed vertica

n200080 [17]

Answer:

The final angular speed is 7.71 rad/s

Explanation:

Given

Cylinder mass, M = 10.6 kg

Cylinder radius, R = 1.00 m

Angular speed, w = 8.00 rad/s.

Mass of putty, m = 0.250-kg

Radius, r = 0.900 m

First, we set up an expression for the initial and final angular momentum of the system.

The moment of inertia of the cylinder is given as I = ½MR²

While the moment of inertia if the putty is mr².

Initial Momentum of the system = Initial momentum of the cylinder =

Li = Iw --- Substitute ½MR² for I

Li = ½MR²w

By

Substituton

Li = ½ * 10.6 * 1² * 8

Li = 42.4kgm²/s

Calculating the final momentum of the system.

First we calculate the final momentum of the cylinder

Li = Iw --- Substitute ½MR² for I

Li = ½MR²wf where wf = final angular speed

By

Substituton

Li = ½ * 10.6 * 1² wf

Li = 5.3w kgm²/s

Then we calculate the final momentum of the putty

Final Momentum of the putty =

L2 = Iwf --- Substitute mr² for I;

L2 = mr²wf --- By Substituton

L2 = 0.25 * 0.9² * wf

L2 = 0.2025wf kgm²/s

Final momentum = Li + L2

Lf = (5.3wf + 0.2025wf) kgm²/s

Lf = 5.5025wf kgm²/s

By conservation of momentum

Li = Lf

Where Li = 42.4kgm²/s and Lf = 5.5025wf kgm²/s

So, we have

5.5025wf kgm²/s = 42.4kgm²/s --- make wf the subject of formula

wf = 42.4/5.5025

wf = 7.71 rad/s

Hence, the final angular speed is 7.71 rad/s

5 0

3 years ago

A battery with an emf of 12.0 V shows a terminal voltage of 11.8 V when operating in a circuit with two lightbulbs rated at 3.0

Blababa [14]

Answer:

$R = 24 ohm$

Explanation:

As we know that terminal voltage and EMF of cell is given as

$V = EMF - ir$

$ir = 12 - 11.8$

$ir = 0.2 Volts$

resistance of two bulbs is given as

$R = \frac{V^2}{P}$

$R = \frac{12^2}{3}$

$R = 48 ohm$

now these two bulbs are connected in parallel

so equivalent resistance is given as

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$

$\frac{1}{R} = \frac{1}{48} + \frac{1}{48}$

so we have

$R = 24 ohm$

3 0

3 years ago

Imagine you want to create a model of Solar system (and beyond) with the Sun

kondaur [170]

Using the scale model of the Sun given;

The diameter of the Milky Way = 4.28 × 10¹¹ m
The diameter of the Earth = 5.44 × 10⁻³ m

<h3>What is the diameter of the Milky Way?</h3>

The diameter of the Milky Way is about 1 × 10¹⁸ km.

The diameter of the Sun is about 1.4 × 10⁶ km

The diameter of the Earth is about 1.27 × 10⁴ km.

Using the scale model of the Sun given, the diameter of the Milky Way = (1 × 10¹⁸ km/1.4 × 10⁶ km) × 0.6 m

The diameter of the Milky Way = 4.28 × 10¹¹ m

Using the scale model of the Sun given, the diameter of the Milky Way = (1.27 × 10⁴ km/1.4 × 10⁶ km) × 0.6 m

The diameter of the Earth = 5.44 × 10⁻³ m

In conclusion, the diameter of the Milky Way is far bigger than the Sun while the diameter of the Sun is about 5400 times bigger than the Earth.

Learn more about the Milky Way, Sun, and Earth at: brainly.com/question/1995133

#SPJ1

5 0

2 years ago