Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance
Where, d = distance
A = amplitude
Put the value into the formula
Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Answer: Georgia Department of Insurance
Explanation: I hope this help :]
Science is an art, u cant make art without creativity
Answer:
1.5m
Explanation:
Velocity=1500m/s
Frequency=1000hz
Wavelength =velocity ➗ frequency
wavelength =1500 ➗ 1000
Wavelength=1.5m
Answer:
option D
Explanation:
given.
horizontal velocity of arrow and a ball given as 50 m/s and 44 m/s respectively from the top of a building over flat ground.
In vertical direction, they are both identical
In vertical direction the initial velocity of arrow and a ball is 0 m/s
Their acceleration due to gravity is same for both arrow and a ball 9.8 m/s²
they will react bottom at the same time
time of flight is same for both
now,
In horizontal direction,
distance = speed × time
Since speed is more for arrow, it will travel more horizontal distance at the same time.
the correct answer is option D
