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harkovskaia [24]
3 years ago
7

What are three benefits of being assertive and what are three tips to help you develop an assertive style of communication? (Sit

e 1)
                 
Physics
1 answer:
uysha [10]3 years ago
5 0
1. The benefits of been assertive include the following:
a. It improve one self image: When you choose to be assertive, you adopt  a self realistic image.
b. It enhance how we understand others: Been assertive helps you to see others in a more realistic context.
c. Promote self awareness and self confidence: Been assertive helps you to develop a greater respect for your own points of view when dealing with issues. 
d. Been assertive is a less stressful way of communicating.

2. The three steps needed to develop assertive communication, include:
a. Learn to 'no' more often: refuse to please everyone and do not behave according to people expectations of you. Be yourself.
b. Your tone must be properly turned when talking: Do not raise your voice or rush a communication. Be calm and cool when talking.
c. Be open to communication: Be willing to discuss issues until a suitable solution is found.
d. Pay attention to non-verbal communication of others: Keep eye contact when communicating and pay attention to other body gestures. 

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Answer:

a) F = 30 N, b)   I = 12 N s , c)  I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

     p₀ = m v

The speed can be found by kinematics

     v = v₀ - g t

     v = 0 - 10 0.4

     v = -4.0 m / s

Final after division

     pf = m₁ v₁f + m₂ v₂f

    p₀ = pf

    M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

     F = mg

     F = 3 10 = 30 N

b) The expression for momentum is

     I = Ft

Before the explosion the only force that acts is the weight

    I = mg t

    I = 3 10 0.4

    I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

    M v = m₁ v₁f + m₂ v₂f

    v₂f = (M v - m₁ v₁f) / m₂

    v₂f = (3 (-4) - 1 6) / 2

   v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

    I = F t = Dp

   F t = pf –po)

   F t= [m₁ v₁f + m₂ v₂f]

   

   I = [1 6 + 2 (-9) -0]

   I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

 ΔI =If – I₀  

ΔI = - 12 – (-12)

ΔI = -0 N s

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