Joe and Max shake hands and say goodbye. Joe walks east 0.50 km to a coffee shop, and Max flags a cab and rides north 3.45 km to
1 answer:
Answer:
3.486 km
Explanation:
Suppose Joe and Max's directions are perfectly perpendicular (east vs north). We can calculate their distance at the destinations using Pythagorean theorem:
where J = 0.5 km and M= 3.45 km are the distances between Joe and Max to their original parting point, respectively. s is the distance between them.
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Answer:
Explanation:
<u>Given:</u>
Frequency = f = 40 Hz
<u>Required:</u>
Time period = T = ?
<u>Formula:</u>
<u>Solution:</u>
T = 1 / 40
T = 0.025 seconds
Hope this helped!
<h3>~AH1807</h3>
The frequency of middle C on a string is
f = 261.6 Hz.
The given linear density is
ρ = 0.02 g/cm = (0.02 x 10⁻³ kg)/(10⁻² m)
= 0.002 kg/m
The length of the string is L = 1 m.
Let T = the tension in the string (N).
The velocity of the standing wave is
In the fundamental mode, the wavelength, λ, is equal to the length, L.
That is
Because v = fλ, therefore
From given information, obtain
T = (0.002 kg/m)*(261.6 1/s)²*(1 m)²
= 136.87 N
Answer: 136.9 N (nearest tenth)
f = 261.6 Hz.
The given linear density is
ρ = 0.02 g/cm = (0.02 x 10⁻³ kg)/(10⁻² m)
= 0.002 kg/m
The length of the string is L = 1 m.
Let T = the tension in the string (N).
The velocity of the standing wave is
In the fundamental mode, the wavelength, λ, is equal to the length, L.
That is
Because v = fλ, therefore
From given information, obtain
T = (0.002 kg/m)*(261.6 1/s)²*(1 m)²
= 136.87 N
Answer: 136.9 N (nearest tenth)