To understand the meaning of the variables in Gauss's law, and the conditions under which the law is applicable. Gauss's law is
1 answer:
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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
Answer:
v=5.86 m/s
Explanation:
Given that,
Length of the string, l = 0.8 m
Maximum tension tolerated by the string, F = 15 N
Mass of the ball, m = 0.35 kg
We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :
v is the maximum speed
Hence, the maximum speed of the ball is 5.86 m/s.