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pentagon [3]
2 years ago
13

The diagram shows two charged objects, X and Y. Answer X: positive Y: positive

Physics
1 answer:
Sveta_85 [38]2 years ago
4 0

The diagram shows two charged objects, X and Y.

Answer

X: positive

Y: positive

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A student pushes a 50-N box across the floor a distance of 15 m. How much work was done to move the box?
irinina [24]

Answer:750

Explanation:

50 times 15

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Small, slowly moving spherical particles experience a drag force given by Stokes' law: Fd = 6πηrv where r is the radius of the p
Dominik [7]

Answer:

Explanation:

At the time of a body achieving terminal velocity, the drag force becomes equal to the weight of the body less the buoyant force by the surrounding medium which can be represented by the following equation

\frac{4\pi\times r^3(d-\rho)}{3} =6\pi\times n\times r\times v

Where r is radius of the body , d is density of the material of the body σ is density of the medium and n is coefficient of viscosity of the medium and v is terminal velocity.

Simplifying

v = \frac{2\times r^2(d-\rho)}{9\times n}

Assuming the value of density of air as 1.225 kg/m³ and putting other given values in the formula we get

v = [tex]\frac{2\times (1.2\times10^{-5})^2(2182-1.225)}{9\times 1.8\times10^{-5}}[/tex]

v = 387 x 10⁻⁵ m/s

Terminal velocity = 387 x 10⁻⁵ m/s

Time taken to fall a distance of 100 m

= \frac{100}{387\times10^{-5}}

= 2.6 x 10⁴ s.

5 0
2 years ago
Bonnie and Clyde are trying to steal the world's largest diamond from a 10 story
german

Answer:

The speed Clyde will be falling at is 33.72.

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3 years ago
In a real machine, the work output is always less than the work input.<br> a. True<br> b. False
Artist 52 [7]
<span>In a real machine, the work output is always less than the work input is true. The answer is letter A. it follows the law of entropy where no energy can be converted completely into work. Under this law, Carnot’s theorem states that Carnot’s engine can perform 100% of work. However, no such engine has ever succeeded the conversion of work into 100%. The greatest efficiency so far is at 80%.  Because there will always be factors that could affect the conversion of work. </span>
4 0
2 years ago
Give two mathematical examples of Newton's third law and how you get the solution​
bagirrra123 [75]

Answer:

1) Any particle moving in a horizontal plane slowed by friction, deceleration = 32 μ

2) The particle moving by acceleration = P/m - 32μ OR The external force = ma + 32μm

Explanation:

* Lets revise Newton’s Third Law:

- For every action there is a reaction, equal in magnitude and opposite

 in direction.

- Examples:

# 1) A particle moving freely against friction in a horizontal plane

- When no external forces acts on the particle, then its equation of

  motion is;

∵ ∑ forces in direction of motion = mass × acceleration

∵ No external force

∵ The friction force (F) = μR, where μ is coefficient of the frictional force

   and R is the normal reaction of the weight of the particle on the

   surface

∵ The frictional force is in opposite direction of the motion

∴ ∑ forces in the direction of motion = 0 - F

∴ 0 - F = mass × acceleration

- Substitute F by μR

∴ - μR = mass × acceleration

∵ R = mg where m is the mass of the particle and g is the acceleration

  of gravity

∴ - μ(mg) = ma ⇒ a is the acceleration of motion

- By divide both sides by m

∴ - μ(g) = a

∵ The acceleration of gravity ≅ 32 feet/sec²

∴ a = - 32 μ

* Any particle moving in a horizontal plane slowed by friction,

 deceleration = 32 μ

# 2) A particle moving under the action of an external force P in a

  horizontal plane.

- When an external force P acts on the particle, then its equation

 of motion is;

∵ ∑ forces in direction of motion = mass × acceleration

∵ The external force = P

∵ The friction force (F) = μR, where μ is coefficient of the frictional force

   and R is the normal reaction of the weight of the particle on the

   surface

∵ The frictional force is in opposite direction of the motion

∴ ∑ forces in the direction of motion = P - F

∴ P - F = mass × acceleration

- Substitute F by μR

∴ P - μR = mass × acceleration

∵ R = mg where m is the mass of the particle and g is the acceleration

  of gravity

∴ P - μ(mg) = ma ⇒ a is the acceleration of motion

∵ The acceleration of gravity ≅ 32 feet/sec²

∴ P - 32μm = ma ⇒ (1)

- divide both side by m

∴ a = (P - 32μm)/m ⇒ divide the 2 terms in the bracket by m

∴ a = P/m - 32μ

* The particle moving by acceleration = P/m - 32μ

- If you want to fin the external force P use equation (1)

∵ P - 32μm = ma ⇒ add 32μm to both sides

∴ P = ma + 32μm

* The external force = ma + 32μm

7 0
3 years ago
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