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KIM [24]
3 years ago
13

What is the difference between charging by contact and charging by induction in terms of electron transfer.

Physics
1 answer:
Veronika [31]3 years ago
5 0

Answer:

the main difference between charging by contact and charging by induction is that in the first case, the two objects are touching, while in the second case, the two objects do not touch

Explanation:

There are three methods of charging an object:

- Charging by friction: this is done by rubbing an object against another object. An example is when a plastic rod is rubbed with a wool cloth. When this is done, electrons are transferred from the wool to the rod, so both objects remain charged at the end of the process

- Charging by contact: this is done by putting in contact a charged object with a neutral, conducting object. In this case, the charges are transferred from the charged object to the neutral object; at the end of the process, the neutral object will also have a net electric charge, so it will be also charged.

- Charging by induction: in this case, we take a charged object, and a neutral object, and we place the two objects close to each other, but without touching. Let's assume that the charged object is negatively charged: in this case, the positive charges in the neutral object are attracted towards the negative charges of the charged object, while the negative charges of the neutral object are repelled away. As a result, the positive and negative charges in the neutral object split apart. If the object is connected to the ground, then negative charges move away, so the neutral object will remain positively charged.

Therefore, the main difference between charging by contact and charging by induction is that in the first case, the two objects are touching, while in the second case, the two objects do not touch.

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D. A bargain hunter purchases a "gold" crown at a flea market. After she gets home, she hangs it from a scale and finds its weig
s2008m [1.1K]

Answer:

Wc = 7.84    weight of crown

Ww = 7.84 - 6.86 = .98       weight of water displaced

Density = 7.84 / .98 = 8     crown is 8 X that of water

Since gold has a density of 19.3 that of water the crown is certainly not 100 percent (if any) gold  

4 0
2 years ago
A proton, moving north, enters a magnetic field. Because of this field, the proton curves downward. We may conclude that the mag
PIT_PIT [208]

Answer:

towards the west

4 0
3 years ago
An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process,
Sedbober [7]

Answer:

a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa

Explanation:

a. Internal energy and the relative specific volume at s_1 are determined  from A-17:u_1=214.07kJ/kg, \ \alpha_r_1=621.2.

The relative specific volume at s_2 is calculated from the compression ratio:

\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825

#from this, the temperature and enthalpy at state 2,s_2 can be determined using interpolations T_2=862K and h_2=890.9kJ/kg. The specific volume at s_1 can then be determined as:

\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg

Specific volume,s_2:

\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg

The pressures at s_2 \ and\  s_3 is:

P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa

.The thermal efficiency=> maximum temperature at s_3 can be obtained from the expansion work at constant pressure during s_2-s_3

\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K

b.Relative SV and enthalpy  at s_3 are obtained for the given temperature with interpolation with data from A-17 :a_r_3=4.553 \ and\  h_3=1909.62kJ/kg

Relative SV at s_4 is

a_r_4=\frac{r}{r_c}\alpha _r_3

==\frac{16}{2}\times4.533\\=36.424

Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;

n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:

MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa

Hence, the mean effective relative pressure is 674.95kPa

3 0
3 years ago
A bat hits a moving baseball. If the bat delivers a net eastward impulse of 1.3 N-s and the ball starts with an initial horizont
Georgia [21]

Answer:

141.30grams

Explanation:

First, denote our known values;

J=1.3N-s(+,east)\\u=-3.8ms^-^1(-,west)\\v=5.4ms^-^1(+,east)\\m=?

Mass is impulse divided by change in velocity:

m=\frac{J}{v-u}\\=\frac{1.3}{5.4--3.8}\\\\=\frac{1.3}{9.2}\\=0.1413Kgs

Hence, the mass of the ball is 141.30grams

3 0
3 years ago
A 1.50 cm high diamond ring is placed 20.0 cm from a concave mirror with radius of curvature 30.00 cm. The magnification is ____
Rama09 [41]

Answer:

Magnification, m = -0.42

Explanation:

It is given that,

Height of diamond ring, h = 1.5 cm

Object distance, u = -20 cm

Radius of curvature of concave mirror, R = 30 cm

Focal length of mirror, f = R/2 = -15 cm (focal length is negative for concave mirror)

Using mirror's formula :

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}, f = focal length of the mirror

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}

\dfrac{1}{v}=\dfrac{1}{-15}+\dfrac{1}{-20}

v = -8.57 cm

The magnification of a mirror is given by,

m=\dfrac{-v}{u}

m=\dfrac{-(-8.57)}{-20}

m = -0.42

So, the magnification of the concave mirror is 0.42. Thew negative sign shows that the image is inverted.

5 0
3 years ago
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