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murzikaleks [220]
3 years ago
14

Newton's third law states that for every action force there is an equal and opposite reaction force. An idiot in your class says

, "Wow that means everything cancels and nothing ever moves, it is all an illluussion! Wowwwwww" What statement best proves to him he is an idiot.
a) The equal and opposite forces act on different objects
b) If there is even a slight imbalance in the third law there will be a net force causing acceleration
C) he's right, all forces cancel, any motion I have ever seen is wrong
Physics
2 answers:
Otrada [13]3 years ago
4 0
A. Because the third laws say that for every action force the is an equal and opposite reaction force
Veronika [31]3 years ago
4 0

\huge \mathfrak{Answer.... }

The Correct Answer is :

B. if there is even a slight imbalance in third law there will be a net force causing acceleration.

A slight difference in the forces can result in acceleration of an object.

\mathrm{✌TeeNForeveR✌}

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D. It oxidises /////
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2 years ago
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What type of wave passes through the spring in the frog toy? Why?
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Is the Frog Toy a piece of literature? If so, the type of wave is the "Wave of Life." Spring brings new life and a new start from the harsh winter.

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3 years ago
The specific weight of sea water is 10.1 kN/m^3. Convert to lbs/in^3.
Viktor [21]

Answer:

0.03719 lbs/in³

Explanation:

Specific weight is given by multiplying the density of an object to the acceleration due to gravity.

\gamma =\rho g\\\Rightarrow \rho=\frac{gamma}{g}\\\Rightarrow \rho=\frac{10.1\times 10^3}{9.81}\\\Rightarrow \rho=1029.562\ kg/m^3

1\ kg=2.20462\ lb

1\ m=39.3701\ in

\\\Rightarrow 1029.562\ kg/m^3=\frac{1029.562\times 2.20462}{39.3701^3}=0.03719\ lbs/in^3

So,

10.1\ kN/m^3=0.03719\ lbs/in^3

8 0
3 years ago
The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pres
12345 [234]

Answer:0.0704 kg

Explanation:

Given

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V=0.025 m^3

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as the volume remains constant therefore

\frac{P_1}{T_1}=\frac{P_2}{T_2}

\frac{311.325}{298}=\frac{P_2}{323}

P_2=337.44 KPa

therefore Gauge pressure is 337.44-101.325=236.117 KPa

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m_1=0.91 kg

Final mass m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}

m_2=0.839

Therefore m_1-m_2=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back

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3 years ago
A spring with a force constant of 5.3 n/m has a relaxed length of 2.60 m. when a mass is attached to the end of the spring and a
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This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant. 
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