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3 years ago

l o which of the following can move from one atom to another A. protons Msideus. B. neutrons C. electrons the nucleus

Physics

1 answer:

Tanzania [10]3 years ago

5 0

Electrons can move from one atom to another.

When a lot of them are doing it at the same time,
you have an electric current.

We asked around here at Brainly, and nobody knows
what an "Msideus" is, but we all know that there aren't
any of them in atoms.

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What is elasticity in polymer generally related to?

Andrej [43]

The elasticity of a polymer is primarily due to the structure of the molecule and the cross-linking between strands. Hydrogen bonding is a contributor to the shape of the molecule, but not a major player in terms of elasticity. We would have to answer "false".
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6 0

3 years ago

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In Millikan's experiment, an oil drop of radius 1.362 μm and density 0.888 g/cm3 is suspended in chamber C when a downward-point

Misha Larkins [42]

Answer:

The charge on the oil drop is 3e

Explanation:

F = qE

Where;

F is the applied force in Newton

E is the electric field potential N/C

q is charge in C

Given;

Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m

density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³

Electric field potential = 1.92 ✕ 10⁵ N/C

F =mg

mass of the oil drop = density, ρ X volume of the oil drop

volume of the oil drop (spherical) = (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³

⇒ volume of the oil drop = 10.584 X 10⁻¹⁸ m³

mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)

⇒ mass of the oil drop = 9.399 X 10⁻¹⁵ kg

⇒ F =mg = 9.399 X 10⁻¹⁵ kg X 9.8 = 9.21 X10⁻¹⁴ N

F = qE

q = F/E

q = (9.21 X10⁻¹⁴)/(1.92 ✕ 10⁵) = 4.797 X 10⁻¹⁹ C

In terms of e

1e = 1.6 X10⁻¹⁹ C

= (4.797 X 10⁻¹⁹ C)/(1.6 X10⁻¹⁹ C) = 3e

Therefore, the charge on the oil drop is 3e

7 0

3 years ago

A piston-cylinder device initially contains 0.08 m3 of nitrogen gas at 150 kPa and 200°C. The nitrogen is now expanded to a pres

Lemur [1.5K]

Answer:

$V_2 = 0.125 m^3$

Work done = = 5 kJ

Explanation:

Given data:

volume of nitrogen $v_1 = 0.08 m^3$

$P_1 = 150 kPa$

$T_1 = 200 degree celcius = 473 Kelvin$

$P_2 = 80 kPa$

Polytropic exponent n = 1.4

$\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}$

putting all value

$\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}$

$\frac{T_2} = 395.23 K = 122.08 DEGREE \ CELCIUS$

polytropic process is given as

$P_1 V_1^n = P_2 V_2^n$

$150\times 0.08^{1.4} = 80 \times V_2^{1.4}$

$V_2 = 0.125 m^3$

work done $= \frac{P_1 V_1 -P_2 V_2}{n-1}$

$= \frac{150 \times 0.8 - 80 \times 0.125}{1.4-1}$

= 5 kJ

4 0

3 years ago

a train traveling at 30m/s comes to a complete stop in 20 sec.how far did the train go during the deceleration

elixir [45]

Answer:

600m

Explanation:

v=30m/s

t=20s

s=VT

20×30=600m

3 0

3 years ago

Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

$q_1 = 21.0\mu C$

$q_1 = 47.0\mu C$

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

$E = \frac{kq}{x}^2$

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

$\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2$

solving for x we get

$\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}$

x = 0.200 m

b)electric field at half way mean x =0.25

$E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}$

E = 3.84*10^{-4} N/C

6 0

3 years ago

Read 2 more answers