Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = 
t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45
)
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = 
- 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
Answer:
A. That enough light of any frequency would cause electrons to flow.
Explanation:
A P E X
The starter motor's potential difference across the headlight bulbs is 38.45V, requiring an additional 39 a from the battery. Voltage, also known as potential difference.
It is sometimes described as the amount of work needed to move a test charge between two sites, expressed as a unit of charge. Volt is the potential difference's SI unit (V). We only take into account the charge between the locations P and Q when current moves between them in an electric circuit. Electric potential difference between two sites is referred to as voltage, also known as electric pressure, electric tension, or (electric) potential difference. an electric field that is static.
Vh = I*Rn
Vh = 39/5.476*5.40v
Vh = 38.45v
Learn more about voltage here
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Force is calculated F=m×a.
If both ships speed up with the same force, but have a different mass, This means that a also has to be different. If F is the same but ship a has a bigger mass(m) than ship b, then the acceleration(a) of ship b has to be bigger so F of each ship is the same. So the ship with the smaller mass will speed up faster.
The answer i believe is A hope it helps
