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3 years ago

9

[25 points] Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. T

he initial velocity of the ball A is 6 meters per second to the right, and the initial velocity of the balloon B is 12 meters per second to the left.
When the two balls hit each other, what will happen if it is a perfectly inelastic collision?

1 answer:

evablogger [386]3 years ago

5 0

6 meters is left because you subtract 12 meters from 6

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3 years ago

What force must the deltoid muscle provide to keep the arm in this position?

ruslelena [56]

Answer:

Deltoid Force, $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

1. The torque about the point in the shoulder for the deltoid muscle, $T_{Deltoid}$

2. The torque of the arm, $T_{arm}$

Assuming the arm is just being stretched and there is no rotation going on,

$T_{Deltoid}$ = 0

$T_{arm}$ = 0

⇒ $T_{Deltoid} = T_{arm}$

$r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}$

Where,

$r_{d}$ is radius of the deltoid

$F_{d}$ is the force of the deltiod

$\alpha_{d}$ is the angle of the deltiod

$r_{a}$ is the radius of the arm

$F_{a}$ is the force of the arm , $F_{a} = mg$ which is the mass of the arm and acceleration due to gravity

$\alpha_{a}$ is the angle of the arm

The force of the deltoid muscle is,

$F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}$

but $F_{a} = mg$ ,

∴ $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

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2 years ago

A spherical shell is rolling without slipping at constant speed on a level floor. What percentage of the shell's total kinetic e

IgorC [24]

Answer:

41.667 per cent of the total kinetic energy is translational kinetic energy.

Explanation:

As the spherical shell is rolling without slipping at constant speed, the system can be considered as conservative due to the absence of non-conservative forces (i.e. drag, friction) and energy equation can be expressed only by the Principle of Energy Conservation, whose total energy is equal to the sum of rotational and translational kinetic energies. That is to say:

$E = K_{t} + K_{r}$

Where:

$E$ - Total energy, measured in joules.

$K_{r}$ - Rotational kinetic energy, measured in joules.

$K_{t}$ - Translational kinetic energy, measured in joules.

The spherical shell can be considered as a rigid body, since there is no information of any deformation due to the motion. Then, rotational and translational components of kinetic energy are described by the following equations:

Rotational kinetic energy

$K_{r} = \frac{1}{2}\cdot I_{g}\cdot \omega^{2}$

Translational kinetic energy

$K_{t} = \frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2}$

Where:

$I_{g}$ - Moment of inertia of the spherical shell with respect to its center of mass, measured in $kg\cdot m^{2}$ .

$\omega$ - Angular speed of the spherical shell, measured in radians per second.

$R$ - Radius of the spherical shell, measured in meters.

After replacing each component and simplifying algebraically, the total energy of the spherical shell is equal to:

$E = \frac{1}{2}\cdot (I_{g} + m\cdot R^{2})\cdot \omega^{2}$

In addition, the moment of inertia of a spherical shell is equal to:

$I_{g} = \frac{2}{3}\cdot m\cdot R^{2}$

Then, total energy is reduced to this expression:

$E = \frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2}$

The fraction of the total kinetic energy that is translational in percentage is given by the following expression:

$\%K_{t} = \frac{K_{t}}{E}\times 100\,\%$

$\%K_{t} = \frac{\frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2} }{\frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2} } \times 100\,\%$

$\%K_{t} = \frac{5}{12}\times 100\,\%$

$\%K_{t} = 41.667\,\%$

41.667 per cent of the total kinetic energy is translational kinetic energy.

7 0

3 years ago