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amid [387]
3 years ago
11

Armando is the CEO for a successful large business. He wants to create a visual aid to represent the relationships between diffe

rent levels of authority in the company. What does Armando want to create?
a hierarchy
an organizational chart
a mission
a key performance indicator
Physics
2 answers:
xeze [42]3 years ago
7 0
<h3><em>your answer will be an organizational chart </em></h3><h3><em>hope this helps have a nice day :)</em></h3>
Butoxors [25]3 years ago
3 0
The Answer Is: An Organizational Chart.
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What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally h
zysi [14]

Complete question:

What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)

Answer:

The peak emf generated by the coil is 15.721 kV

Explanation:

Given;

Radius of coil, r = 0.250 m

Number of turns, N = 500-turn

time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s

magnetic field strength, B = 0.425 T

Induced peak emf = NABω

where;

A is the area of the coil

A = πr²

ω is angular velocity

ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s =  60 rev/s

Induced peak emf = NABω

                               = 500 x (π x 0.25²) x 0.425 x 376.738

                               = 15721.16 V

                               = 15.721 kV

Therefore, the peak emf generated by the coil is 15.721 kV

5 0
3 years ago
A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block
MariettaO [177]

Answer:

v₀ = 0.5058 m/s

Explanation:

From the question, for the block to hit the bottle, the elastic potential energy of the spring at the bottle (x = 0.08 m) should be equal to the sum of the elastic potential energy of the spring at x = 0.05 m and the kinetic energy of block at x = 0.05 m

Now, the potential energy of the block at x = 0.08 m is ½kx²

where;

k is the spring constant given by; k = ω²m

ω is the angular velocity of the oscillation

m is the mass of the block.

Thus, potential energy of the spring at the bottle(x = 0.08 m) is;

U = ½ω²m(0.08m)²

Also, potential energy of the spring at the bottle(x = 0.05 m) is;

U = ½ω²m(0.05m)²

and the kinetic energy of the block at x = 0.05 m is;

K = ½mv₀²

Thus;

½ω²m(0.08)² = ½ω²m(0.05)² + ½mv₀²

Inspecting this, ½m will cancel out to give;

ω²(0.08)² = ω²(0.05)² + v₀²

Making v₀ the subject, we have;

v₀ = ω√((0.08)² - (0.05)²)

So,

v₀ = 8.1√((0.08)² - (0.05)²)

v₀ = 0.5058 m/s

7 0
3 years ago
The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.40 the weight of the object
Ann [662]

The frictional force of an object is the product of the normal force and coefficient of kinetic friction. Here the frictional force acting on the object  is 16.4 N.

<h3>What is frictional force?</h3>

Frictional force is a kind of force acting on a body to resist it from motion. Thus, the direction of the force will be in negative with the magnitude. Frictional force is the product of coefficient of friction and the normal force.

The normal  force acting on the object of mass 4.2 Kg is N = mg

N = 4.2 Kg × 9.8 m/s² = 41.16 N

Frictional force = ц N

                         = 0.40 × 41.16 N

                         = 16.4 N.

Therefore, the frictional force acting between the surface of the object and the floor is 16.4 N

To find more on frictional force, refer here:

brainly.com/question/1714663

#SPJ1

Your question is incomplete. But your complete question probably was:

The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.40 the weight of the object is 4.2 kg. What is the frictional force of the object?

3 0
1 year ago
Block A is accelerating with Block B at a rate of 0.800 m/s2 along a frictionless surface. It suddenly encounters a surface that
Natalija [7]

Answer:

a=0.6\ m/s^2

Explanation:

The attached figure shows the whole description. Considering the applied force is 100 N.

The acceleration of both blocks A and B, a=0.8\ m/s^2

Firstly calculating the mass m using the second law of motion as :

F = ma

m is the mass

m=\dfrac{F}{a}

m=\dfrac{100\ N}{0.8\ m/s^2}

m = 125 kg

It suddenly encounters a surface that supplies 25.0 N a friction, F' = 25 N

(F-F')=ma

(100-25)=125\times a

a=\dfrac{75}{125}=0.6\ m/s^2

So, the new acceleration of the block is 0.6\ m/s^2. Hence, this is the required solution.

7 0
3 years ago
What can you caculate if you know the frequency and speed of a wave
Mila [183]

Frequency and speed = wavelength.

4 0
4 years ago
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