Why do many people view controls negatively?

1 answer:

5 0

Many People Are Averse to Management "Control" ... and then subsequent decisions about what to do is the essence of management coordination

You might be interested in

From laboratory measurements, we know that a particular spectral line formed by hydrogen appears at a wavelength of 121.6 nanome

lina2011 [118]

Answer:

b) The star is moving away from us.

Explanation:

If an object moves toward us, the light waves it emits are compressed - the wavelength of the light will be shorter, making the light bluer. On the other hand, if an object moves away from us, the light waves are stretched, making it redder. If from laboratory measurements we know that a specific hydrogen spectral line appears at the wavelength of 121.6 nanometers (nm) and the spectrum of a particular star shows the same hydrogen line appearing at the wavelength of 121.8 nm, we can conclude that the star is moving away from npos, since the wavelength related to that star is more expanded.

7 0

3 years ago

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has

Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = $\sqrt{6gL}$