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valkas [14]
3 years ago
15

State clearly how the pith ball electroscope may be used to test for the kind of charge on a body

Physics
1 answer:
PilotLPTM [1.2K]3 years ago
5 0
An object with a known charge is first brought close to the ball so as to induce the opposite charge. Lets say we brought a positive object, so the ball will be negatively charged. Afterwards, the object with the unknown charge is brought closer. If the ball is attracted, then the object is positively charged and if it is repelled then the object is negatively charged.
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HELP ASP PLEASED AND THANKS NO LINKS PLEASED AND NO FILES
SOVA2 [1]

Answer:

just search up a ven-diagram and then try to draw it or trace it then use it for ur question

Explanation:

6 0
3 years ago
An 800 N man climbs 5 m up a ladder. How much gravitational potential energy does he gain?
Artemon [7]

Answer:

4000J

Explanation:

Given parameters:

Weight of the man  = 800N

Height of ladder  = 5m

Unknown:

Gravitational potential energy gained  = ?

Solution:

The gravitational potential energy is due to the position of a body.

 Gravitational potential energy = weight x height

Now insert the parameters;

 Gravitational potential energy  = 800 x 5  = 4000J

5 0
3 years ago
What happens when tectonic plates move away from each other? Molten matter cools and sinks towards the core. Lava travels away f
satela [25.4K]
Molten Mattet Seeps through the crust and forms new land.
8 0
3 years ago
What would be the escape speed for a craft launched from a space elevator at a height of 54,000 km?
Natasha_Volkova [10]

Answer: 3.63 km/s

Explanation:

The escape velocity equation for a craft launched from the Earth surface is:

V_{e}=\sqrt{\frac{2GM}{R}}

Where:

V_{e} is the escape velocity

G=6.67(10)^{-11} Nm^{2}/kg^{2} is the Universal Gravitational constant

M=5.976(10)^{24}kg is the mass of the Earth

R=6371 km=6371000 m is the Earth's radius

However, in this situation the craft would be launched at a height h=54000 km=54000000 m over the Eart's surface with a space elevator. Hence, we have to add this height to the equation:

V_{e}=\sqrt{\frac{2GM}{R+h}}

V_{e}=\sqrt{\frac{2(6.67(10)^{-11} Nm^{2}/kg^{2})(5.976(10)^{24}kg)}{6371000 m+54000000 m}}

Finally:

V_{e}=3633.86 m/s \approx 3.63 km/s

7 0
2 years ago
In the diagram below, what is the property of the wave indicated by the letter A? a.Crest
Ugo [173]
Do you have a picture of the diagram that I could view?
4 0
3 years ago
Read 2 more answers
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