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3 years ago

A boat makes a 120-mile trip downstream in 3 hours but makes the return trip in 4 hours. What is the rate of the current?

2 answers:

8 0

With the current 40 miles per hour
Against the current 30 miles per hour
bv means boat velocity with NO current
cv means current velocity
A) bv + cv = 40 mph
B) bv - cv = 30 mph
Adding both equations:
2bv = 70
boat velocity with NO CURRENT = 35 miles per hour

Using equationA
bv + cv = 40
35 + cv = 40

current velocity = 5 miles per hour

Alchen [17]3 years ago

5 0

Answer:

Speed of current = 5 mph

Step-by-step explanation:

Let u be the speed of boat and v be the speed of current.

A boat makes a 120-mile trip downstream in 3 hours.

120 = 3(u+v)

u + v = 40

It makes the return trip in 4 hours.

120 = 4(u-v)

u - v= 30

Adding both equations

u + v + u - v = 40+30

2u = 70

u = 35 mph

35 + v = 40

v = 5 mph

Speed of current = 5 mph

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segment BC is an are of a circle with radius 30.0 cm, and point P is at the center of curvature of the arc Segment DA is an arc

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The magnetic field is $B = 3.87 *10^{-6} \ T$

Direction is inward

Step-by-step explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

The radius of BC is $r_{BC} = 30 \ cm = 0.30 \ m$

The radius of DA is $r _{DA} = 20 \ cm = 0.20 \ m$

The length of CD is $r_{CD} = 10 cm = 0.1 \ m$

The length of AB is $r_{AB} = 10 cm = 0.1 \ m$

The current is $I = 11.4A$

The magnetic field is mathematically represented as

$B = B_{BC} + B_{DA} + B_{CD} + B_{AB}$

$B = \frac{\mu_o I}{4 \pi } [ \frac{\theta_{DA}}{r_{DA}} + \frac{\theta_{BC}}{r_{BC}}+ \frac{\theta_{CD}}{r_{CD}}+\frac{\theta_{AB}}{r_{AB}}]$

Where

$\theta_{BC} = \theta_{DA} = \frac{2\pi}{3}$

Where $\frac{2 \pi}{3} = 120^o$

$\theta_{CD} = \theta_{AB} = 0^o$

$B = \frac{\mu_o I}{4 \pi } [ \frac{\frac{2\pi}{3} }{0.20} - \frac{\frac{2\pi}{3}}{0.30}}+ \frac{0}{0.10}+\frac{0}{0.10}]$

$B = 3.87 *10^{-6} \ T$

The direction is into the page

This because the magnitude of the magnetic field due to arc BC whose direction is outward is less than that of DA whose direction is inward

This is because according to Fleming's Left Hand Rule the direction of current is perpendicular to the direction of magnetic field so since current in arc BC and DA are moving in opposite direction their magnetic field will also be moving in opposite direction

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