Which type of reaction does this diagram represent? A small ball heads toward a large circle labeled superscript 235 upper U. An
2 answers:
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Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so = A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Answer:
a) vboat = 5.95 m/s b) vriver= 1.05 m/s
Explanation:
a) As observed from the shore, the speed of the boats can be expressed as the vector sum, of the boat speed relative to the water and the river speed relative to the shore, as follows:
vb₁s = vb₁w + vrs
In one case, the boat is moving in the same direction as the water:
vb₁s = vb₁w + vrs = 7.0 m/s (1)
For the other boat, it is clear that is moving in an opposite direction:
vb₂s = vb₂w - vrs = 4.9 m/s (2)
As we know that vb₁w = vb₂w, adding both sides, we can remove the river speed from the equation, as follows:
vb₁w = vb₂w =
b) Replacing this value in (1) and solving for vriver, we have:
vriver = 7.0 m/s - 5.95 m/s = 1.05 m/s
(we could have arrived to the same result subtracting both sides in (1), and (2))
Answer:
3000 kg.m/s
Explanation:
Momentum, p is a product of mass and velocity hence
p=mv where m is mass and v is velocity.
Change in momentum is given by where subscripts f and i represent final and initial respectively. Since the lorry finally comes to rest then the final velocity is zero. Substituting the given figures then
Change in momentum= 6000(0-0.5)=-3000 kg.m/s