That particular strike was very roughly 2.4 km (1.5 miles) away from them.
That's if you use 340 m/s (1120 ft/sec) for the speed of sound.
But the air in the region for several thousand feet around a thunderstorm
is doing weird things to sounds that pass through it, so you can't use any
exact number for the speed of sound in a stormy area.
The only thing you can be absolutely sure of is that Johnny and his friends
need to round up their equipment and get in the house. NOW !
Answer:
d = 2021.6 km
Explanation:
We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them
Airplane 1
Height y₁ = 800m
Angle θ = 25°
cos 25 = x / r
sin 25 = z / r
x₁ = r cos 20
z₁ = r sin 25
x₁ = 18 103 cos 25 = 16,314 103 m
= 16314 m
z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m
2 plane
Height y₂ = 1100 m
Angle θ = 20°
x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m
z₂ = 20 103 without 25 = 8.452 103 m = 8452 m
The distance between the planes using the Pythagorean Theorem is
d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2
Let's calculate
d² = (18126-16314)² + (1100-800)² + (8452-7607)²
d² = 3,283 106 +9 104 + 7,140 105
d² = (328.3 + 9 + 71.40) 10⁴
d = √(408.7 10⁴)
d = 20,216 10² m
d = 2021.6 km
<span>D) The sun's rays will never be directly overhead. The latitude of 23 ½ degrees north is known as the Tropic of Cancer. Above this imaginary line the sun's rays hit earth with decreased angles.</span>
Answer:
0.1 m
Explanation:
It is given that,
Mass of the object, m = 350 g = 0.35 kg
Spring constant of the spring, k = 5.2 N/m
Amplitude of the oscillation, A = 10 cm = 0.1 m
Frequency of a spring mass system is given by :
Time period:
Answer:
a) 0.0625 I_1
b) 3.16 m
Explanation:
<u>Concepts and Principles </u>
The intensity at a distance r from a point source that emits waves of power P is given as:
I=P/4π*r^2 (1)
<u>Given Data</u>
f (frequency of the tuning fork) = 250 Hz
I_1 is the intensity at the source a distance r_1 = I m from the source.
<u>Required Data</u>
- In part (a), we are asked to determine the intensity I_2 a distance r_2 = 4 in from the source.
- In part (b), we are asked to determine the distance from the tuning fork at which the intensity is a tenth of the intensity at the source.
<u>solution:</u>
(a)
According to Equation (1), the intensity a distance r is inversely proportional to the distance from the source squared:
I∝1/r^2
Set the proportionality:
I_1/I_2=(r_2/r_1)^2 (2)
Solve for I_2 :
I_2=I_1(r_2/r_1)^2
I_2=0.0625 I_1
(b)
Solve Equation (2) for r_2:
r_2=(√I_1/I_2)*r_1
where I_2 = (1/10)*I_1:
r_2=(√I_1/1/10*I_1)*r_1
=3.16 m