Ignoring air resistance, the Kinetic energy before hitting the ground will be equal to the potential energy of the Piton at the top of the rock.
So we have 1/2 MV^2 = MGH
V^2 = 2GH
V = âš2GH
V = âš( 2 * 9.8 * 325)
V = âš 6370
V = 79.81 m/s
Although the sample is not shown in this question, we can conclude that it would be reasonably easy for David to provide evidence of the color, consistency, temperature, and texture of the soil.
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<h3>What are these properties an example of?</h3>
These are all examples of the physical properties of a sample. Since we cannot see the sample that David is using, it would be safest to assume that he would have no trouble providing evidence as to the physical properties of the soil, the:
- Color
- Consistency
- Temperature
- Texture
are all examples of this.
Therefore, we can confirm that David can provide evidence of the color, consistency, temperature, and texture of the soil.
To learn more about physical properties visit:
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Number a is a correct one
Answer:
Zero
Explanation:
Because using
Deta X= dsinစ x n(lambda)
But we know that for central maxima
n is zero
So after substituting
Deta x = 0
Answer:
6 cm long
Explanation:
F = 4110N
Vo(speed of sound) = 344m/s
Mass = 7.25g = 0.00725kg
L = 62.0cm = 0.62m
Speed of a wave in string is
V = √(F / μ)
V = speed of the wave
F = force of tension acting on the string
μ = mass per unit density
F(n) = n (v / 2L)
L = string length
μ = mass / length
μ = 0.00725 / 0.62
μ = 0.0116 ≅ 0.0117kg/m
V = √(F / μ)
V = √(4110 / 0.0117)
v = 592.69m/s
Second overtone n = 3 since it's the third harmonic
F(n) = n * (v / 2L)
F₃ = 3 * [592.69 / (2 * 0.62)
F₃ = 1778.07 / 1.24 = 1433.927Hz
The frequency for standing wave in a stopped pipe
f = n (v / 4L)
Since it's the first fundamental, n = 1
1433.93 = 344 / 4L
4L = 344 / 1433.93
4L = 0.2399
L = 0.0599
L = 0.06cm
L = 6cm
The pipe should be 6 cm long