Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m
Naturally we assume that 10000 km/hr is initial velocity (same as being shot from a cannon), and no air resistance. With so high a velocity, the effect of diminishing gravity with increasing radius must be taken into account, so you use an energy solution. M is earth mass, r is earth radius.
KE/m = (9000000/3600)^2/2 = 3858025 J/kg
ΔPE/m = (PE(at height) - PE(at surface))/m = -GM/(r+h) + GM/r
KE/m = ΔPE/m
KE/m - GM/r = -GM/(r+h)
h = -GM / (KE/m - GM/r) - r = 335665.44 m
(Using G = 6.673E-11 Nm^2/kg^2, M = 5.9742E24 kg, r = 6378100 m)
Answer:
the tempature and the map and climate
Explanation
Answer:
The charged particle follows a spiral path in a magnetic field.
Explanation:
A charge immersed in a region with an electric field experiences a force that acts along the same direction of the electric field. In particular:
- The force has the same direction as the electric field if the charge is positive
- The force has the opposite direction as the electric field if the charge is negative
Therefore, a charge moving in an electric field is accelerated along the direction of the electric field.
On the other hand, a charge in motion in a region with a magnetic field experiences a force that acts perpendicular to the direction of the field. This means that a charge in motion in a magnetic field will acquire a circular motion in the plane perpendicular to the direction of the magnetic field.
As a result, if the particle has also a original motion outside this plane, its final motion will consist of:
- A uniform motion along that direction, +
- A circular motion along the plane perpendicular to the field
So, the resultant motion of the particle will be a spiral path. So the correct answer is
The charged particle follows a spiral path in a magnetic field.
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