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2 years ago

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A 0.020-kg bullet traveling at a speed of 300 m/s embeds in a 1.0-kg wooden block resting on a horizontal surface. The block sli

des horizontally 4.0 m on a surface before stopping. What is the coefficient of friction between the block and the surface?

1 answer:

lys-0071 [83]2 years ago

5 0

Answer:

The coefficient of friction between the block and the surface is μk=0.441

Explanation:

Let

mb = bullet mass = 0.02 kg

vb = bullet speed = 300 m/s

mw = mass of the wood block = 1 kg

d = distance covered by the block plus the bullet = 4 m

u = speed of the bullet plus the block

g = gravity = 9.8 m/s2

μk = coefficient of friction, variable unknown

We will use the following expression to calculate the coefficient of friction:

μk = $\frac{(\frac{mb*vb}{mb+mw})^{2} }{2*g*d}$

Replacing known values:

μk= $\frac{(\frac{(0.02kg)*(300m/s)}{0.02kg+1kg})^{2} }{2*9.8m/s^{2}*4m }=0.441$

The coefficient of friction is a dimensionless value

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melisa1 [442]

Taking the vertical component of the displacement
1.1 - 0.2 = 0.9 mile
The horizontal component of the displacement
-0.3 mile

The magnitude of the displacement is
√[ (0.9)² + (-0.3) ] = 0.95 mile

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3 years ago

A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl

Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=157^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.$

Therfore,

$\lambda^{'}-\lambda=4.64 p.m.$

Here, the photon's incident wavelength is $\lamda=7.33pm$

So,

$\lambda^{'}=7.33+4.64=11.97 p.m$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

here, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda$ and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therfore,

$\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.$

This is the de Broglie wavelength of the electron after scattering.

8 0

3 years ago

Vector A has a magnitude of 63 units and points west, while vector B has the same magnitude and points due south. Find the magni

ozzi

Given :

Vector A has a magnitude of 63 units and points west, while vector B has the same magnitude and points due south.

To Find :

The magnitude and direction of

a) A + B .

b) A - B.

Solution :

Let , direction in north is given by +j and east is given by +i .

So , $A=-63i$ and $B=63j$

Now , A + B is given by :

$A+B=-63i+63j$

$| A+B | = 63\sqrt{2}$

Direction of A+B is 45° north of west .

Also , for A-B :

$A-B=-63i-63j$

$|A-B|=63\sqrt{2}$

Direction of A-B is 45° south of west .

( When two vector of same magnitude which are perpendicular to each other are added or subtracted the resultant is always 45° from each of them)

Hence , this is the required solution .

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2 years ago

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antoniya [11.8K]

Answer:

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Explanation:

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2 years ago

The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi

Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

$\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$

Where,

B= Magnetic Field

l = length

$\mu_0$ = Vacuum permeability

$\epsilon_0$ = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

$B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}$

Recall that the speed of light is equivalent to

$c^2 = \frac{1}{\mu_0 \epsilon_0}$

Then replacing,

$B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}$

$B = \frac{r}{2C^2} \frac{dE}{dt}$

Our values are given as

$dE = 2150N/C$

$dt = 5s$

$C = 3*10^8m/s$

$D = 0.440m \rightarrow r = 0.220m$

Replacing we have,

$B = \frac{r}{2C^2} \frac{dE}{dt}$

$B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}$

$B =5.25*10^{-16}T$

Therefore the magnetic field around this circular area is $B =5.25*10^{-16}T$

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2 years ago