Answer:
the shooting angle ia 18.4º
Explanation:
For resolution of this exercise we use projectile launch expressions, let's see the scope
R = Vo² sin (2θ) / g
sin 2θ = g R / Vo²
sin 2θ = 9.8 75/35²
2θ = sin⁻¹ (0.6)
θ = 18.4º
To know how for the arrow the tree branch we calculate the height of the arrow at this point
X2 = 75/2 = 37.5 m
We calculate the time to reach this point since the speed is constant on the X axis
X = Vox t
t2 = X2 / Vox = X2 / (Vo cosθ)
t2 = 37.5 / (35 cos 18.4)
t2 = 1.13 s
With this time we calculate the height at this point
Y = Voy t - ½ g t²
Y = 35 sin 18.4 1.13 - ½ 9.8 1,13²
Y = 6.23 m
With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch
Answer:
a. λ = 647.2 nm
b. I₀ 9.36 x 10⁻⁵
Explanation:
Given:
β = 56.0 rad , θ = 3.09 ° , γ = 0.170 mm = 0.170 x 10⁻³ m
a.
The wavelength of the radiation can be find using
β = 2 π / γ * sin θ
λ = [ 2π * γ * sin θ ] / β
λ = [ 2π * 0.107 x 10⁻³m * sin (3.09°) ] / 56.0 rad
λ = 647.14 x 10⁻⁹ m ⇒ λ = 647.2 nm
b.
The intensity of the central maximum I₀
I = I₀ (4 / β² ) * sin ( β / 2)²
I = I₀ (4 / 56.0²) * [ sin (56.0 /2) ]²
I = I₀ 9.36 x 10⁻⁵
Answer:
Explanation:
When the car is under an accelerating force and hits a tree, the instant force received by the tree is the same force that is accelerating the car.
The accelerating force can be calculated using Newton's second law:
Where m is the mass of the car and a is the acceleration.
From the calculation, the value of the acceleration is 5.8 m/s^2.
<h3>What is uniform acceleration?</h3>
The term uniform acceleration refers to a situation in which the velocity increases by equal amounts in equal time intervals.
Given the fact that the car started from rest and reached a velocity of 780.34 mph or 348.84 m/s in 1 minute of 60 seconds;
v = u + at
a = v/t
a = 348.84 m/s/ 60 seconds
a = 5.8 m/s^2
Learn more about acceleration:brainly.com/question/12550364?
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