The northward components of the resultant displacement is 40.96 m and the westward components of the resultant displacement of the bird from its nest is 28.68 m.
<h3>
Displacement of the bird</h3>
The displacement of the bird is the change in the position of the bird.
<h3>Vertical component of the bird's displacement </h3>
Vy₁ = -25 m x sin(55)
Vy₁ = -20.48 m
Vy₂ = 75 m x sin(55)
Vy₂ = 61.44 m
Total vertical displacement = 61.44 m - 20.48 m = 40.96 m
<h3>Horizontal component of the bird's displacement </h3>
Vx₁ = -25 m x cos(55)
Vx₁ = -14.34 m
Vx₂ = 75 m x cos(55)
Vx₂ = 43.02 m
Total horizontal displacement = 43.02 m - 14.34 m = 28.68 m
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Answer: The distance is 723.4km
Explanation:
The velocity of the transverse waves is 8.9km/s
The velocity of the longitudinal wave is 5.1 km/s
The transverse one reaches 68 seconds before the longitudinal.
if the distance is X, we know that:
X/(9.8km/s) = T1
X/(5.1km/s) = T2
T2 = T1 + 68s
Where T1 and T2 are the time that each wave needs to reach the sesmograph.
We replace the third equation into the second and get:
X/(9.8km/s) = T1
X/(5.1km/s) = T1 + 68s
Now, we can replace T1 from the first equation into the second one:
X/(5.1km/s) = X/(9.8km/s) + 68s
Now we can solve it for X and find the distance.
X/(5.1km/s) - X/(9.8km/s) = 68s
X(1/(5.1km/s) - 1/(9.8km/s)) = X*0.094s/km= 68s
X = 68s/0.094s/km = 723.4 km
X - rays are highly penetrating electromagnetic radiations that are formed when the cathode rays strikes a dense metal.
It has many characteristics such as :
- travels with the speed of light.
Answer b protons and electrons
