What is Newtrons law

Chemistry

2 answers:

Nutka1998 [239]3 years ago

6 0

Answer:

nuytans law was had a law of moton pls brainly =3

Explanation:

spayn [35]3 years ago

5 0

Answer:

The focus of Lesson 1 is Newton's first law of motion - sometimes referred to as the law of inertia. Newton's first law of motion is often stated as. An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. A force is a push or a pull that acts upon an object as a results of its interaction with another object. ... These two forces are called action and reaction forces and are the subject of Newton's third law of motion. Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction.

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Misha Larkins [42]

Your answer is D, OH-! correct me if I’m wrong.

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3 years ago

Methylamine (ch3nh2) is a weak base. at equilibrium, which expression equals the base dissociation constant (kb) of methylamine?

Molodets [167]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Below are the choices that can be found elsewhere:

a. [CH3NH-] * [OH-] / [CH3NH3+]
<span>b. [CH3NH3+] / ([CH3NH2] * [OH-]) </span>
<span>c. [CH3NH3+] * [OH-] / [CH3NH2] </span>
<span>d. [CH3NH3+] * [OH-] / [CH3NH2] * [H2O]
</span>
The answer is C.

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3 years ago

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konstantin123 [22]

He thought they needed to be idolized

8 0

3 years ago

You are given 10ml (M) 20 Naoh solution in a conical flask and asked to titrate with (M) 20 Hcl and (M) 20 H2so4 separately. cal

Solnce55 [7]

Answer:

$n_{HCl}=0.2molHCl\\n_{H_2SO_4}=0.1molH_2SO_4$

Explanation:

Hello!

In this case, since the reactions between NaOH and the acids are:

$NaOH+HCl\rightarrow NaCl+H_2O\\\\2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

Whereas we can see the 1:1 and 2:1 mole ratios between NaOH and HCl and H2SO4 respectively. In such a way, at the equivalence point we realize that:

$n_{HCl}=n_{NaOH}=V_{NaOH}M_{NaOH}=0.01L*20mol/L=0.2molHCl\\\\2n_{H_2SO_4}=n_{NaOH}\\\\n_{H_2SO_4}=\frac{1}{2} V_{NaOH}M_{NaOH}=\frac{0.01L*20mol/L}{2} =0.1molH_2SO_4$