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tensa zangetsu [6.8K]
2 years ago
10

The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0. Exp

ress your answer in terms of any or all of m, F, and t.
Physics
1 answer:
olganol [36]2 years ago
6 0

Answer:

Ft

Explanation:

We are given that

Initial velocity=u=0

We have to find the magnitude of p of the momentum of the particle at time t.

Let mass of particle=m

Applied force=F

Acceleration, a=\frac{F}{m}

Final velocity , v=a+ut

Substitute the values

v=0+\frac{F}{m}t=\frac{F}{m}t

We know that

Momentum, p=mv

Using the formula

p=m\times \frac{F}{m}t=Ft

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2 years ago
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A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of 1.055\times 10^{- 7} in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV =250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, m_{p} = 1.67\times 10^{- 27} kg

The initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This energy of proton is \simeq 250 GeV

Thus the speed of the proton, v\simeq c

Now, the time taken to cover 1 km = 1000 m of the distance:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

Now, in accordance to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus the increase in wave packet's width is relatively quite small.

Hence, we can say that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

3 0
2 years ago
A ship sends out a 1200 Hz sound wave, which has a wavelength of 120 cm in the water. What would happen if that ship sent out a
OlgaM077 [116]

Answer:

the wave length becomes doubled or becomes two times the initial wavelength = 240 cm

Explanation:

From wave,

v = λf................ Equation 1

Where v = velocity of the wave, λ = wavelength of the wave, f = frequency of the wave.

Given: f = 1200 Hz, λ = 120 cm = 1.2 m

Substitute into equation 1

v = 1200(1.2)

v = 1440 m/s.

When the ship sent out a 600 Hz sound wave,

make λ the subject of formula in equation 1

λ = v/f............. Equation 2

Given: f = 600 Hz, v = 1440 m/s

Substitute into equation 2

λ = 1440/600

λ = 2.4 m or 240 cm.

When the ship sent out a 600 Hz sound wave instead, the wave length becomes doubled or becomes two times the initial wavelength = 240 cm

3 0
2 years ago
A circular pipe of 25-mm outside diameter is placed in an airstream at 25C and 1-atm pressure. The air moves in cross flow over
kifflom [539]

Answer:

f_D = =3.24 N/m

Explanation:

data given

properties of air\nu\ of air =19.31*10^{-6} m2/s

\rho = 1.048 kg/m3

k = 0.0288 W/m.K

WE KNOW THAT

Reynold's number is given as

Re =\frac{VD}{\nu}

      = \frac{ 15*0.025}{19.31*10^{-6}}

      = 1.941 *10^4

drag coffecient is given as

C_D = \frac{f_D}{A_f\frac{\rho v^2}{2}}

solving for f_D

f_D = C_D A_f*\frac{\rho v^2}{2}

     =C_D D*\frac{\rho v^2}{2}

Drag coffecient for smooth circular cylinder is 1.1

therefore Drag force is

f_D = 1.1*0.025 *\frac{1.048*15^2}{2}

f_D = =3.24 N/m

4 0
2 years ago
What is the best approximate value for the elastic potential energy (EPE) of the spring elongated by 3.0 meters?
DIA [1.3K]

The elastic potential energy of the spring is 6.8 J

Explanation:

The elastic potential energy of a compressed/stretched spring is given by the equation:

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the elongation of the spring

The spring constant of the spring in this problem can be found by keeping in mind the relationship between force (F) and elongation (x) (Hooke's law):

F=kx

By looking at the graph and comparing it with the formula, we realize that the slope of the force-elongation graph corresponds to the spring constant. Therefore in this case,

k=\frac{15.0-0}{10.0-0}=1.5 N/m

Therefore when the spring has a elongation of x=3.0 m, its potential energy is

E=\frac{1}{2}(1.5)(3.0)^2=6.8 J

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

3 0
2 years ago
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