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2 years ago

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An internet radio broadcast isA. not excludable and not rival in consumption.B. excludable and not rival in consumption.C. exclu

dable and rival in consumption.D. not excludable and rival in consumption.

1 answer:

rodikova [14]2 years ago

8 0

Answer:

A. not excludable and not rival in consumption.

Explanation:

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What’s Chemical energy

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Answer:

Chemical energy is energy that chemical substance to undergoes a chemical reaction to make a new substance.

Explanation:

<h3>hope this makes sense and helps</h3>

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Biologists use ball-and-stick models to study complex molecules by representing atoms with balls and bonds with sticks. DNA is a

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Mathew throws a ball straight up into the air. It rises for a period of time and then begins to drop. At which points in the bal

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2 years ago

You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 35.0 m above its l

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This question can be solved by using the equations of motion.

a) The initial speed of the arrow is was "9.81 m/s".

b) It took the arrow "1.13 s" to reach a height of 17.5 m.

a)

We will use the second equation of motion to find out the initial speed of the arrow.

$h= v_it + \frac{1}{2}gt^2\\$

where,

vi = initial speed = ?

h = height = 35 m

t = time interval = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}$

<u>vi = 9.81 m/s</u>

b)

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

$h= v_it + \frac{1}{2}gt^2\\$

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 17.5 m

vi = initial speed = 9.81 m/s

t = time = ?

Therefore,

$17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0$

solving this quadratic equation using the quadratic formula, we get:

t = -3.13 s (OR) t = 1.13 s

Since time can not have a negative value.

Therefore,

<u>t = 1.13 s</u>

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

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