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Zina [86]
4 years ago
14

A rod attracts a positively charged hanging ball. the rod is?a) negativeb) positivec) neutrald) either negative or neutrale) eit

her positive or neutral
Physics
1 answer:
makvit [3.9K]4 years ago
8 0

Answer:

The correct option is a

Explanation:

This question seeks to test a general rule in physics (on charges) which states that <u>like charges repel but unlike charges attract</u>. This means that, a negatively charged substance will repel or not attract another negatively charged material and the same applies to a positively charged substance also. However, a <u>negatively charged substance will attract a positively charged material and vice versa,</u> hence only a negatively charged rod will attract a positively charged hanging ball.

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What is an example of light energy converted into electrical energy?
irakobra [83]

Answer:

Solar panel

Explanation:

You can often find solar panels on top of rooftops of buildings or apartments (usually in less developed countries), as because it consumes the light energy from sunlight and converts it into electrical energy which can be later used to run electrical energy in that specific building.

3 0
4 years ago
In modern quantum–based atomic theory, what do we call the particular space around the nucleus in which an electron moves?
faust18 [17]
electron cloud hope this helps
8 0
4 years ago
A particle starts with a speed of 12m/s and moves in a straight line in such a way that its speed increases by 4m/s every second
nikitadnepr [17]

Answer:

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6 0
3 years ago
The star Rho1 Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a
s344n2d4d5 [400]

Answer:

82780.42123 m/s

14.45 days

Explanation:

m = Mass of the planet

M = Mass of the star = 0.85\times 1.989\times 10^{30}\ kg=1.69065\times 10^{30}\ kg

r = Radius of orbit of planet = 0.11\times 149.6\times 10^{9}\ m=16.456\times 10^{9}\ m

v = Orbital speed

The kinetic and potential energy balance is given by

\frac{GMm}{r^2}=\frac{mv^2}{r}\\\Rightarrow v=\sqrt{\dfrac{Gm}{r}}\\\Rightarrow v=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.69065\times 10^{30}}{16.456\times 10^{9}}}\\\Rightarrow v=82780.42123\ m/s

The orbital speed of the star is 82780.42123 m/s

The orbital period is given by

t=\frac{2\pi r}{v}\\\Rightarrow t=\dfrac{2\pi \times 16.456\times 10^{9}}{82780.42123}\\\Rightarrow t=1249040.48419\ seconds=\dfrac{1249040.48419}{24\times 60\times 60}=14.45\ days  

The orbital period is 14.45 days

5 0
3 years ago
Use the data provided to calculate the gravitational
nydimaria [60]

Answer:

3 kg:   14.7 kJ

6 kg:   29.4 kJ

9 kg:  44.1 kJ

Explanation:

I just took the quiz

8 0
4 years ago
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