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brilliants [131]
3 years ago
7

The chemical equation provided shows iron rusting to form iron oxide. Use the drop-down menu to choose the coefficients that wil

l balance the chemical equation.
()Fe + ()O2 → ()Fe2O3

Physics
2 answers:
attashe74 [19]3 years ago
7 0

Answer:

here is the answer

Explanation:

have a good day

Brrunno [24]3 years ago
3 0

Answer:

4 Fe + 3 O_2 \rightarrow 2 Fe_2 O_3

Explanation:

The original equation is:

Fe + O_2 \rightarrow Fe_2 O_3

We notice that:

- we have 1 atom of Fe on the left, and 2 atoms of Fe on the right

- we have 2 atoms of O on the left, and 3 atoms of O on the right

Therefore, the equation is not balanced.

In order to balance it, we can add:

- a coefficient 3 in front of O_2

- a coefficient 2 in front of Fe_2 O_3

So we have:

Fe + 3 O_2 \rightarrow 2Fe_2O_3

Now the oxygen is balanced, but the iron it not balanced yet, since we have 1 Fe on the left and 4 on the right. Therefore, we should add a coefficient 4 on the Fe on the left:

4 Fe + 3 O_2 \rightarrow 2 Fe_2 O_3

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The antiseptics used for cuts and wounds in human flesh have low surface tension. Why?
lakkis [162]
Because liquids with low surface tension spread easily, and therefore are easier to apply to cuts and wounds. 
4 0
3 years ago
A 100-kg block being released from rest from a height of 1.0 m. It then takes it 1.40 s to reach the floor. What is the mass m o
dimulka [17.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The mass of the other block is  m_1 = 81.14 \ kg

Explanation:

From the question we are told that

   Mass of the first block is  m_1 = 100 \  kg

   The height is  s =  1.0 \  m

   The time it takes it is  t = 1.40 \  s

 Generally from kinematic equation

       s =  ut + \frac{1}{2} at^2

Here u  is the initial velocity which zero given that it was at rest initially

So

     s =  0 * t + \frac{1}{2} at^2

=>  s =   \frac{1}{2} at^2

=> 1 =   \frac{1}{2}*  a *  (1.40 )^2

=>  a = 1.0204 \  m/s^2

Generally from the diagram the resultant force due to the weight of the first object and the tension on the string is  mathematically represented as

      mg - T = ma

=>   T  =  m g - ma

=>   T  =  m(g - a)

=>   T  =  877.96 \  N

Generally from the diagram the resultant force due to the weight of the second object and the tension on the string is  mathematically represented as  

     T - m_1g  =  m_1 a

=>   877.96  =  m_1 (a + g)

=>   877.96  =  m_1 (1.0204  + 9.8 )

=>   m_1 = 81.14 \ kg

5 0
3 years ago
when there is a constant unbalanced force applied in the opposite direction as the motion, speed will __________ and acceleratio
Nutka1998 [239]
brainly.com/question/20066572
6 0
2 years ago
Why are lighter cars allowed to drive on motorways at higher speeds than heavier ones?
daser333 [38]

Answer:

say a lighter car is going 80mph and a heavier car is going the same speed who do you think will have more damage i will say the heavier car

Explanation:

3 0
3 years ago
A 3.00-kg block starts from rest at the top of a 33.0° incline and slides 2.00 m down the incline in 1.80 s. (a) Find the accele
ElenaW [278]

(a) 1.23 m/s^2

Let's analyze the motion along the direction of the incline. We have:

- distance covered: d = 2.00 m

- time taken: t = 1.80 s

- initial velocity: u = 0

- acceleration: a

We can use the following SUVAT equation:

d = ut + \frac{1}{2}at^2

Since u=0 (the block starts from rest), it becomes

d=\frac{1}{2}at^2

So by solving the equation for a, we find the acceleration:

a=\frac{2d}{t^2}=\frac{2(2.00 m)}{(1.80 s)^2}=1.23 m/s^2

(b) 0.50

There are two forces acting on the block along the direction of the incline:

- The component of the weight parallel to the surface of the incline:

W_p = mg sin \theta

where

m = 3.00 kg is the mass of the block

g = 9.8 m/s^2 is the acceleration due to gravity

\theta=33.0^{\circ} is the angle of the incline

This force is directed down along the slope

- The frictional force, given by

F_f = - \mu mg cos \theta

where

\mu is the coefficient of kinetic friction

According to Newton's second law, the resultant of the forces is equal to the product between mass and acceleration:

W-F_f = ma\\mg sin \theta - \mu mg cos \theta = ma

Solving for \mu, we find

\mu = \frac{g sin \theta - a}{g cos \theta}=\frac{(9.8 m/s^2)sin 33.0^{\circ} - 1.23 m/s^2}{(9.8 m/s^2) cos 33.0^{\circ}}=0.50

(c) 12.3 N

The frictional force acting on the block is given by

F_f = \mu mg cos \theta

where

\mu = 0.50 is the coefficient of kinetic friction

m = 3.00 kg is the mass of the block

g = 9.8 m/s^2 is the acceleration of gravity

\theta=33.0^{\circ} is the angle of the incline

Substituting, we find

F_f = (0.50)(3.00 kg)(9.8 m/s^2) cos 33.0^{\circ} =12.3 N

(d) 6.26 m/s

The motion along the surface of the incline is an accelerated motion, so we can use the following SUVAT equation

v^2 - u^2 = 2ad

where

v is the final speed of the block

u = 0 is the initial speed

a = 1.23 m/s^2 is the acceleration

d = 2.00 m is the distance covered

Solving the equation for v, we find the speed of the block after 2.00 m:

v=\sqrt{u^2 + 2ad}=\sqrt{0^2+2(9.8 m/s^2)(2.00 m)}=6.26 m/s

5 0
3 years ago
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