The amplitude of wave-c is 1 meter.
The speed of all of the waves is (12meters/2sec)= 6 m/s.
The period of wave-a is 1/2 second.
Answer:
80 ft/s
Explanation:
Given:
Δy = 100 ft
v₀ = 0 ft/s
a = 32.2 ft/s²
Find: v
v² = v₀² + 2aΔx
v² = (0 ft/s)² + 2 (32.2 ft/s²) (100 ft)
v = 80.2 ft/s
Rounded, the speed when it reaches the ground is 80 ft/s.
Answer:
The speed of wave in the second string is 55.3 m/s.
Explanation:
Given that,
Speed of wave in first string= 58 m/s
We need to calculate the wave speed
Using formula of speed for first string
...(I)
For second string
...(II)
Divided equation (II) by equation (I)
Here, Tension is same in both string
So,
The linear density of the second string
Now, Put the value of linear density of second string
Hence, The speed of wave in the second string is 55.3 m/s.
Answer:
(a) 8Ω (b) Ratio = Parra/P8 ohm = 1
Explanation:
Solution
Recall that,
An high-fidelity amplifier has one output for a speaker of resistance of = 8 Ω
Now,
(a) How can two 8-Ω speakers be arranged, when one = 4-Ω speaker, and one =12-Ω speaker
The Upper arm is : 8 ohm, 8 ohm
The Lower arm is : 12 ohm, 4 ohm
The Requirement is = (16 x 16)/(16 + 16) = 8 ohm
(b) compare your arrangement power output of with the power output of a single 8-Ω speaker
The Ratio = Parra/P8 ohm = 1
Answer:
it will float because the density of water is 1g/cm³ and is therefore less dense
Explanation:
