Answer:
1) The speed of sound increases
2) 440 Hz
3) 29°C
4) 17°C
5) 434 Hz
6) 12 m/s
7) 17.3 m
Explanation:
1) The speed of sound increases
2) V = f×λ
f = V/λ = 343/0.78 = 439.744 ≈ 440 Hz
3) V = f×λ
512 × 0.68 = 348.16 m/s
348.16 - 331 = 17.16
T = 17.16/0.6 = 28.6 ≈ 29°C
4) Increase in speed = 350 - 340 = 10
Increase in temperature = 10/0.6 = 16.67° ≈ 17°C
5) f = V/λ = 343/0.79 = 434 Hz
6) 331 + 0.6×30 - (331 × 0.6 ×10) = 12 m/s
7) V = 331 + 0.6×25 = 346m/s
λ = 346/20 = 17.3 m
Answer:
When there is wind it takes longer
Explanation:
With no wind, the round trip time is

When we have a constant wind speed w

comparing the reciprocal times;

This means that t1 is smaller than t2, ergo, it takes longer with wind
Answer:
D
Explanation:
because if the solvent is more than the solvent then we can't resolve it.
so our product will be suspended
Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.