annakoslovsky3040
21.10.2020 •
Physics
A cannon fires a shell straight upward; 1.6 s after it is launched, the shell is moving upward with a speed of 19 m/s. Assuming air resistance is negligible, find the speed (magnitude of velocity) of the shell at launch and 5.5 s after the launch.
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Ответ:
-19.259m/s
Explanation:
Given;
Final velocity = 19m/s
time t = 1.6s
u is the initial velocity
g is the acceleration due to gravity = 9.81m/s²
Using the equation of motion to first get the initial velocity of the shell:
v = u-gt
19 = u - (9.81)(1.6)
19 = u - 15.696
u = 19+15.696
u = 34.696m/s
The initial velocity of the shell is 34.696m/s
Next is to find the speed of the shell 5.5s after the launch
Using the equation of motion:
v = u-gt
v = 34.696-9.81(5.5)
v = 34.696 - 53.955
v = -19.259m/s
The negative value of the velocity shows that the velocity is travelling in the downward direction
Ответ:
For example, say a bowling ball weighs 3.0kg and is travelling at a speed of 3.0m/s. Its momentum would be 3.0×3.0=9.0 kg·m/s.
Now say we have a baseball weighing 0.20kg and it is travelling at a speed of 47.0m/s. Its momentum would be 0.20×47.0=9.4 kg·m/s, which is more than that of the bowling ball.