All categories

1 year ago

Acrostic poem for transformation

1 answer:

5 0

An acrostic poem for transformation simply refers to those simple poems conveying transformation messages in which the first letter of each line forms a word or phrase vertically.

A poem can be defined as a a piece of writing in which the expression of feelings and ideas is given intensity by particular attention to diction.

So therefore, an acrostic poem for transformation simply refers to those simple poems conveying transformation messages in which the first letter of each line forms a word or phrase vertically.

Complete question:

What do you understand by acrostic poem for transformation?

Learn more about poem:

brainly.com/question/9861

#SPJ1

You might be interested in

A 06-C charge and a .07-C charge are apart at 3 m apart. What force attracts them?

Burka [1]

Answer:

the force of attraction between the two charges is 4.2 x 10⁹ N.

Explanation:

Given;

the magnitude of first charge, q₁ = 0.06 C

the magnitude of the second charge, q₂ = 0.07 C

distance between the two charges, r = 3 m

The force of attraction between the two charges is calculated as ;

$F = \frac{Kq_1q_2}{r^2}$

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/C²

$F = \frac{Kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(0.06)(0.07)}{3^2} \\\\F = 4.2 \times 10^{6} \ N$

Therefore, the force of attraction between the two charges is 4.2 x 10⁹ N.

5 0

2 years ago

An 80kg astronaut traveled to the moon, where gravity is one-sixth (116) as

PIT_PIT [208]

Answer:

Wmoon = 131 [N]

Explanation:

We know that the weight of a body is equal to the product of mass by gravitational acceleration.

Since we are told that the gravitational acceleration of the moon is equal to one-sixth of the acceleration of Earth's gravitation. Then we must multiply the value of Earth's gravitation by one-sixth.

$w_{moon}=\frac{1}{6} *m*g\\w_{moon}=\frac{1}{6} *80*9.81\\w_{moon}=130.8 [N] = 131 [N]$

7 0

2 years ago

What is the net work doneon the object over the distance shown?

GuDViN [60]

$A)F_0d$

Explanation

If you graph the force on an object as a function of the position of that object, then the area under the curve will equal the work done on that object, so we need to find the area under the function to find the work

Step 1

find the area under the function.

Area:

$\text{Area}=rec\tan gle_{green}+triangle_{gren}-triangle_{red}$ $\begin{gathered} \text{the area of a rectangle is given by} \\ A_{rec}=lenght\cdot widht \\ \text{and} \\ \text{the area of a triangle is given by:} \\ A_{tr}=\frac{base\cdot height}{2} \end{gathered}$

$\begin{gathered} \text{Area}=rec\tan gle_{green}+triangle_{gren}-triangle_{red} \\ \text{replace} \\ \text{Area}=(F_0\cdot d)+\frac{(F_0\cdot d)}{2}-\frac{(F_0\cdot d)}{2} \\ \text{Area}=(F_0\cdot d) \\ Area=F_0d \end{gathered}$

therefore, the answer is

$A)F_0d$

I hope this helps you

4 0

8 months ago

What is point c called?

FromTheMoon [43]

Then every line segment has one and only one mid-point.

（っ＾▿＾）

5 0

2 years ago

Read 2 more answers

You are in a planet where the acceleration due to gravity is known to be 3.28 m/s^2. You drop a ball and record that the ball ta

notsponge [240]

B) 7.87 m/s

The gravitational pull is the rate of change of velocity which is the acceleration. Formula for acceleration is;
$a = \frac{final \: velocity - initial \: velocity}{time \: taken}$

Given:

• Initial velocity = 0m/s; I dropped the ball, and didn't throw it, so it was at rest firstly
• Time taken = 2.40s
• Acceleration = 3.28m/s^2

We're require to find the final velocity, at which the ball hit the ground with. Ignoring air resistance, keep in mind that the velocity of an object increases as it comes closer to the ground.

$3.28 = \frac{final \: velocity - 0}{2.40}$
$3.28 \times 2.40 = final \: velocity$
$final \: velocity = 7.872 \frac{m}{ {s} }$

5 0

3 years ago