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Svetach [21]
3 years ago
14

Two identical balls are at rest and side by side at the top of a hill. You let one ball, A, start rolling down the hill. A littl

e later you start the second ball, B, down the hill by giving it a shove. The second ball rolls down the hill along a line parallel to the path of the first ball and passes it. At the instant ball B passes ball A:
a. it has the same position and the same velocity as A.b. it has the same position and the same acceleration as A.c. it has the same velocity and the same acceleration as A.d. it has the same displacement and the same velocity as A.e. it has the same position, displacement and velocity as A.
Physics
1 answer:
ICE Princess25 [194]3 years ago
3 0

Answer:

Option b. it has the same position and the same acceleration as A

Explanation:

Let's analyze every statement:

a. it has the same position and the same velocity as A

In the instant where B passes A, they Do have the same position. Velocity however, cannot be the same because if they were, ball B would never pass ball A. So, this is false.

b. it has the same position and the same acceleration as A

As we said in the previous option, the position is the same. The acceleration is gravity for both balls, so this is true.

c. it has the same velocity and the same acceleration as A

Acceleration is the same but velocities are not, so this is false.

d. it has the same displacement and the same velocity as A

The distance they have traveled is the same, so the displacement is the same, but the velocity is not, so this is false.

e. it has the same position, displacement and velocity as A

The position and displacement is the same but not velocity, so this is false.

Only option b is true.

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3 years ago
An aluminum wire is 5.87 m long and has a diameter of 6.07 mm.
antoniya [11.8K]

Answer: 5.72 x 10-3Ω

Explanation:

Hi, to answer this question, first we have to calculate the cross sectional area of the cable:

Diameter (D)=6.07 mm

Since: 1000mm = 1m

6.07mm/ 1000mm/m = 0.00607 meters

Area of a circle : π (d/2)^2

A = π (0.00607/2)^2= 0.000028937 m2

Resistance formula:

Resistance (R) = P(resistivity) L (length)÷A (cross sectional area )

Replacing with the values given:

R = (2.82x10-8 x 5.87) / 0.000028937

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What are the similarities between strong nuclear force and weak nuclear force
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3 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.

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^^^^^^^^^^^^^^^^^^^^^^^^^^^
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