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Ksivusya [100]
3 years ago
14

An example of Newton's third law of motion is...

Physics
1 answer:
kipiarov [429]3 years ago
3 0
Newton’s third law states that for every action there is an equal and opposite reaction therefore, the third answer is right
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Physics help, thank you guys so much!
katrin2010 [14]

Answer:

 Δt = 5.85 s

Explanation:

For this exercise let's use Faraday's Law

           emf = -  \frac{d \phi}{dt} -  d fi / dt

           \phi = B. A

           \phi = B A cos θ

The bold are vectors. It indicates that the area of the body is A = 0.046 m², the magnetic field B = 1.4 T, also iindicate that the normal to the area is parallel to the field, therefore the angle θ = 0 and cos 0 =1.

suppose a linear change of the magnetic field

            emf = - A \frac{B_f - B_o}{ \Delta t}

             Dt = - A  \frac{B_f - B_o}{emf}

the final field before a fault is zero

       

let's calculate

            Δt = - 0.046 (0- 1.4) / 0.011

            Δt = 5.85 s

4 0
3 years ago
Help Plsss
sattari [20]
On the Newtonian theory of gravity, gravitation affects anything with mass. Assuming that none of the answer choices is the only thing that exists in the universe, all of the answer choices are subject to the law of universal gravitation (hence “universal”).

Satellites, water, frogs, and stars all have mass as they are all composed of matter. Thus, all four answer choices should be circled.
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3 years ago
Which law says that the induced voltage is directly proportional to the magnetic flux?
Vsevolod [243]
law of electromagnetic induction hope this helps 
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3 years ago
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Which is a legacy of the enlightenment?
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Education that result in understanding and the spread knowledge physics
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A person carries a plank of wood 1.6 m long with one hand pushing down on it at one end with a force F1 and the other hand holdi
Elis [28]

Answer: 115.52\ N

Explanation:

Given

Length of plank is 1.6 m

Force F_1 is applied on the left side of plank

Force F_2 is applied 43 cm from the left end O.

Mass of the plank is m=13.7\ kg

for equilibrium

Net torque must be zero. Taking torque about left side of the plank

\Rightarrow mg\times 0.8-F_2\times 0.43=0\\\\\Rightarrow F_2=\dfrac{13.7\times 9.8\times 0.8}{0.43}\\\\\Rightarrow F_2=249.78\ N

Net vertical force must be zero on the plank

\Rightarrow F_1+W-F_2=0\\\Rightarrow F_1=F_2-W\\\Rightarrow F_1=249.78-13.7\times 9.8\\\Rightarrow F_1=115.52\ N

8 0
2 years ago
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