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Alenkinab [10]
4 years ago
9

List and describe the top 3 qualities of an effective leader?

Physics
1 answer:
s2008m [1.1K]4 years ago
7 0

BEING HONEST , OPTIMISTIC,DISCIPLINED, PUNCTUAL ,TRUTHFUL AND GOOD ORATOR(SPEAKER) , AND HOPEFUL ARE OTHER BASIC QUALITIES AN EFFECTIVE LEADER MUST POSSESS. ALSO,HE MUST NEVER DIFFERENCE BETWEEN ANY OF HIS FOLLOWERS OR DO PARTIALLY AT ANY TIME . HE SHOULD TREAT EVERY ONE EQUALLY.

Explanation:

You might be interested in
Two small objects are suspended from threads. When the objects are moved close together, they attract one another. What of the f
creativ13 [48]

Answer:d

Explanation:

All the given situations are possible because

(a)When particles are oppositely charged then they attract each other

(b)One is Positively charged and other is uncharged: Charged particle will induce charges of opposite nature to attract the other particle

(c)Negatively charged particles will induce the positive charge in the uncharged particle to attract the initially uncharged particle.

                 

4 0
3 years ago
A tug-of-war game is played by five c/hildren: three on one team and two on the other. How much force will the two child team ha
dem82 [27]

Answer:

no, 3 porces is more tha 2 so the power between the 3 should be more than 2

Explanation:

4 0
3 years ago
A solar panel measures 0.57 m by 1.3 m. In direct sunlight, the panel delivers 6.6 A at 6 V. If the intensity of sunlight is 300
zheka24 [161]

Answer:

efficiency of solar panel is 18%

Explanation:

Energy received by the sunlight is given as

Power = intensity \times area

here we know the dimensions of the plate as

L = 0.57 m

W = 1.3 m

now we have

Area = (0.57)(1.3) m^2

A = 0.741 m^2

now the power received by the sun light is given as

P_1 = 0.741(300) = 222.3 W

now the output power due to solar panel is given as

P_{out} = V i

P_{out} = (6.6 A)(6 V)

P_{out} = 39.6 Watt

now the efficiency is given as

\eta = \frac{P_{out}}{P_{in}}

\eta = \frac{39.6}{222.3}

\eta = 0.18

8 0
3 years ago
A fl oor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and
N76 [4]

Answer:

59.4 meters

Explanation:

The correct question statement is :

A floor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 4.5 s, in order to buff an especially scuff ed area of the floor. How far (in meters) does a spot on the outer edge of the disk move during this time?

Solution:

We know for a circle of radius r and θ angle by an arc of length S at the center,

S=rθ

This gives

θ=S/r

also we know angular velocity

ω=θ/t    where t is time

or

θ=ωt

and we know

1 revolution =2π radians

From this we have

angular velocity ω = 1.4 revolutions per sec =  1.4×2π radians /sec = 1.4×3.14×2×= 8.8 radians / sec

Putting values of ω and time t in

θ=ωt

we have

θ= 8.8 rad / sec × 4.5 sec

θ= 396 radians

We are given radius r = 15 cm = 15 ×0.01 m=0.15 m (because 1 m= 100 cm and hence, 1 cm = 0.01 m)

put this value of θ and r  in

S=rθ

we have

S= 396 radians ×0.15 m=59.4 m

7 0
4 years ago
Read 2 more answers
A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision
marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

x \approx 103,46x10^{9}m.

B) The space time coordinate t of the collision in Earth's reference frame is

t=377,29s

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

                          x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                                y'=y

                                z'=z

                          t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

                       x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                           y=y'

                           z=z'

                        t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

First we calculate the expression in the denominator

                            \frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}

                                \sqrt{1-\frac{v^{2}}{c^{2}}} =0,714

then we calculate t

                      t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                      t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}

                      t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}

                      t=\frac{190s+79,388s}{0,714}

finally we get that

                                     t=377,29s

then we calculate x

                         x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                         x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+39872396914m}{0,714}}

                         x=\frac{73872396914m}{0,714}}

                         x=103462740775,91m

finally we get that

                                     x \approx 103,46x10^{9} m

5 0
3 years ago
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