Answer:
X = 15.88 m
Explanation:
Given:
Initial Velocity V₀ = 13.4 m/s
θ = 30.1 °
g = 9.8 m/s²
To Find horizontal distance let "X" we have to time t first.
so from motion 2nd equation at Height h = 0
h = V₀y t + 1/2 (-g) t ² (ay = -g)
0 = 13.4 sin 30.1° t - 0.5 x 9.81 x t² (V₀y = V₀ Sin θ)
⇒ t = 1.37 s
Now For Horizontal distance X, ax =0m/s²
X = V₀x t + 1/2 (ax) t ²
X = 13.4 m × cos 30.1° x 1.37 s + 0
X = 15.88 m
Answer:
<h2>T(Period) = 1.33s</h2><h2>f(Frequency) = 0.75Hz (cycles/second)</h2>
Explanation:
<h2>Given:</h2><h2 /><h2>λ = 4.0m</h2><h2>Amplitude = 25m</h2><h2>d = 24m</h2><h2>s = 8.0s</h2><h2 /><h2>Required:</h2><h2>f = ?</h2><h2>T = ?</h2><h2 /><h2>Analysis:</h2><h2>v = λf</h2><h2>f =N/t</h2><h2>T = 1/f</h2><h2 /><h2>v = d/t</h2><h2 /><h2>Solve:</h2><h2>v = d/t = 24/8.0 → v = 3.0m/s</h2><h2>v =λf → f = v/λ = 3.0/4.0 → f = 0.75Hz</h2><h2>T = 1/f = 1/0.75 → T = 1.33s</h2><h2 /><h2>Hopes this helps. Mark as brainlest plz!</h2>
Answer:
v₁f = 0.5714 m/s (→)
v₂f = 2.5714 m/s (→)
e = 1
It was a perfectly elastic collision.
Explanation:
m₁ = m
m₂ = 6m₁ = 6m
v₁i = 4 m/s
v₂i = 2 m/s
v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i + ((2m₂) / (m₁ + m₂)) v₂i
v₁f = ((m – 6m) / (m + 6m)) * (4) + ((2*6m) / (m + 6m)) * (2)
v₁f = 0.5714 m/s (→)
v₂f = ((2m₁) / (m₁ + m₂)) v₁i + ((m₂ – m₁) / (m₁ + m₂)) v₂i
v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)
v₂f = 2.5714 m/s (→)
e = - (v₁f - v₂f) / (v₁i - v₂i) ⇒ e = - (0.5714 - 2.5714) / (4 - 2) = 1
It was a perfectly elastic collision.
