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3 years ago

Which type of reaction is this formula an example of? CH3OH + O2→CO2 + 2H2O + heat

Physics

1 answer:

alexdok [17]3 years ago

8 0

Explanation:

This is an exothermic reaction. An exothermic reaction is one which releases heat to the surroundings. When CH3OH reacts with O2, heat is released on the product side of the reaction.

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ElenaW [278]

Answer:

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3 years ago

If you apply an unbalanced force to this object it will ______. Mention all that will apply: change velocity, accelerate, change

Papessa [141]

Answer:

All of them: change velocity, accelerate, change position

Explanation:

We can answer this question by using Newton's second law:

F = ma

where

F is the net force on the object

m is the mass of the object

a is the acceleration

We notice that when there is an unbalanced force on the object, $F\neq 0$ , and therefore

$a\neq 0$

whcih means that the object will accelerate.

But acceleration is the rate of change of velocity, v:

$a=\frac{\Delta v}{\Delta t}$

And so,

$\Delta v \neq 0$

which means that the object will change velocity.

If the object is changing velocity, this means that it is also moving: therefore, the position of the object must be changing, so also the option "change position" is correct.

6 0

3 years ago

What are all the Newton’s laws of motion?

balandron [24]

1st Law Of Motion: An object will often remain at rest or continue moving unless acted upon.
2nd Law Of Motion: Force is equal to mass times acceleration
3rd Law Of Motion: For every action there is an equal or opposite reaction

3 0

3 years ago

Can somebody helppp me pls thank you:)))))

Elena L [17]

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4 years ago

Read 2 more answers

A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130

konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

$W=Fd$ (1)

F: applied force = 130N

d: distance = 5.0m

$W=(130N)(5.0m)=650J$

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

$W_f=F_fd=\mu Mgd$ (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

$W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J$

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

$W = 0J$

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

$\Delta K=W_T$ (3)