What do you mean? What’s the question?
Answer:
5156.37 ft
Explanation:
Given data:
weight ( W ) = 243,000 Ib
Motion of shuttle ; from 189 mi/hr to 101 mi/hr
dv/dt = -0.0002 V^2
I/v * dv/dt = -0.0002 ds
<em>Convert mi/hr to ft/s</em> ( 1 mi/hr = 1.467 ft/s)
189 mi/hr = 277.263 ft/s
101 mi/hr = 148.167 ft/s
After Integrating
In ( 148.167 / 277.263 ) = -0.0002 ( S1 - S2 )
S1 - S2 = -0.627 / -0.0002
S1 - S2 = 3135 ft/s
Now from 101 mi/hr to 35 mi/hr
dv/dt = ( - 0.0002 V^2 + 3.5 )
ds = V*dv / ( -0.0002 v^2 - 3.5 )
given : 35 mi/hr = 51.345 ft/s
101 mi/hr = 148.167 ft/s
Integrate
S3 - S2 = - In( 0.0002 v^2 + 3.5 ) / 0.0002 * 2 ]
= 1644.75 ft/s
S4 - S3 = 376.62 ft/s
attached below is the remaining part of the solution
Total distance travelled = 3135 + 1644.75 + 376.62 = 5156.37 ft
Answer:
0.266666667mi/1min
Explanation:
12 divided by 45 = 0.266666667
Answer:
4.5 x 10¹⁴ Hz
666.7 nm
1.8 x 10⁵ J
The color of the emitted light is red
Explanation:
E = energy of photons of light = 2.961 x 10⁻¹⁹ J
f = frequency of the photon
Energy of photons is given as
E = h f
2.961 x 10⁻¹⁹ = (6.63 x 10⁻³⁴) f
f = 4.5 x 10¹⁴ Hz
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of photon
Using the equation
c = f λ
3 x 10⁸ = (4.5 x 10¹⁴) λ
λ = 0.6667 x 10⁻⁶ m
λ = 666.7 x 10⁻⁹ m
λ = 666.7 nm
n = number of photons in 1 mole = 6.023 x 10²³
U = energy of 1 mole of photons
Energy of 1 mole of photons is given as
U = n E
U = (6.023 x 10²³) (2.961 x 10⁻¹⁹)
U = 1.8 x 10⁵ J
The color of the emitted light is red
Answer:
C = 771.35 J/kg°C
Explanation:
Here, e consider the conservation of energy equation. The conservation of energy principle states that:
Heat Given by Metal Piece = Heat Absorbed by Water + Heat Absorbed by Container
Since,
Heat Given or Absorbed by a material = m C ΔT
Therefore,
m₁CΔT₁ = m₂CΔT₂ + m₃C₃ΔT₃
where,
m₁ = Mass of Metal Piece = 2.3 kg
C = Specific Heat of Metal = ?
ΔT₁ = Change in temperature of metal piece = 165°C - 18°C = 147°C
m₂ = Mass of Metal Container = 3.8 kg
ΔT₂ = Change in temperature of metal piece = 18°C - 15°C = 3°C
m₃ = Mass of Water = 20 kg
C₃ = Specific Heat of Water = 4200 J/kg°C
ΔT₃ = Change in temperature of water = 18°C - 15°C = 3°C
Therefore,
(2.3 kg)(C)(147°C) = (3.8 kg)(C)(3°C) + (20 kg)(4186 J/kg°C)(3°C)
C[(2.3 kg)(147°C) - (3.8 kg)(3°C)] = 252000 J
C = 252000 J/326.7 kg°C
<u>C = 771.35 J/kg°C</u>
