The temperature of a lead fishing weight rises from 26∘c to 38∘c as it absorbs 11.3 j of heat. what is the mass of the fishing w
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Answer:
The acceleration is
Explanation:
From the question we are told that
The lift up speed is
The distance covered for the take off run is
Generally from kinematic equation we have that
Here u is the initial speed of the aircraft with value 0 m/ s give that the aircraft started from rest
So
=>
Answer:
At t = 15.0 s the magnitude of the velocity is 58.31 m/s
Explanation:
The given parameters are;
V₀ₓ = 30 m/s
= 100 m/s
The time of flight of the projectile = 25 s
For projectile motion;
Vₓ = V₀ × cos(θ₀)
The magnitude of the velocity V = √(V₀ₓ² + ²)
We have the magnitude of the initial velocity = √(30² + 100²) = 10·√109 m/s
cos(θ₀) = V₀ₓ/V₀ = 30/(10·√109) = 3/√109
θ₀ = cos⁻¹(3/√109) = 73.3°
The components of the velocity after time t is given by the relations;
Vₓ = V₀ × cos(θ₀) = 30 m/s
= V₀ × sin(θ₀) - g×t
When = 0, we have;
0 = V₀ × sin(θ₀) - g×t
g×t = V₀ × sin(θ₀) = 10·√109×0.958 = 100 m/s
t = 100/g = 100/10 = 10 s
The time to reach maximum height = 10 s
At 15.0 seconds, we have;
= V₀ × sin(θ₀) - g×t = 10·√109×0.958 - 10×15 = -50 m/s
Therefore, the projectile is returning at 50 m/s
The magnitude of the velocity =√(30² + 50²) = 10·√34 m/s = 58.31 m/s.