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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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The independent variable:
ValentinkaMS [17]

Answer:

A. Is the one that the experimenter manipulates directly

Explanation:

The independent variable is the one that is manipulated during an experiment by the experimenter.

The dependent variable is the one that is effected by the independent variable in an experiment.

3 0
3 years ago
The sun is part of which type of galaxy?
jeka94
I believe the answer is D. The milky way is a spiral galaxy, You can tell just by looking at it.
4 0
3 years ago
How does kinetic energy affect the stopping distance of a vehicle traveling at 30 mph compared to the same vehicle traveling at
Maru [420]
Well you of course have different kinetic energies with the two speeds.
Kinetic energy = (1/2)*mass*velocity^2
The vehicle's mass is the same in both cases, so we can ignore that as well as 1/2 since it's a constant.
So we have (30)^2 vs (60^2)
which is 900 vs 3600
So having 60 mph compared to 30 mph is 4 times the kinetic energy.
5 0
3 years ago
The force F required to compress a spring a distance x is given by F 2 F0 5 kx where k is the spring constant and F0 is the prel
IrinaVladis [17]

Answer:

a)W=8.333lbf.ft

b)W=0.0107 Btu.

Explanation:

<u>Complete question</u>

The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.

Solution

Preload = F₀=0 lbf

Spring constant k= 200 lbf/in

Initial length of spring x₁=0

Final length of spring x₂= 1 in

At any point, the force during deflection of a spring is given by;

F= F₀× kx  where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.

W=\int\limits^2_1 {} \, Fds \\\\\\W=\int\limits^2_1( {F_0+kx} \,) dx \\\\\\W=\int\limits^a_b {kx} \, dx ; F_0=0\\\\\\W=k\int\limits^2_1 {x} \, dx \\\\\\W=k*\frac{1}{2} (x_2^{2}-x_1^{2}  )\\\\\\W=200*\frac{1}{2} (1^2-0)\\\\\\W=100.lbf.in\\\\

Change to lbf.ft by dividing the value by 12 because 1ft=12 in

100/12 = 8.333 lbf.ft

work required to compress the spring, W=8.333lbf.ft

The work required to compress the spring in Btu will be;

1 Btu= 778 lbf.ft

?= 8.333 lbf.ft----------------cross multiply

(8.333*1)/ 778 =0.0107 Btu.

6 0
3 years ago
A steel ball with mass m is suspended from the ceiling at the bottom end of a light, 17.0-m-long rope. The ball swings back and
iogann1982 [59]

Answer:

1. 18.25 m/s

2. 0 m/s

Explanation:

1.So the centripetal acceleration of the ball at this lowest point must be, taking gravity into account

a_c = \frac{T - mg}{m} = \frac{3mg - mg}{m} = 2g

The speed at this point would then be

v^2 = a_c r = 2gr = 2*9.8*17 = 333.2

v = \sqrt{333.2} = 18.25 m/s

2. Similarly, if T = mg, then the centripetal acceleration must be

a_c = \frac{T - mg}{m} = \frac{mg - mg}{m} = 0

As the ball has no centripetal acceleration, its speed must also be 0 as well.

6 0
3 years ago
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