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gogolik [260]
3 years ago
14

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?
Physics
1 answer:
lys-0071 [83]3 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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A string of length L, fixed at both ends, is capable of vibrating at 309 Hz in its first harmonic. However, when a finger is pla
Scorpion4ik [409]

Answer:

 i = 0.3326 L

Explanation:

A fixed string at both ends presents a phenomenon of standing waves, two waves with the same frequency that are added together. The expression to describe these waves is

    2 L = n λ           n = 1, 2, 3…

The first harmonic or leather for n = 1

Wave speed is related to wavelength and frequency

     v = λ f

     λ = v / f

Let's replace in the first equation

    2 L = 1 (v / f₁)

For the shortest length L = L-l

   2 (L- l) = 1 (v / f₂)

These two equations form our equation system, let's eliminate v

    v = 2L f₁

    v = 2 (L-l) f₂

    2L f₁ = 2 (L-l) f₂

    L- l = L f₁ / f₂

    l = L - L f₁ / f₂

    l = L (1- f₁ / f₂)

.

Let's calculate

    l / L = (1- 309/463)

    i / L = 0.3326

4 0
3 years ago
In the Hydrogen atom, the energy spacing between the is 4.07 x 101 J (Joules). When an is the frequency of the photons emitted?
agasfer [191]

Answer:

The frequency of the photon is 3.069\times10^{14}\ Hz.

Explanation:

Given that,

Energy E=4.07\times10^{-19}\ J

We need to calculate the energy

Using relation of energy

E_{4}-E_{2}=\Delta E

Where, \Delta E =  energy spacing

4h\nu-2h\nu=4.07\times10^{-19}

\nu=\dfrac{4.07\times10^{-19}}{2h}

Put the value of h into the formula

\nu=\dfrac{4.07\times10^{-19}}{2\times6.63\times10^{-34}}

\nu=3.069\times10^{14}\ Hz

Hence, The frequency of the photon is 3.069\times10^{14}\ Hz.

4 0
3 years ago
Is "A firecracker that has not been lit kinetic, or potential energy?
love history [14]
A firecracker before been lit has potential energy in it. It is chemical potential energy which is due to the explosives in it.When it is lit, it gets converted into heat,light and kinetic energy.
7 0
3 years ago
A solid disk of radius 1.4 cm and mass 430 g is attached by a wire to one of its circular faces. It is twisted through an angle
Neporo4naja [7]

Answer:

    f= 4,186  10²  Hz

Explanation:

El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por

             w = √ k/I

donde ka es constante de torsion de hilo e I es el momento de inercia del disco

El  momento de inercia de indican que giran un eje que pasa                 por enronqueces

           I= ½ M R2  

reduzcamos las cantidades al sistema SI

         R= 1,4 cm = 0,014  m

         M= 430 g = 0,430 kg

substituimos

           w= √ (2 k/M R2)

calculemos  

           w = RA ( 2 370 / (0,430  0,014 2)

           w = 2,963 103 rad/s

la velocidad angular esta relacionada con la frecuencia por

            w =2pi f

            f= w/2π

            f= 2,963 10³/ (2π)

            f= 4,186  10²  Hz

5 0
3 years ago
Gravity is affected by...
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Mass and distance
force /pull of gravity decreases with the increase in separation between the two bodies
the amount of gravity an object possesses is proportional to the mass of that object.
4 0
3 years ago
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