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3 years ago

• The net force acting on an object equals the applied force plus the force of friction.

Physics

1 answer:

Andrew [12]3 years ago

6 0

Answer:

false

Explanation:

ke sath hi hai j clin oncol biol to u h

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A boy pushes a cart with a constant velocity of 0.5m/s by applying a force of 60 N. What is the total frictional force acting on

Nutka1998 [239]

Answer:

60N

Explanation:

in this case the minimum amount of force required must be equal to the friction Force. i.e Newton's first law of motion.

therefore the maximum amount of frictional force is equal to the applied force which is 60N.

because of the net force acting on the object is zero the object is in constant motion . i.e equal and opposite force must be applied so that the object is in constant velocity therefore the total frictional force must be 60N

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3 years ago

What is the force needed to give a .25 kg arrow an acceleration of 196m/s2?

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49N is the force needed to give a .25 kg arrow an acceleration of 196m/s2. F =ma â‡’ =( 0.25kg)(196m/s2) = 49N if the arrow is shot horizontally where the applied force is entirely in the x-direction.

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On what criteria is the Fujita scale for Tornadoes based?

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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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2 years ago