Explanation:
Initial time, t₁ = 2:30 pm
Final time, t₂ = 2:30:45
We need to find the motion of students in terms of time. Final time is 45 seconds more than the initial time.
Change in time,
Hence, this is the required solution.
If the average speed of an orbiting space shuttle is 27 800 km/h, determine the time required for it to circle Earth. Assume tha
1 answer:
Answer:
t = 1.51 hours
Explanation:
given,
Speed of space shuttle. v = 27800 Km/h
Radius of earth, R = 6380 Km
height of shuttle above earth, h = 320 Km
Total radius of the shuttle orbit
r' = R + h
r' = 6380 + 320
r' = 6700 Km
distance, d = 2 π r
d = 2 π x 6700
t = 1.51 hours
Time require by the shuttle to circle the earth is equal to 1.51 hr.
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Answer:
The speed at the end of the track = 27 m/s
The acceleration = 1.2 m/s²
Please find the Δx vs Δt, v vs Δt, a vs Δt
Explanation:
We have;
x = u·t + 1/2·a·t²
Where;
x = The distance = 300 m
u = The initial velocity = 0 m/s (Ball at rest)
t = The time taken = 22.4 s
Therefore;
300 = 0 + 1/2×a×22.4²
a = 2×300/22.4² = 1.19579 ≈ 1.2 m/s²
v = u + a×t
∴ v = 0 + 1.2 × 22.4 = 26.88 ≈ 27 m/s
Part of the table of values is as follows;
t, x, v
0, 0, 0
0.4, 0.095663, 0.478316
0.8, 0.382653, 0.956632
1.2, 0.860969, 1.434948
1.6, 1.530611, 1.913264
2, 2.39158, 2.39158
2.4, 3.443875, 2.869896
2.8, 4.687497, 3.348212
3.2, 6.122445, 3.826528
3.6, 7.748719, 4.304844
