Matthew is waterskiing. As the boat starts moving, he is at an angle of 8.0° to the right of the boat. The boat applies 250 newt
1 answer:
The work done is B.
Explanation:
The work done by a force on an object is given by
where
F is the magnitude of the force
d is the displacement of the object
is the angle between the direction of the force and of the displacement
For the boat in this problem, we have:
F = 250 N (force applied)
d = 50 m (displacement)
Substituting, we find the work done:
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Answer:
During those 3.00 seconds before stopping, the car travels a distance of 6 m.
Explanation:
The simple rule of three is a tool that is used to quickly solve problems, where three pieces of information must be known, and one of them operates as an unknown to be known.
Two magnitudes are directly proportional if one magnitude increases the other also does it, and if the magnitude decreases the other in the same way.
Being a, b and c known data and x the unknown, the value that we want to know, the rule of three when the magnitudes are directly proportional is applied as follows:
a ⇒ b
c ⇒ x
So:
In this case, knowing that a truck travels at 2 m/s, the rule of three applies as follows: if in 1 second the truck travels 2 m, in 3 seconds how much distance does it travel?
distance= 6 m
<u><em> During those 3.00 seconds before stopping, the car travels a distance of 6 m.</em></u>
Answer:
a) θ₁ = 0.487º , b) t = 0.400 s , x = 11.73 ft
Explanation:
For this exercise let's use the projectile launch relationships.
The initial height is I = 5 ft and the final height y = 4 ft
y = y₀ + t - ½ g t²
The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft
x = v₀ₓ t
t = x / v₀ₓ
We replace
y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²
v_{oy} = v₀ sin θ
v₀ₓ = vo cos θ
y –y₀ = x tan θ - ½ g x² / v₀² cos² θ
5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)
1 = 120 tan θ - 0.0213 sec² θ
Let's use the trigonometry relationship
Sec² θ = 1 - tan² θ
1 = 120 tan θ - 0.0213 (1 –tan²θ)
0.0213 tan²θ + 120 tanθ -1.0213 = 0
We change variables
u = tan θ
u² + 5633.8 u - 48.03 = 0
We solve the second degree equation
u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2
u = [- 5633.8 ± 5633.82] / 2
u₁ = 0.0085
u₂= -5633.81
u = tan θ
θ = tan⁻¹ u
For u₁
θ₁ = tan⁻¹ 0.0085
θ₁ = 0.487º
For u₂
θ₂ = -89.99º
The launch angle must be 0.487º
b) let's look for the time it takes for the arrow to arrive
x = v₀ₓ t
t = x / v₀ cos θ
t = 120 / (300 cos 0.487)
t = 0.400 s
The deer must be at a distance of
v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s
x = v t
x = 29.33 0.4
x = 11.73 ft