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3 years ago

Newton’s third law of motion describes A. action and reaction forces. B. balanced forces. C. centripetal forces. D. net force.

Physics

2 answers:

IgorC [24]3 years ago

6 0

<span>A. action and reaction forces.</span>

skelet666 [1.2K]3 years ago

5 0

The correct answer is A. Action and reaction forces

Explanation:

Isaac Newton was an important English physicist, astronomer, and mathematician mainly known in the field of physics for his tree laws related to motion. In the case of the third law, this law states every action of movement implies a reaction that is expressed as an opposite force in equal magnitude that acts against the first object that is moving, for example when you push an object there with your hand is a first force you apply to the object you are pushing, but also the object applies an opposite proportional force in your hand which shows action and reaction forces. Therefore Newton's third law of motion describes action and reaction forces.

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Answer:

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2 years ago

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The projectile launcher shown below will give the object on the right an initial horizontal speed of 5.9 m/s. While the other ob

Crazy boy [7]

Answer:

104.3 cm or 179.7

Explanation:

First find time that it takes for the object to hit the ground

$\sqrt{(2H)/g} -> \sqrt{(2 x 179)/ 9.8} = 6.04s\\$ *

Then find xf of projectile $xf= 5.9(6.04) = 37.7\\\\$

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2 years ago

Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.8 kg and contains

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Answer:

C = 771.35 J/kg°C

Explanation:

Here, e consider the conservation of energy equation. The conservation of energy principle states that:

Heat Given by Metal Piece = Heat Absorbed by Water + Heat Absorbed by Container

Since,

Heat Given or Absorbed by a material = m C ΔT

Therefore,

m₁CΔT₁ = m₂CΔT₂ + m₃C₃ΔT₃

where,

m₁ = Mass of Metal Piece = 2.3 kg

C = Specific Heat of Metal = ?

ΔT₁ = Change in temperature of metal piece = 165°C - 18°C = 147°C

m₂ = Mass of Metal Container = 3.8 kg

ΔT₂ = Change in temperature of metal piece = 18°C - 15°C = 3°C

m₃ = Mass of Water = 20 kg

C₃ = Specific Heat of Water = 4200 J/kg°C

ΔT₃ = Change in temperature of water = 18°C - 15°C = 3°C

Therefore,

(2.3 kg)(C)(147°C) = (3.8 kg)(C)(3°C) + (20 kg)(4186 J/kg°C)(3°C)

C[(2.3 kg)(147°C) - (3.8 kg)(3°C)] = 252000 J

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2 years ago

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Evgen [1.6K]

Equation 1 :
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Equation 2:m2 : m2a= F - T

Adding 1 and 2a=0.073

Placing in equation 1

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2 years ago

Point charges q1 = 14 µC and q2 = −60 µC are fixed at r1 = (5.0î − 4.0ĵ) m and r2 = (9.0î + 7.5ĵ) m. What is the force (in N) of

Lostsunrise [7]

Answer:

The force on q₁ due to q₂ is (0.00973i + 0.02798j) N

Explanation:

F₂₁ = $\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}$

Where;

F₂₁ is the vector force on q₁ due to q₂

K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²

r₂₁ is the unit vector

|r₂₁| is the magnitude of the unit vector

|q₁| is the absolute charge on point charge one

|q₂| is the absolute charge on point charge two

r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)

|r₂₁| = $\sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}$

(|r₂₁|)² = 148.25

$F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }$

= 0.050938(0.19107i + 0.54933j) N

= (0.00973i + 0.02798j) N

Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N

7 0

3 years ago