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Answer:
33.6 m
Explanation:
Given:
v₀ = 0 m/s
a = 47.41 m/s²
t = 1.19 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (0 m/s) (1.19 s) + ½ (47.41 m/s²) (1.19 s)²
Δx = 33.6 m
<em>Resultant angle; θ = 25.59° </em>
This question is dealing with bearings and distance.
We are told that from point A, the camel walks 20 km at 15° in the south of east direction.
Thus, d_s,e = 20 km
Resolving along the horizontal east direction gives; d_e = 20 cos 15
d_e = 19.32 km
Also, resolving along the vertical south direction gives; d_s = 20 sin 15
d_s = 5.18 km
Net vertical distance; d_vert = 8km - 5.18km = 2.72 km
Net horizontal distance; d_hor = 25km - 19.32 km = 5.68 km
Now, the resultant angle is given by;
tan θ = d_vert/d_hor
tan θ = 2.72/5.68
tan θ = 0.4789
θ = tan^(-1) 0.4789
θ = 25.59°
Read more at; brainly.com/question/22518031
Answer:
We know that the speed of sound is 343 m/s in air
we are also given the distance of the boat from the shore
From the provided data, we can easily find the time taken by the sound to reach the shore using the second equation of motion
s = ut + 1/2 at²
since the acceleration of sound is 0:
s = ut + 1/2 (0)t²
s = ut <em>(here, u is the speed of sound , s is the distance travelled and t is the time taken)</em>
Replacing the variables in the equation with the values we know
1200 = 343 * t
t = 1200 / 343
t = 3.5 seconds (approx)
Therefore, the sound of the gun will be heard at the shore, 3.5 seconds after being fired