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MissTica
3 years ago
5

A river flows from south to north at 5.4 km/hr. on the west bank of this river, a boat launches and travels perpendicular to the

current with a velocity of 7.6 km/hr due east. if the river is 1.4 km wide at this point, how far downstream does the boat land on the east bank of the river relative to the point it started at on the west bank?
Physics
1 answer:
Aleks [24]3 years ago
7 0
The motion of the boat is basically a uniform motion on both directions, north-south (NS) and east-west (EW), with two constant velocities: v_{SN}=5.4~km/h and v_{EW}=7.6~km/h. 
From the law of motion on the EW direction we can find the total time of the motion:
S_{EW}(t) = v_{EW} t
since we know S_{EW}=1.4~km, we find
t= \frac{S_{EW}}{v_{EW}} =0.18~h
and then, we can find how far the boat went on the NS direction during this time:
S_{NS}=v_{NS} t =5.4~km/h \cdot 0.18~h = 1~km
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Answer:

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Explanation:

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A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antin
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Answer:

a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

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ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so V_{max1} = A × ω

V_{max1} = 2.2×10⁻³ × 2722.69 m/s

V_{max1} =  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 )

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so

V_{max2} = A' × ω

V_{max2} = 1.555×10⁻³ × 2722.69

V_{max2} = 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

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