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:
v = -14 m/s
Explanation:
Given that,
Initial location of the ball, X₁ = 10 m
Final position of the ball, X₂ = -25 m
Time taken to travel is, t = 2.5 s
The average velocity of the ball is given by the formula,
V = X₂ - X₁ / t m/s
Substituting the values in the above equation,
V = -25 - 10 / 2.5
= -14 m/s
The negative sign in the velocity indicates that ball rolls in the opposite direction.
Hence, the average velocity of the ball is v = -14 m/s
Answer:
a. 2v₀/a b. 2v₀/a
Explanation:
a. Since you are moving with a constant velocity v₀, the distance, s you cover in time = t max is s = v₀t.
Since the dragster starts from rest with an acceleration, a, using
s' = ut + 1/2at² where u = 0 and s' = distance moved by dragster
s' = 0t + 1/2at²
s' = 1/2at²
Since the distance moved by me and the dragster must be the same,
s = s'
v₀t. = 1/2at²
v₀t. - 1/2at² = 0
t(v₀ - 1/2at) = 0
t= 0 or v₀ - 1/2at = 0
t= 0 or v₀ = 1/2at
t= 0 or t = 2v₀/a
So the maximum time tmax = 2v₀/a
b. Since the distance covered by me to meet the dragster is s = v₀t in time, t = tmax which is also my distance from the dragster when it started. So, my distance from the dragster when it started is s = v₀(2v₀/a)
= 2v₀/a
Answer:
brainly getting too strict smh...
Explanation: