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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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Read 2 more answers
A wave pulse travels along a string at a speed of 200 cm/s. What will be the speed if:
34kurt

Answer:

a) v_f = \sqrt{\frac{2TL}{m}} = \sqrt{2} v = \sqrt{2}2m/s =2.83 m/s

The velocity increase by a factor of \sqrt{2}

b) v_f = \sqrt{\frac{TL}{4m}} = \frac{1}{2} v = \frac{1}{2} *2m/s =1 m/s

The velocity decrease by a factor of 2.

c) v_f = \sqrt{\frac{4TL}{m}} = 2} v = 2 *2m/s =4m/s

The velocity increase by a factor of 2

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The velocity not changes.

Explanation:

For this case we know that the velocity is v = 200 cm/s = 2m/s

v_f represent the final velocity after the changes specified,

Part a

The formula for the speed of a wave in a string is given by:

v = \sqrt{\frac{T}{\rho}}

And the linear density is defined as:

\rho = \frac{m}{L}

And if we replace this we got:

v = \sqrt{\frac{TL}{m}}

If the tension mass is doubled we have this:

v_f = \sqrt{\frac{2TL}{m}} = \sqrt{2} v = \sqrt{2}2m/s =2.83 m/s

The velocity increase by a factor of \sqrt{2}

Part b

If we mass is quadrupled we have this:

v_f = \sqrt{\frac{TL}{4m}} = \frac{1}{2} v = \frac{1}{2} *2m/s =1 m/s

The velocity decrease by a factor of 2.

Part c

If the length is quadrupled we have this:

v_f = \sqrt{\frac{4TL}{m}} = 2} v = 2 *2m/s =4m/s

The velocity increase by a factor of 2

Part d

For this case we know that the mass and the length are both quadrupled and we got:

v_f = \sqrt{\frac{4TL}{4m}} = v = 2m/s

The velocity not changes.

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Answer:

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Explanation:

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