The sweater has a tendency to attract electrons.
The leather jacket has a lower tendency to attract electrons than the sweater.
Explanation:
The sweater and the leather jackets are made up of distinct fabrics that based on their minutest particles called an atom.
An atom is made up of sub-atomic particles of protons, neutrons and electrons.
- Electrons occupies the bulk volume of the atom and they are easily lost in atoms that are big. They are negatively charged.
- Protons are positively charged and are very difficult to lose. They occupy the tiny nucleus with neutrons.
- A body that becomes negatively charged will be said to have a hihg tendency to attract electrons. Normally atoms are electrically neutral. When additional electrons are added to them, they become negatively charged.
- In this case, the sweater has a high affinity for electrons and it will attract the ones on the leather jacket.
- The leather jacket has a low tendency to attract electrons than the sweater and it will lose some of its electrons to the sweater.
Learn more:
Protons, neutrons and electrons brainly.com/question/2757829
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I'll be happy to solve the problem using the information that
you gave in the question, but I have to tell you that this wave
is not infrared light.
If it was a wave of infrared, then its speed would be close
to 300,000,000 m/s, not 6 m/s, and its wavelength would be
less than 0.001 meter, not 12 meters.
For the wave you described . . .
Frequency = (speed) / (wavelength)
= (6 m/s) / (12 m)
= 0.5 / sec
= 0.5 Hz .
(If it were an infrared wave, then its frequency would be
greater than 300,000,000,000 Hz.)
As we know that electrostatic force between two charges is given as
here we know that electrostatic repulsion force is balanced by the gravitational force between them
so here force of attraction due to gravitation is given as
here we can assume that both will have equal charge of magnitude "q"
now we have
now we have
Explanation:
Given:
v₀ = 0 m/s
a = 2.50 m/s²
t = 4 s
Find: v
v = at + v₀
v = (2.50 m/s²) (4 s) + 0 m/s
v = 10 m/s
Answer:
Explanation:
Using the magnification formula.
Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)
M = v/u = H1/H2
v/u = H1/H2...1
3) Given the radius of curvature of the concave lens R = 20cm
Focal length F = R/2
f = 20/2
f = 10cm
Object distance u = 5cm
Object height H2= 5cm
To get the image distance v, we will use the mirror formula
1/f = 1/u+1/v
1/v = 1/10-1/5
1/v = (1-2)/10
1/v =-1/10
v = -10cm
Using the magnification formula
(10)/5 = H1/5
10 = H1
H1 = 10cm
Image height of the peg is 10cm
4) If u = 15cm
1/v = 1/f-1/u
1/v = 1/10-1/15
1/v = 3-2/30
1/v = 1/30
v = 30cm
30/15 = H1/5
15H1 = 150
H1/= 10cm
5) if u = 20cm
1/v = 1/f-1/u
1/v = 1/10-1/20
1/v = 2-1/20
1/v = 1/20
v = 20cm
20/20 = H1/5
20H1 = 100
H1 = 5cm
6) If u = 30cm
1/v = 1/f-1/u
1/v = 1/10-1/30
1/v = 3-1/30
1/v = 2/30
v = 30/2 cm
v =>15cm
15/30 = Hi/5
30H1 = 75
H1 = 75/30
H1 = 2.5cm
