A 200kg bucket of cement<br>

Suppose a light source is emitting red light at a wavelength of 700 nm and another light source is emitting ultraviolet light at

klasskru [66]

Answer:

b) twice the energy of each photon of the red light.

Explanation:

$\lambda$ = Wavelength

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

c = Speed of light = $3\times 10^8\ m/s$

Energy of a photon is given by

$E=h\nu\\\Rightarrow E=h\dfrac{c}{\lambda}$

Let $\lambda_1$ = 700 nm

$\lambda_2=350\\\Rightarrow \lambda_2=\dfrac{\lambda_1}{2}$

For red light

$E_1=\dfrac{hc}{\lambda_1}$

For UV light

$E_2=\dfrac{hc}{\dfrac{\lambda_1}{2}}$

Dividing the equations

$\dfrac{E_1}{E_2}=\dfrac{\dfrac{hc}{\lambda_1}}{\dfrac{hc}{\dfrac{\lambda_1}{2}}}\\\Rightarrow \dfrac{E_1}{E_2}=\dfrac{1}{2}\\\Rightarrow E_2=2E_1$

Hence, the answer is b) twice the energy of each photon of the red light.

7 0

3 years ago

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A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1

Archy [21]

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

$t_{s}$ = 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

3 0

3 years ago

What do we measure energy in

Goshia [24]

Energy is measured by joule(j)

3 0

3 years ago

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As Clinton walks he pushes his shoe against the track.

Valentin [98]

Answer:A:The track pushes back on Clinton's shoe with the same force.

Explanation:According to Newton's third law of motion, for every action force there is an equal and opposite reaction force. In this case, the action force is Clinton's shoe pushing on the track. As this happens, there is an equal and opposite reaction force in which the track pushes back on Clinton's shoe with the same force.

4 0

3 years ago

You have a pulley 10.4 cm in diameter and with a mass of 2.3 kg. You get to wondering whether the pulley is uniform. That is, is

madreJ [45]

Answer:

Explanation:

Given

Diameter of Pulley=10.4 cm

mass of Pulley(m)=2.3 kg

mass of book $(m_0)=1.7 kg$