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iragen [17]
3 years ago
7

A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a

ir drag coefficient cD is given by cD= 0.001 m-1. What is the maximum height that the ball reaches? Compare this to the maximum height achieved when air drag is neglected.
Physics
1 answer:
stira [4]3 years ago
3 0

Answer:

Explanation:

Given

acceleration is given by

a=-g-c_Dv^2

where \ddot{y}=a

\dot{y}=v

Also acceleration is given by

a=v\frac{\mathrm{d} v}{\mathrm{d} s}

ds=\frac{v}{a}dv

\int ds=\int \frac{v}{-g-0.001v^2}dv

\Rightarrow Let -g-0.001v^2=t

-0.001\times 2vdv=dt

vdv=-\frac{dt}{0.002}

at\ v_0=50\ m/s,\ t=-g-0.001(50)^2

t=-g-2.5

at v=0,\ t=-g

\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}

\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}

s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}

s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})

s=113.608\ m

when air drag is neglected maximum height reached is

h=\frac{v_0^2}{2g}

h=\frac{50^2}{2\times 9.8}

h=127.55\ m

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An LED operation at 850 nm center wavelength has a spectral width of 45 nm. What is the pulse spreading in ns/km
Papessa [141]

Answer:

\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}

Explanation:

From the given information, the LED is operating with a given wavelength of 850 nm or 0.85 μm.

Hence, the material dispersion is \dfrac {d \tau _{mat}}{d \lambda } \simeq (80 \ ps / (nm.km) \ )

Now, using the pulse spread formula:

\dfrac{\sigma_{mat}}{L} = \dfrac{d \tau _{mat} }{d \lambda} \sigma \lambda

\dfrac{\sigma_{mat}}{L} = (80 \ ps/ ( m.km) \ )  \times (45 \ nm)

Thus, the pulse spreading as a result of  material dispersion is:\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}

3 0
2 years ago
Two identical objects in outer space have a head-on collision and stick together. If, before the collision, one had been moving
julia-pushkina [17]

Answer:

1.5 m/s

Explanation:

Momentum is conserved and conservation of momentum is

p₁ + p₂ = p'₁ + p'₂

or

m₁v₁ + m₂v₂ = m₁v'₁  + m₂v'₂

In our problem, after collision v'₁ will be equal to v'₂.

Since objects are identical m₁ = m₂

m(v₁+ v₂) = 2m x v'₁

(2m/s + 1m/s) = 2v'₁

v'₁ = v'₂ = 1.5 m/s

5 0
3 years ago
HELP ;
mestny [16]

Answer:

The frequency would double.

Explanation:

Given:

Speed of wave (v) = constant.

Frequency of wave initially (f₁) = 2 Hz

Initial wavelength of the wave (λ₁) = 1 m

Final wavelength of the wave (λ₂) = 0.5 m

Final frequency of the wave (f₂) = ?

We know that the product of wavelength and frequency of the wave is equal to the speed of the wave.

Therefore, framing in equation form, we have:

Wavelength × Frequency = Speed

\lambda\times f=v

It is given that speed of the wave remains the same. So, the product must always be a constant.

Therefore,

\lambda\times f=constant\ or\ \\\lambda_1\times f_1=\lambda_2\times f_2

Now, plug in the given values and solve for 'f₂'. This gives,

1\times 2=0.5\times f_2\\\\f_2=\frac{2}{0.5}=4\ Hz

Therefore, the final frequency is 4 Hz which is double of the initial frequency.

f₂ = 2f₁ = 2 × 2 = 4 Hz

So, the second option is correct.

7 0
3 years ago
Read 2 more answers
A baseball is hit with a bat. The direction of the ball is completely reversed and its speed is doubled. If the actual contact w
alexdok [17]
Before the impact, let the velocity of the baseball was v m/s.

After being hit by the bat its velocity is -2v
So, change in velocity, Deltav=v-(-2v)=3v
Acceleration is defined as the rate of change in velocity, i.e. actual change in velocity divided by the time taken to change it. Time taken to change velocity is the time of actual contact of the bat and ball, i.e. 0.31 s.

a=(Deltav)/(Deltat)
=(3v)/0.37
Therefore, a/v=3/0.31=9.7 s^-1
So, the ratio of acceleration of the baseball to its original velocity is 9.7.
8 0
3 years ago
Read 2 more answers
An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field
lawyer [7]

Answer:

a) ΔV = 25.59 V, b)  ΔV = 25.59 V,  c)  v = 7 10⁴ m / s,  v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

          Em₀ = U = e DV

final point. Where they hit the target

          Em_f = K = ½ m v2

energy is conserved

          Em₀ = Em_f

         e ΔV = ½ m v²

         ΔV = \frac{1}{2} mv²/e     (1)

If the speed of light is c and this is 100% then 1% is

         v = 1% c = c / 100

         v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

         ΔV = \frac{1}{2}  \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }

         ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

             K_e = K_p

              K_p = ½ m v_e²

              K_p = \frac{1}{2}  9.1 10⁻³¹ (3 10⁶)²

              K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

              ΔV = Kp / M

              ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

              ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

             e ΔV = ½ M v²

             v² = 2 e ΔV / M

             v² = \frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }

              v² = 49,035 10⁸

               v = 7 10⁴ m / s

Relation

        v/c = \frac{7 \ 10^4 }{ 3 \ 10^8}

        v/c= 2.33 10⁻⁴

8 0
3 years ago
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