It is determined by the nature of the green light. Because lasers create light at almost a single frequency, green laser light would appear as a thin line of pure green. Other sources of "green" light emit light at a variety of frequencies, including yellow and blue, resulting in a strong green band in the center that fades into blue-green and yellow-green at the borders.
For example, here’s a graph of the spectrum of a green LED, showing the color range: Attachment #1
and here’s a graph of the transmission spectra of several standard photographic filters, including green: Attachment #2
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The gravitational potential energy will increase by 423.36 J
<h3>How to determine the potential energy at ground level</h3>
- Mass (m) = 72 kg
- Acceleration due to gravity (g) = 9.8 m/s²
- Height (h) = 0 m
- Potential energy at ground level (PE₁) =?
PE = mgh
PE₁ = 72 × 9.8 × 0
PE₁ = 0 J
<h3>How to determine the potential energy at 60 cm (0.6 m)</h3>
- Mass (m) = 72 kg
- Acceleration due to gravity (g) = 9.8 m/s²
- Height (h) = 0.6 m
- Potential energy at 60 cm (0.6 m) (PE₂) =?
PE = mgh
PE₂ = 72 × 9.8 × 0.6
PE₂= 423.36 J
<h3>How to determine the change in potential energy </h3>
- Potential energy at ground level (PE₁) = 0 J
- Potential energy at 60 cm (0.6 m) (PE₂) = 423.36 J
- Change in potential energy =?
Change in potential energy = PE₂ - PE₁
Change in potential energy = 423.36 - 0
Change in potential energy = 423.36 J
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Answer:
V = 331.59m/s
Explanation:
First we need to calculate the time taken for the shell fire to hit the ground using the equation of motion.
S = ut + 1/2at²
Given height of the cliff S = 80m
initial velocity u = 0m/s²
a = g = 9.81m/s²
Substitute
80 = 0+1/2(9.81)t²
80 = 4.905t²
t² = 80/4.905
t² = 16.31
t = √16.31
t = 4.04s
Next is to get the vertical velocity
Vy = u + gt
Vy = 0+(9.81)(4.04)
Vy = 39.6324
Also calculate the horizontal velocity
Vx = 1330/4.04
Vx = 329.21m/s
Find the magnitude of the velocity to calculate speed of the shell as it hits the ground.
V² = Vx²+Vy²
V² = 329.21²+39.63²
V² = 329.21²+39.63²
V² = 108,379.2241+1,570.5369
V² = 109,949.761
V = √ 109,949.761
V = 331.59m/s
Hence the speed of the shell as it hits the ground is 331.59m/s
I think the answer would be physics.
Answer:
The thermometer makes use of a physical property of a thermometric substance which changes continuously with temperature. The physical property is referred to as thermometric property.
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Thermometric Properties Used In Various Thermometers.