10/70×360°
=51.4°
hope thus helps
answer
v = 4.2
set up an equation
we can use the formula for kinetic energy since we know mass and kinetic energy
K = 
v = 
values
K = 0.013 J
m = 1.5g = 0.0015 kg
plug in values
v = 
v = 4.2
Answer:
34.8 and 55.2º
Explanation:
This is a projectile launching exercise, as we are told that the range of the arrow must be equal to its range and = 31 m let's use the equation
The scope equation is
R = v₀² sin 2θ /g
sin 2 θ = R g / v₀²
sin 2 θ = 31 9.8 / 18²
2 θ = sin⁻¹ 0.93765
θ = 34.8º
At the launch of projectiles we have two complementary angles with the same range in this case 34.8 and (90-34.8) = 55.2º
Answer:
The horizontal distance the pumpkin will travel after it slips from the eagle is 17.02 m
Explanation:
Given;
height above the ground, h = 16.4 m
speed of the eagle, v = 9.3 m/s
The time it will take the pumpkin to fall at the given height is calculated as;

The horizontal distance traveled at this time is given by;
x = vt
x = (9.3)(1.83)
x = 17.02 m
Therefore, the horizontal distance the pumpkin will travel after it slips from the eagle is 17.02 m
Answer:
Muscovite mica. Flourite. Olivine.