The masses of two bodies their distance apart and the gravitational contents G is in every day case i.e. close to the earth surface the gravitational field is considered to be constant.
Answer:
Positive
<h3>This type of charging is called charging by induction </h3>
- In this the uncharged body gets the opposite type of charge.
when a negatively charged rod is bring near uncharged spherical conductor it attracts positive charge and repels negative charges
so In left side all positive charges appears and in right side all negetive
<em><u>Types </u></em><em><u>of</u></em><em><u> </u></em><em><u>method</u></em><em><u>s</u></em><em><u> </u></em><em><u>of </u></em><em><u>charg</u></em><em><u>ing</u></em><em><u> </u></em>
- <em><u>Charging </u></em><em><u>by </u></em><em><u>contact </u></em>
- <em><u>charging </u></em><em><u>by </u></em><em><u>rubbing </u></em>
- <em><u>charging </u></em><em><u>by </u></em><em><u>induction</u></em><em><u>.</u></em>
Explanation:
The hydraulic system works on the principle of Pascal's law which says that the pressure in an enclosed fluid is uniform in all the directions. ... As the pressure is same in all the direction, the smaller piston feels a smaller force and a large piston feels a large force.
The space station completes 2 revolutions each minute, so that it traverses a distance of 2<em>π</em> (100 m) = 200<em>π</em> m each minute, giving it a linear/tangential speed of
<em>v</em> = (200<em>π</em> m) / (60 s) ≈ 10.472 m/s
(a) The astronaut would experience an acceleration of
<em>a</em> = <em>v</em> ² / (100 m) ≈ 1.09662 m/s² ≈ 0.1119<em>g</em> ≈ 0.11<em>g</em>
<em />
(b) Now you want to find the period <em>T</em> such that <em>a</em> = <em>g</em>. This would mean the astronaut has a tangential speed of
<em>v</em> = (200<em>π</em> m) / <em>T</em>
so that her centripetal/radial acceleration would match <em>g</em> :
<em>a</em> = <em>g</em> = ((200<em>π</em> m) / <em>T </em>)² / (100 m)
Solve for <em>T</em> :
(100 m) <em>g</em> = (400<em>π</em> ² m²) / <em>T</em> ²
<em>T</em> ² = (400<em>π</em> ² m²) / ((100 m) <em>g</em>) = (4<em>π</em> ² m)/<em>g</em>
<em>T</em> = √((4<em>π</em> ² m) / (9.8 m/s²)) ≈ 2<em>π</em> √(0.102 s²) ≈ 2.007 s ≈ 2.0 s
