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3 years ago

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What’s the difference between gravitational force & gravitational field strength?

2 answers:

Lelu [443]3 years ago

7 0

what’s the difference between gravitational force & gravitational field strength?

Zepler [3.9K]3 years ago

5 0

Answer:

Gravitational field strength is the force experienced by a unit mass. Gravitational force is the amount of force acting on a body. It is the product of field strength times the mass under consideration. Gravitational pull is just a more colloquial name for gravitational force.

Explanation:

hope it helps u

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Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin

Zarrin [17]

Answer:

$\rm 9.186\times 10^{-7}\ C.$

Explanation:

<u>Given:</u>

Diameter of the plates of the capacitor, D = 21 cm = 0.21 m.
Distance of separation between the plates, d = 1.0 cm = 0.01 m.
Minimum value of electric field that produces spark, $\rm E=3\times 10^6\ N/C.$

When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

$\rm E=\dfrac{\sigma}{\epsilon_o}.$

where,

$\rm \sigma$ = surface charge density of the plate of the capacitor = $\dfrac qA$ .

$\rm q$ = magnitude of the charge on each of the plate.
$\rm A$ = surface area of each of the plate = $\rm \pi \times (Radius)^2=\pi \times\left ( \dfrac{D}{2}\right )^2= \pi \times \left ( \dfrac{0.21}{2}\right )^2=3.46\times 10^{-2}\ m^2.$
$\epsilon_o$ = electrical permittivity of free space, having value = $8.85\times 10^{-12}\rm \ C^2N^{-1}m^{-2}.$

For the minimum value of electric field that produces spark,

$\rm E = \dfrac{q}{A\epsilon_o}\\\Rightarrow q = E\ A\epsilon_o\\=3\times 10^6\times 3.46\times 10^{-2}\times 8.85\times 10^{-12}\\=9.186\times 10^{-7}\ C.$

It is the maximum value of the magnitude of charge which can be added up to each of the plates of the capacitor.

4 0

2 years ago

The early workers in spectroscopy (Fraunhofer with the solar spectrum, Bunsen and Kirchhoff with laboratory spectra) discovered

Anestetic [448]

The hot gases produce their own characteristic pattern of spectral lines, which remain fixed as the temperature increases moderately.

<h3><u>Explanation: </u></h3>

A continuous light spectrum emitted by excited atoms of a hot gas with dark spaces in between due to scattered light of specific wavelengths is termed as an atomic spectrum. A hot gas has excited electrons and produces an emission spectrum; the scattered light forming dark bands are called spectral lines.

Fraunhofer closely observed sunlight by expanding the spectrum and a huge number of dark spectral lines were seen. "Robert Bunsen and Gustav Kirchhoff" discovered that when certain chemicals were burnt using a Bunsen burner, atomic spectra with spectral lines were seen. Atomic spectral pattern is thus a unique characteristic of any gas and can be used to independently identify presence of elements.

The spectrum change does not depend greatly on increasing temperatures and hence no significant change is observed in the emitted spectrum with moderate increase in temperature.

8 0

3 years ago

What is the instantaneous speed of the monkey at time t=5s?

bezimeni [28]

I’m assuming we’re suppose to get some kind of graph but, Instantaneous speed is the speed that is happening right now. Like driving a car at 15k/h. The instantaneous speed of the car 15k/h. On the graph, at 5s. Wherever the line is, will tell you what the speed is.

6 0

3 years ago

The current through a 10 ohm resistor connected to a 120 volt power supply is

Leno4ka [110]

Answer:I=12 A

Explanation:

Given

Resistance $R=10 \Omega$

Voltage $V=120 V$

According to ohm's law current through a conductor is directly proportional to the voltage applied.

$V\propto I$

$V=IR$

where V=Voltage

I=Current

R=resistance

$I=\frac{V}{R}$

$I=\frac{120}{10}$

$I=12 A$

4 0

2 years ago

A 0.500-kg stone is moving in a vertical circular path attached to a string that is 75.0 cm long. The stone is moving around the

Semenov [28]

Answer:

B. 7.07 m/s

Explanation:

The velocity of the stone when it leaves the circular path is its tangential velocity, $v$ , which is given by

$v=\omega r$

where $\omega$ is the angular speed and $r$ is the radius of the circular path.

$\omega$ is given by

$\omega = 2\pi f$

where $f$ is the frequency of revolution.

Thus

$v=2\pi fr$

Using values from the question,

$v=2\pi\times 1.50\times0.75$

<em>Note the conversion of 75 cm to 0.75 m</em>

$v=2\times3.14\times 1.50\times0.75 = 9.42\times0.75 = 7.065=7.07$

6 0

3 years ago