A small fish is dropped by a pelican that is rising steadily at 0.500 m/s. How far below the pelican is the fish after 2.50 s?
1 answer:
Refer to the diagram shown below.
h = original height of the pelican when the fish is dropped (not relevant).
S = distance traveled by the fish as a function of time, measured upward.
u = 0.5 m/s, the upward velocity with which the fish is dropped.
g = 9.8 m/s², the acceleration due to gravity.
Use the following equation:
S = ut + (1/2)gt²
S = (0.5 m/s)*(2.5 s) + 0.5*(-9.8 m/s²)*(2.5 s)²
= -29.375 m
The negative sign means that the fish drops by 29.375 m from the original height of h.
Answer: The fish is 29.375 m below where the pelican dropped it after 2.5 s.
h = original height of the pelican when the fish is dropped (not relevant).
S = distance traveled by the fish as a function of time, measured upward.
u = 0.5 m/s, the upward velocity with which the fish is dropped.
g = 9.8 m/s², the acceleration due to gravity.
Use the following equation:
S = ut + (1/2)gt²
S = (0.5 m/s)*(2.5 s) + 0.5*(-9.8 m/s²)*(2.5 s)²
= -29.375 m
The negative sign means that the fish drops by 29.375 m from the original height of h.
Answer: The fish is 29.375 m below where the pelican dropped it after 2.5 s.
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Answer:
The value of the average convection coefficient is 20 W/Km².
Explanation:
Given that,
For first object,
Characteristic length = 0.5 m
Surface temperature = 400 K
Atmospheric temperature = 300 K
Velocity = 25 m/s
Air velocity = 5 m/s
Characteristic length of second object = 2.5 m
We have same shape and density of both objects so the reynold number will be same,
We need to calculate the value of the average convection coefficient
Using formula of reynold number for both objects
Here,
Put the value into the formula
Hence, The value of the average convection coefficient is 20 W/Km².
Hope this helps you.
