Complete question is;
In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight 125 cm above the water level to a point 185 cm from his foot at the edge of the pool. Where does the spot of light hit the bottom of the pool, relative to the edge, if the pool is 215 cm deep
Answer:
356.2 cm
Explanation:
First of all, let's find the angle of incidence from the distances
;
tan θ1 = l1/h1
tan θ1 = 185/125
tan θ1 = 1.48
θ1 = tan^(-1) 1.48
θ1 = 55.95°
For the refraction from air into water, we have
n_air × sin θ1 = n_water × sin θ2,
sin θ2 = (n_air × sin θ1)/n_water
Where,
n_water is refractive index of water = 1.33
n_air is refractive index of air = 1
Thus;
sin θ2 = (1 × sin 55.95)/1.33
sin θ2 = 0.62297
θ2 = sin^(-1) 0.62297
θ2 = 38.53°
So, the horizontal distance from the edge of the pool from
l is;
l = l1 + l2 = l1 + h2 tan θ2
where;
l1 = 185cm
h2 = 215 cm
So,
l = 185 + 215(tan 38.53)
l = 185 + 171.2045
l = 356.2045 cm ≈ 356.2 cm
Answer:
a. The horizontal component of acceleration a₁ = 0.68 m/s²
The vertical component of acceleration a₂ = -0.11 m/s²
b. -9.19° = 350.81° from the the positive x-axis
Explanation:
The initial velocity v₁ of the fish is v₁ = 4.00i + 1.00j m/s. Its final velocity after accelerating for t = 19.0 s is v₂ = 17.0i - 1.00j m/s
a. The acceleration a = (v₂ - v₁)/t = [17.0i - 1.00j - (4.00i + 1.00j)]/19 = [(17.0 -4.0)i - (-1.0 -1.0)j]/19 = (13.0i - 2.0j)/19 = 0.68i - 0.11j m/s²
The horizontal component of acceleration a₁ = 0.68 m/s²
The vertical component of acceleration a₂ = -0.11 m/s²
b. The direction of the acceleration relative to the unit vector i,
tanθ = a₂/a₁ = -0.11/0.68 = -0.1618
θ = tan⁻¹(-0.1618) = -9.19° ⇒ 360 + (-9.19) = 350.81° from the the positive x-axis
Answer:
Explanation:
Given that,
Force force acting on the Skater's rocket is 220 N
Resistive force acting on it, F' = 30 N
Mass of Shelby, m = 60 kg
We need to find the acceleration of Shelby. Net force acting on Shelby is given by :
Net force = 220-30
= 190 N
The formula of net force is :
F = ma
a is acceleration of Shelby
So, the acceleration of Shelby is 
