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irakobra [83]
3 years ago
13

If you measured all the energy related to motion and all the stored energy in the particles of a substance, which would you be m

easuring? heat capacity heat thermal energy
Physics
1 answer:
Nady [450]3 years ago
3 0

If you measured all the energy related to motion and all the stored energy in the particles of a substance, you would be measuring the thermal energy of the particles. If there is movement of the particles, they are also releasing energy in the form of heat.

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A thin cylindrical shell and a solid cylinder have the same mass and radius. The two are released side by side and roll down, wi
Vitek1552 [10]

Answer:

Explanation:

Let the velocity be v

Total energy at the bottom

= rotational + linear kinetic energy

= 1/2 Iω² + 1/2 mv² ( I moment of inertia of shell  = mr² )

= 1/2 mr²ω² + 1/2 mv² ( v = ω r )

= 1/2 mv² +1/2 mv²

= mv²

mv² = mgh ( conservation of energy )

v² = gh

v = √gh

= √9.8 x 1.8

= 4.2 m /s

8 0
3 years ago
A student measures the time it takes for two reactions to be completed reaction eight is completed and 57 seconds and reaction b
Maslowich
The answer to this problem is 1200
8 0
3 years ago
A tightrope walker more easily balances on a tightwire if his pole
cestrela7 [59]
B) droops.

Why?
To maintain balance, you do not need something short so you're balanced well... You need something long and droopy to maintain balance. The pole should be held by your waist and it should be light.

Hope this helps!~
4 0
3 years ago
The rotational speed of earth is similar to?​
m_a_m_a [10]
Actually, the speed of the earth is the same everywhere, taking the angular speed as the valid measure of the speed
8 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}

Hence, v0 = 6.86 m/s

4 0
3 years ago
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