If Katie swims from one end of the pool, to the other side, and then swims back to her original spot, her average velocity is ha
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Answer:
a)Initial speed of the projectile = 196.2 m/s
b)Maximum altitude = 490.5 m
c) Range of projectile = 3398.28 m
d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m
Explanation:
Time of flight of a projectile is given by the expression,
Here θ = 30° and t = 20 s
a)
Initial speed of the projectile = 196.2 m/s
b) Maximum altitude is given by
Maximum altitude = 490.5 m
c) Range of projectile is given by
Range of projectile = 3398.28 m
d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s
Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s
We have equation of motion s = ut + 0.5 at²
Horizontal motion
u = 169.91 m/s
a = 0 m/s²
t = 15 s
Substituting
s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m
Vertical motion
u = 98.1 m/s
a = -9.81 m/s²
t = 15 s
Substituting
s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m
Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m
Answer:
Frictional force, F = 45.9 N
Explanation:
It is given that,
Weight of the box, W = 150 N
Acceleration,
The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.
It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,
Frictional force is given by :
F = 45.9 N
So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.