Answer:
<em>The frequency of of the note = 131 Hz.</em>
Explanation:
<em>Frequency:</em><em> Frequency can be defined as the number of complete oscillation completed by a wave in one seconds. The S.I unit of frequency is Hertz ( Hz)</em>
v = λf ............................ Equation 1
Making f the subject of the equation,
f = v/λ .......................... Equation 2
Where v = Speed, λ = wavelength, f = frequency
<em>Given: v = 343 m/s, λ = 2.62 m.</em>
<em>Substituting these values into equation 2</em>
<em>f = 343/2.62</em>
<em>f = 131 Hz</em>
<em>Thus the frequency of of the note = 131 Hz.</em>
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Answer:
The value of A is 1.5m/s^2 and B is 0.5m/s^³
Explanation:
The mass of the rocket = 2540 kg.
Given velocity, v(t)=At + Bt^2
Given t =0
a= 1.50 m/s^2
Now, velocity V(t) = A*t + B*t²
If, V(0) = 0, V(1) = 2
a(t) = dV/dt = A+2B × t
a(0) = 1.5m/s^²
1.5m/s^² = A + 2B × 0
A = 1.5m/s^2
now,
V(1) = 2 = A× 1 + B× 1^²
1.5× 1 +B× 1 = 2m/s
B = 2-1.5
B = 0.5m/s^³
Now Check V(t) = A× t + B × t^²
So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² × 1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s
Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)
Answer:
<h3>Gold's natural state has a definite shape and a definite volume. Gold's natural state is solid.</h3>
<h3>In nature, gold most often occurs in metallic form, alloyed with silver.</h3>
