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ikadub [295]
3 years ago
14

Is a charge of 5.8×10^-18 is possible

Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
6 0
I think yes. possible. Because electron charge is much smaller than that..
You might be interested in
Determine the specific heat of a certain metal if a 450 gram sample of it loses 34 500 Joules of heat as its temperature starts
slamgirl [31]

Answer:

c = 0.4356 J/gK

Explanation:

Given the following data;

Mass = 450 grams

Initial temperature, T1 = 150°C

Final temperature, T2 = 53°C

Quantity of heat = 34500 Joules

To find the specific heat capacity of the metal;

Heat capacity is given by the formula;

Q = mcdt

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 53 - 150

dt = -97°C

Converting the temperature in Celsius to Kelvin, we have;

dt = 273 + (-97) = 176 Kelvin

Making c the subject of formula, we have;

c = \frac {Q}{mdt}

Substituting into the equation, we have;

c = \frac {34500}{450*176}

c = \frac {34500}{79200}

c = 0.4356 J/gK

6 0
3 years ago
What happens when an object is moved against gravity, such as rolling a toy car up a
Anarel [89]

Answer:

gravitational force remain constant

7 0
3 years ago
Me and my little sister need help!!
dezoksy [38]
#3 Is insulators
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4 0
3 years ago
Read 2 more answers
A test charge of +3 µC is at a point P where the electric field due to the other charges is directed to the right and has a magn
umka2103 [35]

Answer:

Electric field will remain same.

Explanation:

Given that

At initial condition ,charge q = +3μ C

Electric field E = 4106 N/c

As we know that

Electric field due to charge q

E=\dfrac{kq}{r^2}  

Now when charge is replaced by new charge q'= -3μ C

From the expression of electric field we can say that electric field will remain same from same quantity of electric charge.

So we can say that electric field will remain same.

5 0
4 years ago
An electron moving along the x axis has an initial speed of 1 × 106 m/s at the origin. Its speed is reduced to 5 × 105 m/s at th
mina [271]

Answer:

-2.13 V

Explanation:

Given parameters are:

v_i = 1 * 10^6 m/s

v_f = 5 * 10^5 m/s

q = -e = 1.602 * 10^{-19} C

m = 9.109 * 10^{-31} kg

By conservation of energy principle, we know that the potential difference between two points is equal to the change in the kinetic energy between these points.

-\Delta U = \Delta KE\\\\-e\Delta V = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2

Hence,

1.602*10^{-19}\Delta V = \frac{1}{2}*9.11*10^{-31}((5*10^5)^2 - (1*10^6)^2)

⇒ \Delta V = -2.13 V

5 0
3 years ago
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