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
Answer:
B
Explanation:
From Newton's law of motion, we have:
V^2 = U^2 + 2gH
Where V and U are final and initial velocity respectively.
H is the height.
For the object to have a sustain a maximum height it means the final velocity of the object is zero.
By computing the height of the object sustain by A, we have:
0^2 = 2^2 -2×10×H
0= 4 -20H
4 = 20H;
H= 0.2m
For object B we have;
0^2 = 1^2 -2×10×H
0 = 1 -20H
H = 1/20= 0.05m
From computing the height sustain by both objects, we see object B is projected at a shorter height into atmosphere than A.
Hence object B will return to the ground first.
Answer:
Explanation:
The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:
Here d is the distance at which the earthquake take place and 
is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:
Answer:
The correct answer is Gamma Rays.
Explanation:
