Answer:
13.6 cm
Explanation:
From Snell's law:
n₁ sin θ₁ = n₂ sin θ₂
In the air, n₁ = 1, and light from the horizon forms a 90° angle with the vertical, so sin θ₁ = sin 90° = 1.
Given n₂ = 4/3:
1 = 4/3 sin θ
sin θ = 3/4
If x is the radius of the circle, then sin θ is:
sin θ = x / √(x² + 12²)
sin θ = x / √(x² + 144)
Substituting:
3/4 = x / √(x² + 144)
9/16 = x² / (x² + 144)
9/16 x² + 81 = x²
81 = 7/16 x²
x ≈ 13.6
The time taken to hit the ground is 3.9 s, the range is 18m and the final velocity is 42.82 m/s
<h3>
Motion Under Gravity</h3>
The motion of an object under gravity is the vertical motion of the object under the influence of acceleration due to gravity.
Given that a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s.
a. how much later does the ball hit the ground?
The time can be calculated by considering the vertical component of the motion with the use of formula below.
h = ut + 1/2gt²
Where
- Initial velocity u = 0 ( vertical velocity )
- Acceleration due to gravity g = 9.8 m/s²
Substitute all the parameters into the formula
75 = 0 + 1/2 × 9.8 × t²
75 = 4.9t²
t² = 75/4.9
t² = 15.30
t = √15.3
t = 3.9 s
b. how far from the building will it land?
The range can be found by using the formula
R = ut
Where u = 4.6 m/s ( horizontal velocity )
R = 4.6 × 3.9
R = 18 m
c. what is the velocity of the ball just before it hits the ground?
The final velocity will be
v = u + gt
v = 4.6 + 9.8 × 3.9
v = 4.6 + 38.22
v = 42.82 m/s
Therefore, the answers are 3.9 s, 18 m and 42.82 m/s
Learn more about Vertical motion here: brainly.com/question/24230984
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Answer:
2442.5 Nm
Explanation:
Tension, T = 8.57 x 10^2 N
length of rope, l = 8.17 m
y = 0.524 m
h = 2.99 m
According to diagram
Sin θ = (2.99 - 0.524) / 8.17
Sin θ = 0.3018
θ = 17.6°
So, torque about the base of the tree is
Torque = T x Cos θ x 2.99
Torque = 8.57 x 100 x Cos 17.6° x 2.99
Torque = 2442.5 Nm
thus, the torque is 2442.5 Nm.
