Answer:
1.95 kg
Explanation:
Momentum is conserved.
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
0 = (74.9) (-0.215) + m (8.25)
m = 1.95
A. A flood would have the most positive effect
The height at time t is given by
h(t) = -4.91t² + 34.3t + 1
When the ball reaches maximum height, its derivative, h'(t) = 0.
That is,
-2(4.91)t+34.3 = 0
-9.82t + 34.3 = 0
t = 3.4929 s
Note that h''(t) = -9.82 (negative) which confirms that h will be maximum.
The maximum height is
hmax = -4.91(3.4929)² + 34.3(3.4929) + 1
= 60.903 m
Answer:
The ball attains maximum height in 3.5 s (nearest tenth).
The ball attains a maximum height of 60.9 m (nearest tenth)
Answer: 18.27°
Explanation:
Given
Index of refraction of blue light, n(b) = 1.64
Wavelength of blue light, λ(b) = 440 nm
Index of refraction of red light, n(r) = 1.595
Wavelength of red light, λ(r) = 670 nm
Angle of incident, θ = 30°
Angle of refraction of red light is
θ(r) = sin^-1 [(n(a)* sin θ) / n(r)], where n(a) = index of refraction of air = 1
So that,
θ(r) = sin^-1 [(1 * sin 30) / 1.595]
θ(r) = sin^-1 (0.5 / 1.595)
θ(r) = sin^-1 0.3135
θ(r) = 18.27°
Answer: I don't know how to do this
Explanation: sorry I am not sure.
