Really long we’ll not long but far in distance
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:
128 Kelvin = 128 - 273.15 = -145.15 Celsius. Temperature conversion chart Sample temperature conversions 103.55 Kelvin to degrees Fahrenheit 39.82 degrees Fahrenheit to Kelvin
Explanation:
hope this helps have a good day
Time required : 3 s
<h3>Further explanation
</h3>
Power is the work done/second.
To do 33 J of work with 11 W of power
P = 11 W
W = 33 J
