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10 months ago

Antonio has been told that he needs to work on his self- awareness. What might he do to most improve this skill

1 answer:

5 0

For Antonio to improve his self-awareness skills he should be able to recognise hie strengths and his weaknesses.

<h3>What is self awareness?</h3>

Self awareness is the act of being able to recognise your feelings and keep your character under control.

To be a self awareness leader, one should be able to have the following qualities:

They should be able to recognize their strengths and challenges.
They should be able to recognise other people's needs and feelings,
They should be able to notice how their behaviour affects the people around them.

Therefore, it can be said for Antonio to improve his self awareness skills, he should be able to recognise his strength and weaknesses and observe how his character affects the people around him.

Learn more about self awareness here:

brainly.com/question/28039248

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A constant current I circulates counterclockwise around a wire loop in the shape of a triangle, lying in x-y plane, with vertice

Blizzard [7]

Answer:

Considering a triangle like the one of the figures, you can obtain the total magnetic force applied on it like the addition of the forces applied to each of the 3 sides.

Been the magnetic force formula:

$F=\int\limits^a_b {I} \, \overline {dx} \times \overline {B}$

For each segment:

Segment 1 (from [0,0,0] to [a,0,0]):

$F=\int\limits^a_0 {I} \, \overline {x}dx \times B \overline {z}=-IBa\overline {y}$

Segment 2 (from [a,0,0] to [0,a,0]):

$F=\int\limits^{(0,a)}_{(a,0)} {I} \, \frac{\sqrt{2} }{2} (\overline {x}+\overline {y})dx \times B \overline {z}=IB2a(\overline {x}+\overline {y})$

Segment 3 (From [0,a,0] to [0,0,0]):

$F=\int\limits^0_a {I} \, \overline {y}dx \times B \overline {z}=-IBa\overline {x}$

Total force on the wire loop:

$F_{T} =F_{1} +F_{2} +F_{3} =-IBa\overline {y}+IB2a(\overline {x}+\overline {y})-IBa\overline {x}=0N$

5 0

2 years ago

What’s better csp or pv ?

BARSIC [14]

Answer:

CSP

Explanation:

CSP systems store energy through Thermal Energy Storage technologies (TES), so power can be used when there isn't enough sunlight. PV systems, however, can't store thermal energy because they use direct sunlight, rather than heat. For this reason, CSP systems are better for energy storage and efficiency.

8 0

1 year ago

Two ships leave a harbor at the same time, traveling on courses that have an angle of 110∘ between them. If the first ship trave

Allushta [10]

Answer:

49.07 miles

Explanation:

Angle between two ships = 110° = θ

First ship speed = 22 mph

Second ship speed = 34 mph

Distance covered by first ship after 1.2 hours = 22×1.2 = 26.4 miles = b

Distance covered by second ship after 1.2 hours = 34×1.2 = 40.8 miles = c

Here the angle between the two sides of a triangle is 110° so from the law of cosines we get

a² = b²+c²-2bc cosθ

⇒a² = 26.4²+40.8²-2×26.4×40.8 cos110

⇒a² = 2408.4

⇒a = 49.07 miles

6 0

3 years ago

Highway transportation officials are trying to determine if the average speed on Archer road is above 45mph, the post speed limi

USPshnik [31]

Answer:

Option D) -0.0707

Explanation:

We are given the following in the question:

Population mean, μ = 45 mp

Sample mean, $\bar{x}$ = 44.9

Sample size, n = 50

Sample standard deviation, s = 10

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{44.9 - 45}{\frac{10}{\sqrt{50}} } = -0.0707$

Thus, the test statistic is

Option D) -0.0707

4 0

3 years ago

Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are

Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

$d_{12} = \sqrt{3^{2}+3^{2} } = \sqrt{18}$ m : distance from q₁ to q₂

(d₁₂)² = 18 m²

$d_{13} =\sqrt{1^{2}+3^{2} } = \sqrt{10}$ m : distance from q₁ to q₃

(d₁₃)² = 10 m²

$d_{14} =\sqrt{3^{2}+2^{2} } = \sqrt{13}$ m : distance from q₁ to q₄

(d₁₄)² = 13 m²

K= 8.98755 × 10⁹ N *m²/C²

q₁= 7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45° , β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

$F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }$

$F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }$

Fn₁= 23.95*10⁻⁹ N

8 0

3 years ago