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2 years ago

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NO LINKS: Which of the following best demonstrates Newton's Third Law?

1 answer:

irina1246 [14]2 years ago

7 0

The best demonstration that applies to Newton's Third Law of motion would be D) When you walk your foot pushes down on the ground while the ground pushes back on your foot.

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Newton's Third Law states that for every action, there is an equal and opposite reaction. This is actually explains that forces come in pairs and forces are an interaction between two objects. As per the correct option given in the question explains Newton’s Third Law.

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When you walk your foot (say object A) pushes down on the ground while the ground (say object Q) pushes back on your foot with the same force but in the opposite direction.

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What is the magnitude of the electric force between a proton and an electron when they are at a distance of 4.09 angstrom from e

Zinaida [17]

Answer:

$F=1.38*10^{-9}N$

Explanation:

According to Coulomb's law, the magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

$F=\frac{kq_1q_2}{d^2}$

Here k is the Coulomb constant. In this case, we have $q_1=-e$ , $q_2=e$ and $d=4.09*10^-10m$ . Replacing the values:

$F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-1.6*10^{-19}C)(1.6*10^{-19}C)}{(4.09*10^{-10})^2}\\F=-1.38*10^{-9}N$

The negative sign indicates that it is an attractive force. So, the magnitude of the electric force is:

$F=1.38*10^{-9}N$

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2 years ago

If the velocity of the particle is uniform, then the path of the particle must be a

lianna [129]

Answer:

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3 years ago

The root-mean-square speed (thermal speed) of the molecules of a gas is 200m/s at 23.0°C. At 227°C the root-mean-square speed (t

777dan777 [17]

Answer:

330 m/s approx

Explanation:

The RMS speed of a gas is proportional to square root of its absolute temperature is

V ( RMS ) ∝ √T

$\frac{V_1}{V_2} =\sqrt{\frac{T_1}{T_2} }$

Here V₁ = 200 , T₁ = 23 +273 = 300K , T₂ = 227 +273 = 500 K

Putting the values

200 / V₂ = $\sqrt{\frac{300}{500} }$

V₂ = 330 m/s approx

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3 years ago

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3 years ago

A 10-meter-long, 150 kg beam extends horizontally from a wall.One end of the beam is fixed to the wall and the other end is atta

kow [346]

Answer:

the magnitude of the tension on the cable necessary to keep the system in equilibrium is 1060.9 N

Explanation:

Given that;

Length L = 10 m

mass of beam m_b = 150 kg; weight W_beam = 150×9.8

mass of sign m = 75 kg

distance of sign hung from the beam from the wall d = 2.50 m

angle ∅ = 60°

g = 9.8 m/s²

Now,

Torque acting at one end of the beam will be;

$T_{net}$ = Tsin∅ × L - mg(d)-W × (L/2)

for equilibrium, $T_{net}$ = 0

therefore, 0 = Tsin∅ × L - mg(d)-W × (L/2)

so we substitute

Tsin(60°) × 10 - 75×9.8(2.50) - 150 × 9.8× (10/2) = 0

Tsin(60°) × 10 - 1837.5 - 7350 = 0

Tsin(60°) × 10 - 9187.5 = 0

Tsin(60°) × 10 = 9187.5

divide both side by 10

Tsin(60°) = 918.75

T × 0.8660 = 918.75

T = 918.75 / 0.8660

T = 1060.9 N

Therefore, the magnitude of the tension on the cable necessary to keep the system in equilibrium is 1060.9 N

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3 years ago