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4 years ago

6

A worker does 500 J of work on a 10 kg box.If the box transfers 375 J of heat to the floor through the friction between the box

and the floor,what is the velocity of the box after the work has been done on it?

1 answer:

kipiarov [429]4 years ago

3 0

The answer is in attachment

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If 600 J of work is required to move 50 coulombs of charge

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7. A neutron at rest decays (breaks apart) into a proton and an electron. Energy is released in the decay, in the form of kineti

kondaur [170]

Answer:

The fraction of the total energy goes into kinetic energy of proton is $5.44 \times 10^{-4}$

Explanation:

Given:

Mass of proton $m_{p} = 1836 m_{e}$

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

Here neutron at rest decays into proton and electron

$n_{1} ^{0}$ ⇄ $P_{1} ^{1} + e_{0} ^{-1} + Q$

Where $Q =$ energy released

The $Q$ value of this reaction is given by

$Q = (\frac{m_{p} +m _{e} }{m_{e} } ) K$

Where $K =$ kinetic energy of reaction

Here we need to find fraction of the total energy released goes into the kinetic energy of the proton

$\frac{K}{Q} = \frac{m_{e} }{m_{p} + m_{e} }$

$\frac{K}{Q} = (\frac{m_{e} }{1836m_{e} + m_{e} } )$

$\frac{K}{Q} = \frac{1}{1837}$

$\frac{K}{Q} = 5.44 \times 10^{-4}$

Therefore, the fraction of the total energy goes into kinetic energy of proton is $5.44 \times 10^{-4}$

6 0

3 years ago

Find electric field at point p which is a distance l away from the both +q and -q

denis-greek [22]

Answer:

$\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }$

Where

q is the charge
r is the distance
$E$ is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }$

F2 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

⇒

F1+F2= $\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

8 0

4 years ago