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3 years ago

What will atoms do in a chemical reaction?

Physics

1 answer:

tankabanditka [31]3 years ago

5 0

Answer:

In a chemical reaction, the atoms and molecules that interact with each other are called reactants. ... No new atoms are created, and no atoms are destroyed. In a chemical reaction, reactants contact each other, bonds between atoms in the reactants are broken, and atoms rearrange and form new bonds to make the products.

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Iron (II) hydroxide, whose chemical formula is shown here, is soluble in different solutions. In which solution would the Iron (

Liula [17]

Answer:

Explanation:

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3 years ago

How much force is needed to stop a 50 kg gymnast if he decelerates at 25 m/s^2

vlada-n [284]

The force can be calculated by multiplying the mass of the gymnast with her acceleration.

Force = 50 kg × 25 m/s2
Force = 1250 N

A force of at least 1250 N can stop the 50-kg gymnast.

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3 years ago

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kondor19780726 [428]

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4 years ago

Read 2 more answers

Consider a machine of mass 70 kg mounted to ground through an isolation system of total stiffness 30,000 N>m, with a measured

kvv77 [185]

Answer:

a)0.0229 m

b)0.393 rad

c)1.57

d)707.6 N

e)0.298 m/s

Explanation:

Given:

Mass of the machine, m=70 kg
Stiffness of the system, k=30000 N/m
Damping ratio=0.2
Damping force, F=450 N
Angular velocity $\omega=13\ \rm rad/s$

a)We know that the amplitude X at steady state is given by

$X=\dfrac{\dfrac{F_0}{m}}{\sqrt{\omega_n^2-\omega^2)^2 +(2\rho \omega_n\omega)^2}}\\$

Where

$\omega_n=\sqrt{\dfrac{k}{m}}\\\\=\sqrt{\dfrac{30000}{70}}\\\\=20.7\ \rm rad/s$

$\omega=13\ \rm rad/s$
$\rho=0.2$

$F_0=450\ \rm N$
$m=70\ \rm kg$

$X=\dfrac{\dfrac{450}{70}}{\sqrt{20.7^2-13^2)^2 +(2\times 0.2\times20.7\times13)^2}}\\[tex]X=0.0229\ \rm m$

b) The phase shift of the motion is given by

$\tan\phi=\dfrac{2\rho \omega_n \omega }{\omega_n^2-\omega^2}\\\\\dfrac{2\times0.2\times20.7\times13 }{20.7^2-\13^2}\\\\\phai=0.393\\$

c)Transmissibility ratio is given by

$T.R.=\sqrt{\dfrac{1+(2\rho r)^2}{(1-r^2)^2+{(2\rho r)^2}}}\\\\T.R.=\sqrt{\dfrac{1+(2\times0.2\times0.628)^2}{(1-0.628^2)^2+{(2\times0.2\times0/628)^2}}}\\\\=1.57$