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love history [14]

2 years ago

14

A/ Can someone help me, how can the third law of motion (Newton) help solve a problem / issue. be specific

1 answer:

mezya [45]2 years ago

6 0

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Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

2 years ago

In Lesson 20, a magnesium strip was used to ignite the thermite reaction. When magnesium is placed in a flame from a small blow

nadezda [96]

Answer:

2 electrons will be needed by unbound oxygen in order to fill its 2nd shell.

Explanation:

The chemical reaction between magnesium and oxygen gives magnesium oxide as a product.The reaction is chemically represented as:

$2Mg(s)+O_2(g)\rightarrow 2MgO(s)$

Magnesium is a metal of group-2 with 2 valence electrons.It has atomic number of 12.

$[Mg]=1s^22s^22p^63s^2$

In order to attain noble gas configuration it will loose two electrons.

$[Mg]^{2+}=1s^22s^22p^6$

$Mg\rightarrow Mg^{2+}+2e^-$ ...[1]

Oxygen is a non metal of group-16 with 6 valence electrons..It has atomic number of 8.

$[O]=1s^22s^22p^4$

In order to attain noble gas configuration it will gain two electrons.

$[O]^{2-}=1s^22s^22p^6$

$O+2e^-\rightarrow O^{2-}$ ..[2]

2 electrons will be needed by unbound oxygen in order to fill its 2nd shell.

6 0

3 years ago

What is the primary evidence used to determine how the Moon formed? O A. Moon craters and Earth craters were caused by the same

Leona [35]

Answer is a
Explanation it is a

7 0

2 years ago

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Which of these best defines weather? (3 points) a Atmospheric conditions over 30-year period Ob Day-to-day condition of the atmo

prisoha [69]

Answer:

b Day-to-day condition of the atmosphere

Explanation:

Weather is short term, Climate is long term

3 0

1 year ago

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A baseball is thrown through the air. It's initial velocity, described as a vector, is → v ( t = 0 ) = 17.1 ˆ i + 14.7 ˆ j m / s

ludmilkaskok [199]

Answer:

a = - 9.8 j ^ m/s²

Explanation:

This is a projectile launch problem, they give us the initial velocity in the two components

v₀ₓ = 17.1 m / s

$v_{oy}$ = 14.7 m / s

They indicate that the only acceleration that exists is the acceleration of gravity, which acts in the direction towards the center of the Earth, in general in a coordinate system it coincides with the direction of the y axis.

a = - g j ^

a = - 9.8 j ^ m /s²

6 0

2 years ago