To solve a problem using the equation for keplerâs third law, enrico must convert the average distance of mars from the sun from
2 answers:
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Answer:
Your friend has to wait 0.26 s after you throw the ball to start running.
Explanation:
The equation that gives the position vector of the ball is as follows:
r = (x0 + v0 · t · cos α, y0 + v0 · t ·sin α + 1/2 · g · t²)
Where:
x0 = initial horizontal positon
v0 = initial velocity
t = time
α = throwing angle
y0 = initial vertical position
g = acceleration due to gravity
The equation of displacement of your friend is as follows:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of your friend at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
Please, see the attached figure for a description of the situation. Notice that the frame of reference is located at the throwing point.
Let´s find the time of flight of the ball. We know that at the final time, the y-component of the vector r has to be -6.00 m (1 m above the ground). Then:
y = y0 + v0 · t ·sin α + 1/2 · g · t²
-6.00 m = 0 m + 9.00 m/s · t · sin 33.0° - 1/2 · 9.8 m/s² · t²
0 = -4.9 m/s² · t² + 9.00 m/s · sin 33.0° · t + 6.00 m
Solving the quadratic equation:
t = 1.71 s
Now that we have the time of flight, we can calculate the x-component of the vector r (the horizontal distance traveled by the ball):
x= x0 + v0 · t · cos α
x = 0m + 9.00 m/s · 1.71 s · cos 33°
x = 12.9 m
Then, your friend will have to run (12.9 m - 11.0 m) 1.9 m to catch the ball 1 m above the ground.
Let´s see, how much time it takes your friend to run that distance:
x = x0 + v0 · t + 1/2 · a · t² (x0 = 0, v0 = 0)
x = 1/2 · a · t²
1.9 m = 1/2 · 1.80 m/s² · t²
Solving for t
t = 1.45 s
Then, since the time of flight of the ball is 1.71 s, your friend has to wait
1.71 s - 1.45 s = 0.26 s after you throw the ball to start running.
Answer:
b) N = 560 N, c) fr = 138.56 N, d) μ = 0.247
Explanation:
a) In the attachment we can see the free body diagram of the system
b) Let's write Newton's second law on the y-axis
N + T_y -W = 0
N = W -T_y
let's use trigonometry for tension
sin θ = T_y / T
cos θ = Tₓ / T
T_y = T sin θ
Tₓ = T cos θ
we substitute
N = W - T sin 30
we calculate
N = 640 - 160 sin 30
N = 560 N
c) as the system goes at constant speed the acceleration is zero
X axis
Tₓ - fr = 0
Tₓ = fr
we substitute and calculate
fr = 160 cos 30
fr = 138.56 N
d) the friction force has the formula
fr = μ N
μ = fr / N
we calculate
μ = 138.56 / 560
μ = 0.247
