The work done on the puck is 96 J
Explanation:
According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.
Mathematically:
where
is the final kinetic energy of the puck, with
m = 2 kg being the mass of the puck
v = 10 m/s is the final speed
is the initial kinetic energy of the puck, with
u = 2 m/s being the initial speed of the puck
Substituting numbers into the equation, we find the work done by the player on the puck:
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Answer:
1.45 K
I had the same question and i got it right.
Answer:
On real life example of a scenario that takes advantage of the inverse relationship between force and time when impulse is constant is when making a serve with a lawn tennis racket
How It is an example of impulse is that when a serve is made by moving the bat slowly, the lawn tennis player uses less force and the ball is in contact with the string for longer a period
When however, the lawn tennis player moves the racket faster, with the strings of the racket highly tensioned he uses more force and the ball also spends less time on the racket to produce the same momentum
Explanation:
The impulse of a force, ΔP is given by the following formula;
ΔP = F × Δt
Where ΔP is constant, we have;
F ∝ 1/Δt
Therefore, for the same impulse, when the force is increased, the time of contact is decreases and vice versa.
The air pressure is most likely lower.
Answer:
angle minimum θ = 41.3º
Explanation:
For this exercise let's use Newton's second law in the condition of static equilibrium
N - W = 0
N = W
The rotational equilibrium condition, where we place the axis of rotation on the wall
We assume that counterclockwise rotations are positive
fr (l sin θ) - N (l cos θ) + W (l/2 cos θ) = 0
the friction force formula is
fr = μ N
fr = μ W
we substitute
μ m g l sin θ - m g l cos θ + mg l /2 cos θ = 0
μ sin θ - cos θ + ½ cos θ= 0
μ sin θ - ½ cos θ = 0
sin θ / cos θ = 1/2 μ
tan θ = 1/2 μ
θ = tan⁻¹ (1 / 2μ)
θ = tan⁻¹ (1 (2 0.57))
θ = 41.3º