Questions 2. Rewrite the following quantities using suitable prefixes 5000 000
1 answer:
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Answer:
(a): the mug hits the floor 0.752m away from the end of the bar.
(b): the speed of the mug at impact are:
V= 4.87 m/s
direction= 70.82º below the horizontal.
Explanation:
Vx= 1.6 m/s
Vy=?
h= 1.1 m
g= 9.8 m/s²
t is the fall time
t=0.47 sec
Vy= g*t
Vy= 4.6 m/s
V= 4.87 m/s
α= tan⁻¹(Vy/Vx)
α= -70.82º
Answer:
v = 9.04 m / s
Explanation:
For this exercise we can use the relation that the work of the non-conservative force (friction) is equal to the variation of the mechanical energy of the system.
W = Em_f - Em₀ (1)
Starting point. Lower slope
Em₀ = K = ½ m v²
highest point. Where is the skier at a height h
Em_f = U = m g h
The work of rubbing
W = -fr x
the negative sign is because the friction force opposes the movement.
Let's set a reference system where the x axis is parallel to the slope and the y axis is perpendicular
let's use trigonometry to break down the weight
cos θ = W_y / W
sin θ = Wₓ / W
W_y = W cos θ
Wₓ = W sin θ
Y axis
N - Wₓ = 0
N = mg sin θ
X axis
fr = m a
the friction force has the expression
fr = μ N
fr = μ mg sin θ
we look for the job
W = - μ mg sin θ x
where x is the distance along the slope
we substitute in 1
-μ mg sin θ x = mg h - ½ m v²
let's use trigonometry to find the distance x
tan 30 = h / x
x = h / tan 30
we substitute
- = m gh - ½ m v²
we use
tan 30 = sin30 / cos30
v² = 2g h + 2 μ g h cos 30
v =
let's calculate
v =
v = 9.04 m / s