Answer:
Explanation:
a) Fundamental frequency
A harmonic is an integral multiple of the fundamental frequency.
b) Wave speed
(i) Calculate the wavelength
In a fundamental vibration, the length of the string is half the wavelength.
(b) Calculate the speed
s
Answer:
required distance is 233.35 m
Explanation:
Given the data in the question;
Sound intensity 
= 1.62 × 10⁻⁶ W/m²
distance r = 165 m
at what distance from the explosion is the sound intensity half this value?
we know that;
Sound intensity is proportional to 1/(distance)²
i.e
∝ 1/r²
Now, let r² be the distance where sound intensity is half, i.e ₂ = ₁/2
Hence,
₂/₁ = r₁²/r₂²
1/2 = (165)²/ r₂²
r₂² = 2 × (165)²
r₂² = 2 × 27225
r₂² = 54450
r₂ = √54450
r₂ = 233.35 m
Therefore, required distance is 233.35 m
The question is incomplete. The mass of the object is 10 gram and travelling at a speed of 2 m/s.
Solution:
It is given that mass of object before explosion is,m = 10 g
Speed of object before explosion, v = 2 m/s
Let 
be the masses of the three fragments.
Let 
be the velocities of the three fragments.
Therefore, according to the law of conservation of momentum,
So the x- component of the velocity of the m2 fragment after the explosion is,
∴ 
For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s
Answer:
3secs
Explanation:
Given the following parameters
height H= 81.3m
Velocity v = 12.4m/s
Required
Time it take to reach the ground
Using the equation of motion
H = ut+1/2gt²
81.3 = 12.4t + 1/2(9.8)t²
81.3 = 12.4t + 4.9t²
4.9t² + 12.4t - 81.3 = 0
Using the general formula to find t
t = -12.4±√12.4²-4(4.9)(-81.3)/2(4.9)
t = -12.4±√153.76+1593.48/2(4.9)
t = -12.4±√1747.24/9.8
t = -12.4+41.8/9.8
t = 29.4/9.8
t = 3secs
Hence it took 3secs to reach the ground
