Answer:
θ₀ = 84.78° (OR) 5.22°
Explanation:
This situation can be treated as projectile motion. The parameters of this projectile motion are:
R = Range of Projectile = 150 m
V₀ = Launch Speed of Projectile = 90 m/s
g = 9.8 m/s²
θ₀ = Launch angle (OR) Angle of Elevation = ?
The formula for range of a projectile is given as:
R = V₀² Sin 2θ₀/g
Sin 2θ₀ = Rg/V₀²
Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²
2θ₀ = Sin⁻¹ (0.18)
θ₀ = 10.45°/2
<u>θ₀ = 5.22°</u>
Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:
θ₀ = 90° - 5.22°
<u>θ₀ = 84.78°</u>
Answer:

Explanation:
using the law of the conservation of energy:


where K is the spring constant, x is the spring compression, N is the normal force of the block,
is the coefficiet of kinetic friction and d is the distance.
Also, by laws of newton, N is calculated by:
N = mg
N = 3.35 kg * 9.81 m/s
N = 32.8635
So, Replacing values on the first equation, we get:

solving for
:

The first two are always the reactants the products come after so they are last
Answer:
B
Explanation:
nothing to do with black holes creating star or related
the answer is d i took the test be cause i go to a k12 onlie school