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umka21 [38]
3 years ago
12

What fraction of all the electrons in a 25 mg water

Physics
1 answer:
mihalych1998 [28]3 years ago
3 0

Answer:

9.11\times 10^{-15}.

Explanation:

The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of  -40 nC is removed from the water droplet.

The charge on one electron, \rm e=-1.6\times 10^{-19}\ C.

Let the N number of electrons have charge -40 nC, such that,

\rm Ne=-40\ nC\\\Rightarrow N=\dfrac{-40\ nC}{e}=\dfrac{-40\times 10^{-9}\ C}{-1.6\times 10^{-19}\ C}=2.5\times 10^{11}.  

Now, mass of one electron = \rm 9.11\times 10^{-31}\ kg.

Therefore, mass of N electrons = \rm N\times 9.11\times 10^{-31}=2.5\times 10^{11}\times 9.11\times 10^{-31}=2.2775\times 10^{-19}\ kg.

It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.

Let it is m times the total mass of the droplet which is 25\ \rm mg = 25\times 10^{-6}\ kg.

Then,

\rm m\times (25\times 10^{-3}\ kg) = 2.2775\times 10^{-19}\ kg.\\m=\dfrac{2.2775\times 10^{-19}\ kg}{25\times 10^{-3}\ kg}=9.11\times 10^{-15}.

It is the required fraction of mass of the droplet.

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Bad White [126]

Answer:

V at C is 3.6 m/s

Explanation:

At A kinetic energy is zero and potential energy=mgh=0.5*9.81*0.6=2.943 J

By conservation of energy.

KE+PE=Constant

At C PE=0.6 J

the KE=2.943-0.6=2.343 J

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3 years ago
A rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s. (a) Find the recoil speed of the rifle. m/s (b) If a
asambeis [7]

Answer:

The recoil speed of the man and rifle is v_{man}=0.016 ms^{-1}.

Explanation:

The expression for the force in terms of mg is as follows;

F=mg

Here, m is the mass and acceleration due to gravity.

Rearrange the expression for mass.

m=\frac{F}{g}

Calculate the combined mass of the man and rifle.

m_{man,rifle}=\frac{650+25}{g}

Put g=9.8 ms^{-2}.

m_{man,rifle}=\frac{650+25}{9.8}

m_{man,rifle}=68.88 kg

The expression for the conservation of momentum is as follows as;

m_{man}u_{man}+m_{bullet}u_{bullet}=m_{man}v_{man}+m_{rifle}v_{man,rifle}

Here, m_{man,rifle} is the mass of the man and rifle,  m_{rifle} is the mass of the rifle,u_{man},u{bullet}  are the initial velocities of the man and bullet and v_{man},v{man,rifle} are the final velocities of the man and rifle and rifle.

It is given in the problem that a rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s.

Convert mass of rifle from gram to kilogram.

m_{bullet}=4.5 g

m_{bullet}=.0045 kg

Put m_{bullet}=.0045 kg,m_{man,rifle}=68.88 kg , u_{man,rifle}=0, v_{bullet}= 240 ms^{-1} and u_{bullet}=0.

m_{man}(0)+m_{bullet}(0)=(68.88)v_{man,rifle}+(.0045)(240)

0=(68.88)v_{man,rifle}+(.0045)(240)

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(68.88)v_{man,rifle}=\frac{-1.08}{68.88}

v_{man,rifle}=-0.016 ms^{-1}  

Therefore, the recoil speed of the man and rifle is v_{man}=0.016 ms^{-1}.

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