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3 years ago

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Immediately after being struck by a hammer, the nail (mass of 50 g) has a velocity of 50 m/s. The total frictional force is 62.5

kM. How far does the nail move before it comes to a stop

1 answer:

Marina86 [1]3 years ago

8 0

Answer:

nail will stop after traveling 1000 m

Explanation:

We have given mass m = 50 gram = 0.05 kg

Frictional force which is used to stop the mass F = 62.5 kN

Initial velocity is given u = 50 m/sec

From newton's law force is equal to F = ma, here m is mass and a is acceleration

So $a=\frac{F}{m}=\frac{62.5\times 10^3}{0.05}=1.25\times 10^6m/sec^2$

As finally nail stops so final velocity v = 0 m/sec

From third equation of motion $v^2=u^2+2as$

So $0^2=50^2-2\times 1.25\times 10^6\times s$

s = 1000 m

So nail will stop after traveling 1000 m

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An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r

bixtya [17]

1) Linear charge density of the shell: $-2.6\mu C/m$

2) x-component of the electric field at r = 8.7 cm: $1.16\cdot 10^6 N/C$ outward

3) y-component of the electric field at r =8.7 cm: 0

4) x-component of the electric field at r = 1.15 cm: $1.28\cdot 10^7 N/C$ outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

1)

The linear charge density of the cylindrical insulating shell can be found by using

$\lambda_2 = \rho A$

where

$\rho = -567\mu C/m^3$ is charge volumetric density

A is the area of the cylindrical shell, which can be written as

$A=\pi(b^2-a^2)$

where

$b=4.7 cm=0.047 m$ is the outer radius

$a=2.7 cm=0.027 m$ is the inner radius

Therefore, we have :

$\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m$

2)

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

$E=E_1+E_2$

where:

$E_1=\frac{\lambda_1}{2\pi r \epsilon_0}$

where

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

$E_2=\frac{\lambda_2}{2\pi r \epsilon_0}$

where

$\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m$

And this field points radially inward, because the charge is negative.

Therefore, the net field is

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C$

in the outward direction.

3)

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

$E_y=0$

4)

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}$

where:

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

$E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C$

5)

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0

3 years ago