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2 years ago

Which sentence describes Newton's third law?

Physics

1 answer:

Elena-2011 [213]2 years ago

7 0

Option D. The image shows a pattern made by magnetic field lines around the magnet.

<h3>What is Newton's third of motion?</h3>

Newton's third of motion states that for every action, their is equal and opposite reaction.

When a magnet is placed in a magnetic field, it creates magnetic field lines because it is the region of space where its influence is felt.

Thus, the image shows a pattern made by magnetic field lines around the magnet.

Learn more about magnetic field here: brainly.com/question/7802337

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8. A 2 kg flower pot weighing 20 N falls from a window ledge.

Alina [70]

The force of the air resistance is 4 N.

The given parameters;

mass of the flower pot, m = 2 kg

weight of the flower pot, W = 20 N

Let the air resistance = F

Apply Newton's second law of motion to determine the force of the air resistance acting upward to oppose the motion of the pot falling downwards.

$\Sigma F = ma\\\\W - F = ma\\\\a = \frac{W - F}{m} \\\\8 = \frac{20 - F}{2} \\\\20 - F = 16\\\\F = 20 - 16\\\\F = 4 \ N$

Thus, the force of the air resistance is 4 N.

Learn more here: brainly.com/question/19887955

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3 years ago

A truck slows from a velocity of 25 m/s to a stop in 70 m. What was the truck’s acceleration

igomit [66]

v^2-u^2=2 x a x d

25^2-0^2=2 x a x 70

625-0=140 x a

625=140a

a=625/140

a=4.46 m/s^2

im not very sure but i think this is how you do this

4 0

2 years ago

Enem 2003 embalagens longa vida brick

tresset_1 [31]

Answer:

Jyftfufhfucyf fyfycyxycydyd

Explanation:

Ufivucjvk

3 0

3 years ago

Particles q1, 92, and q3 are in a straight line.

NNADVOKAT [17]

The net force on q₃ will be 17.51 N. The net force is the algebraic sum of the two forces on the pleading q₃

<h3 /><h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The force,by the charge q₁ on the q₃;

$\rm F_{31}} = \frac{Kq_1q_3}{r^2} \\\\ \rm F_{31}} = \frac{9 \times 10^9 \times -28.1 \times 10^{-6}\times \times -47.9 \times 10^{-6}}{(0.600)^2} \\\\ F_{31}} =33.64 \ N$

The force,by the charge q₂ on the q₃;

$\rm F_{32}} = \frac{Kq_2q_3}{r^2} \\\\ \rm F_{32}} = \frac{9 \times 10^9 \times 25.5 \times 10^{-6}\times \times -74.9\times 10^{-6}}{(0.300)^2} \\\\ F_{32}} =-19.09 \ N$

The net force is the sum of the two forces;

$\rm F_{net}=F_{32}+F_{31}\\\\\ \rm F_{net}=36.6-19.9 \\\\ \rm F_{net}=17.51 \ N$

Hence, the net force on q₃ will be 17.51 N.

To learn more about Columb's law, refer to the link;

brainly.com/question/1616890

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8 0

2 years ago

Find an expression for the electric field E⃗ at the center of the semicircle. Hint: A small piece of arc length Δs spans a small

Hatshy [7]

Answer:

Electric Field a the centre $E=\dfrac{Q}{2\pi^2\epsilon_0 R^2}(-\vec j)$

Explanation:

<u>Given:</u>

Total charge on the semicircle =Q

Radius of the semicircle=R

Let consider a elemental charge on the semicircle at an angle $\theta\\$ with the horizontal. Since the The charge distribution is symmetrical about y axis so the there will symmetrical charge on other side. The horizontal component will cancel out each other and vertical component will get added so let dE be the electric Field due to elemental charge