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3 years ago

What is a major division between gravity and the electromagnetic force on one hand and the nuclear forces on the other?

Physics

1 answer:

alekssr [168]3 years ago

6 0

Answer:

1) Gravity and the electromagnetic force have infinite ranges while the nuclear forces have very small ranges.

Explanation:

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A pulley has a mechanical advantage of 1.

sertanlavr [38]

Answer:

There is no mechanical advantage

Explanation:

The mechanical advantage is possible only when the force needed to lift a load is lesser than the weight of the load.

For example, is we have a mechanical advantage of 2, the force needed to lift will be 1/2 of the weight of the load, and if we have a mechanical advantage of 4, the force needed will be 1/4 of the weight of the load.

In the attached image there are clear examples of mechanical advantage with pulleys.

7 0

3 years ago

What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

Wittaler [7]

Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

$\cos \theta = \frac{adjacent}{hypotenuse}$

$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

$v = \frac{2 \pi (1.97x10^{6} m)}{24h}$

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

$T = 24h . \frac{3600s}{1h}$ ⇒ $84600s$

$v = \frac{2 \pi (1.97x10^{6} m)}{84600s}$

$v = 146.31m/s$

Finally, equation 1 can be used:

$a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}$

$a = 0.010m/s^{2}$

Hence, the centripetal acceleration is $0.010m/s^{2}$

To given the centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ it is gotten:

$\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}$

6 0

3 years ago

A 4 kg particle moves at a constant speed of 2.5 m/s around acircle of radius 2 m. What is its angular momentum about the center

blagie [28]

Explanation:

Given that,

Mass of the particle, m = 4 kg

Speed of the particle, v = 2.5 m/s

The radius of the circle, r = 2 m

We need to find the angular momentum about the center of the circle. The formula for the angular momentum is given by :

$L=mvr$

Substitute all the values,

$L=4\times 2.5\times 2\\\\L=20\ kg{\cdot}m^2s$

So, the angular momentum of the particle is 20 kg-m² s.

6 0

3 years ago

Mechanical (sound) waves to Earth from satellites. How is this possible? aves are unable to travel through a vacuum, such as thr

Anuta_ua [19.1K]

Sound waves are not able to travel in a vacuum, sound requires a medium such as air or a solid for example. Satellites are indeed in space, but radio waves are not sound waves they are a form of light. Light can travel in a vacuum so the messages that satellites beam back to Earth are light waves not sound waves, that is why it is possible.

6 0

3 years ago