Laboratory measurements show hydrogen produces a spectral line at a wavelength of 486.1 nanometers (nm). A particular star's spe
2 answers:
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Answer:
(a). The velocity of the object is -2.496 m/s.
(b). The total distance of the object travels during the fall is 23.80 m.
Explanation:
Given that,
Time = 1.95 s
Distance = 23.5 m
(a). We need to calculate the velocity
Using equation of motion
Put the value into the formula
(b). We need to calculate the total distance the object travels during the fall
Using equation of motion
Put the value in the equation
The total time is
We need to calculate the distance
Using equation of motion
Put the value into the formula
Hence, (a). The velocity of the object is -2.496 m/s.
(b). The total distance of the object travels during the fall is 23.80 m.
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.