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4 years ago

A substance that speeds up a chemical reaction without undergoing a permanent change itself is a(n)

Physics

1 answer:

ratelena [41]4 years ago

8 0

<h3>Answer:</h3>

Catalyst

<h3>Explanation:</h3>

To answer the following question we need to ask ourselves the following questions;

What is a catalyst?

A catalyst is any substance that increases the rate at which a chemical reaction occurs.

How does it perform its function?

Catalysts increase the rate of chemical reaction without be consumed.
They do so by reducing the amount of activation energy of the reactants. By so doing the rate of formation of products increases.

Do catalysts undergo any change after the reaction?

Catalysts do not undergo permanent changes during the reaction. They remain the same at the end of the reaction/

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What date on the cosmic calendar did the milky way galaxy appear?

lord [1]

<span><span>Big Bang- January 1
</span><span>Origin of Milky Way Galaxy- May 1
</span><span>Origin of the solar system- September 9
</span><span>Formation of the Earth- September 14
</span><span>Origin of life on Earth<span>- September 25</span></span></span>

5 0

3 years ago

A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how

GalinKa [24]

Answer: 473.640 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=48.1 m/s$ is the cannonball's initial velocity

$\theta=0$ because we are told the cannonball is shot horizontally

$t$ is the time since the cannonball is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=1.5m$ is the initial height of the cannonball

$y=0$ is the final height of the cannonball (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it $X_{max}$ and this occurs when $y=0$ .

So, firstly we will find the time with (2):

$0=1.5 m+48. 1 m/s sin(0\°) t-(4.9 m/s^{2})t^2$ (3)

Rearranging the equation:

$0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m$ (4)

$-(4.9 m/s^{2})t^2+(48. 1 m/s) t+1.5 m=0$ (5)

This is a <u>quadratic equation</u> (also called <u>equation of the second degree</u>) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula: