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denis-greek [22]
3 years ago
5

What determines the path that an object in projectile motion follows?

Chemistry
2 answers:
mafiozo [28]3 years ago
8 0

Answer:

gravity and air resistance

Explanation:

As the ball is moving in projectile motion it has to overcome the Earth’s force of gravity which is acting on the ball and pulling it downwards. Also the ball has to overcome the resistance offered by the air particles which tends to pull the ball in the backward direction opposing its forward motion.

Andre45 [30]3 years ago
5 0

Projectile is like a cannon shooting a ball so it will be in "free fall"

While in free fall you are subject to gravity pulling the ball down.

Also since it's flying through the air there is air resistance.


So GRAVITY AND AIR RESISTANCE


hope it helped


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PolarNik [594]

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8 0
3 years ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔
Vilka [71]

Answer : The correct option is, (C) 1.1

Solution :  Given,

Initial moles of N_2O_4 = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

The given equilibrium reaction is,

                           N_2O_4(g)\rightleftharpoons 2NO_2(g)

Initially                      c                 0

At equilibrium   (c-c\alpha)           2c\alpha

The expression of K_c will be,

K_c=\frac{[NO_2]^2}{[N_2O_4]}

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

where,

\alpha = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}

K_c=1.066\aprrox 1.1

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0
2 years ago
The right line is a 90° clockwise rotation of the left line about the origin. Click the 90° clockwise button. Are these lines th
inessss [21]

Answer:

Switch the coordinates and change the sign of the second one by multiplying it by negative 1.

Explanation:

Here are some examples and a more general way to understand the problem.

Consider the point (1,1), a 90 degree rotation clockwise about the origin would move it into the 4th quadrant.

The new point is (1,-1) , similarly (-4,2)-> (2,4), (-4,3)-> (3,4)

We take a point p= (x,y) the the result of rotation p 90 clockwise about the orgin is a new point p'=(x',y')= (-y, x). .

In the case of p=(1,0) the new point is p'= (0, -1)

One can use a matrix where the first row is cos(a), sin(a) and the second row is

-sin(a) cos(a) for any clockwise rotation of a degrees about the origin.

If we let a=90 degrees we have

[0 1] as the first row and [-1 0] as the second row. So the matrix is:

|0 1|

|-1 0|

Call that matrix M

So a point p= (x,y) can be multiplied by M as follows Mp=p' where p' is the rotated point.

If p=(-4,2) then Mp

is M(-4,2) which after matrix multiplication means x'=0*-4+1*2=2 and y'=-1*-4+0*2=4

So p'=(2,4)

Try it with (1,0)

x'=1*0+0*1=0

y'=-1*1+0*1=-1

so p'=(0,-1) and (1,0)->(0,-1)

How about the point on the y axis (0,1), it should go to the point (1,0)

0*1+1*1=1 and -1*0+0*1 gives you the pont (1,0) ( we don't see the negative sign because -0 is just 0)

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3 years ago
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