Answer:
105.8 m
46 m/s
Explanation:
From the time the rocket is launched to the time it reaches its maximum height:
v = 0 m/s
a = -10 m/s²
t = 9.2 s / 2 = 4.6 s
Find: Δy and v₀
Δy = vt − ½ at²
Δy = (0 m/s) (4.6 s) − ½ (-10 m/s²) (4.6 s)²
Δy = 105.8 m
v = at + v₀
0 m/s = (-10 m/s²) (4.6 s) + v₀
v₀ = 46 m/s
Explanation:
V=u+at
where,
v=final speed
u=initial speed,(starting speed)
a=acceleration
t=time
- v=u+at = 6=2+a*2
6=2+2a
2a=6-2
2a=4
a=4/2 = 2
a =2
2. to find time taken
v=u+at
25=5*2t
2t=25-5
2t=20
t=20/2
t=10sec
3. finding final speed
v=u+at
v=4+10*2
=4+20
v=24m/sec
5.v=u+at
=5+8*10
=5+80
V=85m/sev
6. v=u+at
8=u+4*2
8=u+8
U=8/8
u=1
these are your missing values
Answer:
The diameter of the camera aperture must be greater than or equal to 1.49m
Explanation:
Let the distance separating two objects, x = 6.0 cm = 0.06 m
The distance between the observer and the two objects, d = 160 km = 160000 m
Let ∅ = minimum angular separation between the two objects that the satellite can resolve
tan( ∅) = x/d
Since there is minimum angular separation, tan( ∅) ≈∅
∅ = x/d
∅ = 0.06/160000
∅ = 3.75 * 10⁻⁷rad
For the satellite to be able to resolve the objects,
D ≥ 1.22λ/∅
λ = 560 nm = 560 * 10⁻⁹
D ≥ 1.22 * (560 * 10⁻⁹)/(3.75 * 10⁻⁷)
D ≥ 149.33 * 10⁻² m
D ≥ 1.49 m
430m. We use the displacement formula x=v*t+1/2*a*t^2. Set upward to be the positive direction, so v=6m/s, a=-g=-9.8m/s^2 (gravity is pulling the object down), and t=10s. The displacement is -h(it's negative because the ending point is the ground, which is negative relative to the starting point). -h=60-4.9*100, h=430 m.