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Illusion [34]
3 years ago
5

A system consists of electrons and protons only. It contains 150 electrons and has a total charge of +22e. What is the mass of t

he system? me=9.11×10−31kg, mp=1.673×10−27kg.
Physics
1 answer:
siniylev [52]3 years ago
8 0

Answer:

Mass of system will be 288.605\times 10^{-28}kg

Explanation:

We have given a system consist 150 electrons

Mass of one electron m_{e}=9.1\times 10^{-31}kg

So mass of 150 electron =150\times 9.1\times 10^{-31}=1365\times 10^{-31}=1.365\times 10^{-28}kg

Charge on system is +22e

So number of proton = 150 +22 = 172

Mass of one proton =1.67\times 10^{-27}kg

So mass of 172 proton =172\times 1.67\times 10^{-27}=287.24\times 10^{-27}=287.24\times 10^{-28}kg

So total mass on the system =287.24\times 10^{-28}+1.365\times 10^{-28}=288.605\times 10^{-28}kg

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3 years ago
Read 2 more answers
If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

s_x=v_x_0 t

t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s

Now for the vertical distance

vy_o=0

than the equation of motion becomes

s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2

Now using this acceleration the value of electric field is calculated as

E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0
3 years ago
A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If
Delvig [45]

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s

The induced emf in the shorter coil is calculated as;

E = NA\frac{\delta B}{\delta t}

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

E = NA\frac{\delta B}{\delta t}

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

8 0
3 years ago
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