California is the third largest state and the only two
bigger states than California are Alaska and Texas so it really depends on how
you want to cross it. There are two routes to cross California depending on how
you plan your visit and places you need to see. Depending on the route you take
crossing California can take from twelve to almost sixteen hours of drive.
Answer:
Explanation:
energy emitted by source per second = .5 J
Eg = 1.43 eV .
Energy converted into radiation = .5 x .12 = .06 J
energy of one photon = 1.43 eV
= 1.43 x 1.6 x 10⁻¹⁹ J
= 2.288 x 10⁻¹⁹ J .
no of photons generated = .06 / 2.288 x 10⁻¹⁹
= 2.6223 x 10¹⁷
wavelength of photon λ = 1275 / 1.43 nm
= 891.6 nm .
momentum of photon = h / λ ; h is plank's constant
= 6.6 x 10⁻³⁴ / 891.6 x 10⁻⁹
= .0074 x 10⁻²⁵ J.s
Total momentum of all the photons generated
= .0074 x 10⁻²⁵ x 2.6223 x 10¹⁷
= .0194 x 10⁻⁸ Js
b ) spectral width in terms of wavelength = 30 nm
frequency width = ?
n = c / λ , n is frequency , c is velocity of light and λ is wavelength
differentiating both sides
dn = c x dλ / λ²
given dλ = 30 nm
λ = 891.6 nm
dn = 3 x 10⁸ x 30 x 10⁻⁹ / ( 891.6 x 10⁻⁹ )²
= 11.3 x 10¹² Hz .
c )
10 nW = 10 x 10⁻⁹ W
= 10⁻⁸ W .
energy of 50 dB
50 dB = 5 B
I / I₀ = 10⁵ ; decibel scale is logarithmic , I is energy of sound having dB = 50 and I₀ = 10⁻¹² W /s
I = I₀ x 10⁵
= 10⁻¹² x 10⁵
= 10⁻⁷ W
= 10 x 10⁻⁸ W
power required
= 10⁻⁸ + 10 x 10⁻⁸ W
= 11 x 10⁻⁸ W.
Answer:
a) y= 3.5 10³ m, b) t = 64 s
Explanation:
a) For this exercise we use the vertical launch kinematics equation
Stage 1
y₁ = y₀ + v₀ t + ½ a t²
y₁ = 0 + 0 + ½ a₁ t²
Let's calculate
y₁ = ½ 16 10²
y₁ = 800 m
At the end of this stage it has a speed
v₁ = vo + a₁ t₁
v₁ = 0 + 16 10
v₁ = 160 m / s
Stage 2
y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²
y₂ = 800 + 150 5 + ½ 11 5²
y₂ = 1092.5 m
Speed is
v₂ = v₁ + a₂ t
v₂ = 160 + 11 5
v₂ = 215 m / s
The rocket continues to follow until the speed reaches zero (v₃ = 0)
v₃² = v₂² - 2 g y₃
0 = v₂² - 2g y₃
y₃ = v₂² / 2g
y₃ = 215²/2 9.8
y₃ = 2358.4 m
The total height is
y = y₃ + y₂
y = 2358.4 + 1092.5
y = 3450.9 m
y= 3.5 10³ m
b) Flight time is the time to go up plus the time to go down
Let's look for the time of stage 3
v₃ = v₂ - g t₃
v₃ = 0
t₃ = v₂ / g
t₃ = 215 / 9.8
t₃ = 21.94 s
The time to climb is
= t₁ + t₂ + t₃
t_{s} = 10+ 5+ 21.94
t_{s} = 36.94 s
The time to descend from the maximum height is
y = v₀ t - ½ g t²
When it starts to slow down it's zero
y = - ½ g t_{b}²
t_{b} = √-2y / g
t_{b} = √(- 2 (-3450.9) /9.8)
t_{b} = 26.54 s
Flight time is the rise time plus the descent date
t = t_{s} + t_{b}
t = 36.94 + 26.54
t =63.84 s
t = 64 s
Answer:
18 m
Explanation:
Given : vo = 0 m/s ; t = 3 s; a = 4 m/s^2 ; d = ? m ; average velocity = ? m/s ; fonal velocity = ? m/s
solving for the final velocity, v
v = a * t
v = 4 m/s^2 * 3 s
v = 12 m / s
Solving for the average velocity. avg v
avg v = (vo + v) / 2
avg v = (0 m / s + 12 m/s) / 2
avg v = 6 m / s
Solving for the distance traveled after 3 s
d = avg v * t
d = 6 m / s * 3 s
d = 18 meters
In the first 3s the car travels 18 meters.
Friction occurs between two contacting surfaces. The coefficient of friction is very much dependent on the roughness of these surfaces. Some of the many ways in which the coefficient can be lessened or decreased are to lubricate the surface or make it shiny by eliminating the spikes which caused the roughness.
