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Eva8 [605]
3 years ago
9

The force required to make an object travel along a circular path is:

Physics
2 answers:
malfutka [58]3 years ago
6 0
<span>The force required to make an object travel along a circular path is:

Centripetal Force.</span>
Makovka662 [10]3 years ago
4 0

Answer:

centripetal force

Explanation:

When an object is moving around a circular path with constant speed, it experiences a force towards its centre. This force is called centripetal force.

The direction of centripetal force is always towards the centre of circular path. The formula to calculate the centripetal force is given by

F = mv^2 / r

Where, m is the mass of the object, v be the speed of the object and r be the radius of circular path.

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Help me please I need help?
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Your answer is c. homologous structures
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PLEASE HELP ASAP:
antoniya [11.8K]

True since coulomb's law states that There is electric force between like charges or opposite charges. The negative sign only shows the nature of the force.

<h3>What is the coulombs law ?</h3>

coulombs formula is given by

F=  \dfrac{K\times q_{1} \times q_{2}}{r^{2} }

Now it states that if two charged particles are separated by the distance r and having same or opposite charge will attract or repel each other.

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Hence coulomb's law states that There is electric force between like charges or opposite charges. The negative sign only shows the nature of the force.

To know more about coulomb's law follow

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Where is the value of kinetic energy equal to zero
bearhunter [10]

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The earth rotates every 86,160 seconds. What is the tangential speed (in m/s) at Livermore (Latitude 37.6819° measured up from e
Lena [83]

Answer:

The tangential speed at Livermore is approximately 284.001 meters per second.

Explanation:

Let suppose that the Earth rotates at constant speed, the tangential speed (v), measured in meters per second, at Livermore (37.6819º N, 121º W) is determined by the following expression:

v = \left(\frac{2\pi}{\Delta t}\right)\cdot R \cdot \sin \phi (1)

Where:

\Delta t - Rotation time, measured in seconds.

R - Radius of the Earth, measured in meters.

\phi - Latitude of the city above the Equator, measured in sexagesimal degrees.

If we know that \Delta t = 86160\,s, R = 6.371\times 10^{6}\,m and \phi = 37.6819^{\circ}, then the tangential speed at Livermore is:

v = \left(\frac{2\pi}{86160\,s} \right)\cdot (6.371\times 10^{6}\,m)\cdot \sin 37.6819^{\circ}

v\approx 284.001\,\frac{m}{s}

The tangential speed at Livermore is approximately 284.001 meters per second.

4 0
3 years ago
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