Object 1 has a mass of 3m and is moving to the right at a velocity of

Physics

1 answer:

andrew11 [14]3 years ago

6 0

Answer:

In a collision, the velocity change is always computed by subtracting the initial velocity value from the final velocity value. If an object is moving in one direction before a collision and rebounds or somehow changes direction, then its velocity after the collision has the opposite direction as before.

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Si se aplica una fuerza de 3n sobre un sistema se genera 15000 cal de calor generandose a su vez un trabajo de 300 j ¿en cuanto

Marizza181 [45]

Answer: +/- 71,65 Calorías

Explanation: Se deduce que la fuerza aplica para aumentar o disminuir la energía del generador.

+/- x Calorías

X = Equivalente de 300 Joules en calorías

Para poder pasar Joules a calorías.

Hacemos una regla de 3 simple

1 Calorías = 4,18 Julios

0,2392 calorías = 1 Julio

( 1 / 4,18 = 0,2392 calorías)

300 Julios = 71,65 calorías

(0,2392 calorías x 300 julios = +/- 71,65 calorías)

5 0

3 years ago

Bob the American has just driven his car across the Canadian border when he sees a

Ludmilka [50]

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it equals 53 miles per hour

4 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 = $\sqrt{6gL}$