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lubasha [3.4K]
3 years ago
15

A man of 60kg moves in a lift of constant velocity 5m/s .What is the reactive force acting on the man by the elevator?

Physics
2 answers:
Flauer [41]3 years ago
4 0

Answer:

588 N

Explanation:

Since the 60 kg is moving at a constant velocity there is no acceleration. In order for the system to be balanced, both the normal force and the force of gravity must be equal. In this case the man has a mass of 60 kg. So to find the force you multiply mass by gravitys constant (9.81). And you end up with an answer of 588.6 but I rounded to 588.

wlad13 [49]3 years ago
3 0

The reactive force acting on the man by the elevator is about 588 Newton

<h3>Further explanation</h3>

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

\large {\boxed {F = ma }

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

<u>Given:</u>

mass of the man = 60 kg

velocity of the lift = 5 m/s

<u>Unknown:</u>

reactive force = N = ?

<u>Solution:</u>

We will use Newton's 2nd law formula for this problem.

\Sigma F = ma

N - w = ma

N - mg = ma

N - 60(9.8) = 60(0) → constant velocity ⇒ a = 0

N = 0 + 588

N = 588 ~ Newton

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441
  • Newton's Law of Motion: brainly.com/question/10431582
  • Example of Newton's Law: brainly.com/question/498822

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

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7 0
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A straight wire of length 4.5 cm moves at a constant speed of 5.2m/s perpendicular to its length and a uniform magnetic field. I
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3 0
3 years ago
Three parallel wires each carry current I in the directions shown in (Figure 1). The separation between adjacent wires is d.
jasenka [17]

(a) The magnitude of the net magnetic force per unit length on the top wire is μI²/πd.

(b) The magnitude of the net magnetic force per unit length on the middle wire is zero.

(c) The magnitude of the net magnetic force per unit length on the bottom wire is 3μI²/4πd.

<h3>Force per unit length</h3>

The magnitude of the net magnetic force per unit length on the top wire is calculated as follows;

F₁/L = (μI₁/2π) x (I₂/d + I₃/d)

F₁/L = (μI/2π) x (I/d + I/d)

F₁/L = (μI/2π) x (2I/d)

F₁/L = μI²/πd

The magnitude of the net magnetic force per unit length on the middle wire is calculated as follows;

F₂/L = (μI₂/2π) x (I₃/d - I₁/d)

F₂/L = (μI/2π) x (I/d -  I/d) = 0

The magnitude of the net magnetic force per unit length on the middle bottom is calculated as follows;

F₃/L = (μI₂/2π) x (I₁/d + I₂/d)

F₃/L =  (μI/2π) x (I/2d + I/d)

F₃/L =  (μI/2π) x (3I/2d)

F₃/L =  3μI²/4πd

Thus, the magnitude of the net magnetic force per unit length on the top wire is μI²/πd.

The magnitude of the net magnetic force per unit length on the middle wire is zero.

The magnitude of the net magnetic force per unit length on the bottom wire is 3μI²/4πd.

Learn more about magnetic force here: brainly.com/question/13277365

#SPJ1

6 0
2 years ago
PLZ HELP!!!!
zavuch27 [327]

Answer:

AAAAAA!!!!!!!

Explanation:

Hope that helped lol!

8 0
2 years ago
Read 2 more answers
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