Ask question

Ask question

All categories

3 years ago

9

A river flows at a rate of 2 km divided by h. A patrol boat travels 54 km upriver and returns in a total time of 9 hr. What is t

he speed of the boat in still water?

1 answer:

dolphi86 [110]3 years ago

4 0

Answer:

12.32 km/h

Explanation:

$V_r$ =Velocity of river = 2 km/h

$V_b$ =Velocity of boat

$V_b-V_r$ = Speed of boat going against river

$V_b+V_r$ = Speed of boat going along river

Distance to travel = 54 km

Total time taken = 9 hours

So,

$\frac{54}{V_b-V_r}+\frac{54}{V_b+V_r}=9\\\Rightarrow \frac{54(V_b+V_r+V_b-V_r)}{V_b^2-V_r^2}=9\\\Rightarrow \frac{54(2V_b)}{V_b^2-V_r^2}=9\\\Rightarrow \frac{54(2V_b)}{9}=V_b^2-V_r^2\\\Rightarrow 12V_b=V_b^2-V_r^2\\\Rightarrow V_b^2-V_r^2-12V_b=0\\\Rightarrow V_b^2-12V_b-4=0$

Solving this quadratic equation we get,

$V_b=\frac{12\pm \sqrt{144+16}}{2}=12.32\ or -0.32$

So, velocity of boat in still water is 12.32 km/h

You might be interested in

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

3 years ago

At what speed do a bicycle and its rider, with a combined mass of 100 kg , have the same momentum as a 1600 kg car traveling at

noname [10]

$Given:\\m_b=100kg\\m_c=1600kg\\v_c=5.2 \frac{m}{s} \\p_b=p_c\\\\Find:\\v_b=?\\\\Solution:\\\\p_b=p_c\\\\p=mv\\\\m_bv_b=m_cv_c\Rightarrow v_b= \frac{m_cv_c}{m_b} \\\\v_b= \frac{1600kg\cdot5.2 \frac{m}{s} }{100kg} =83.2 \frac{m}{s}$

3 0

3 years ago

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef

navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

$\sum F_x = \sum F_y = 0$

$\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }$

So; On the L.H.S

$\mathbf{mg sin \theta =f_s}$

$\mathbf{mg sin \theta =\mu_s N}$

On the R.H.S

N = mg cos θ

Equating both force component together, we have:

$\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}$

$\mathbf{sin \theta =\mu_s \ \ cos \theta}$

$\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}$

From trigonometry rule:

$\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}$

∴

$\mathbf{\mu_s =\tan \theta}}$

$\mathbf{\mu_s =\tan 35^0}}$

$\mathbf{\mu_s = 0.700}}$

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

8 0

3 years ago

Six friends are at a pizzeria. They want to order enough pizza so that each person can eat at least 2/5 . start fraction, 2, d

Lyrx [107]

Friend #1 gets at least 2/5 of a pizza.

Friend #2 gets at least 2/5 .

Friend #3 gets at least 2/5 .

Friend #4 gets at least 2/5 .

Friend #5 gets at least 2/5 .

Friend #6 gets at least 2/5 .

Sum . . . . . . . . . at least 12/5 of a pizza.

Simplify . . . . . . at least 2.4 pizzas.

-- If pizzas can be bought by the half, they should order at least <em>2-1/2 pizzas.</em>

-- If only whole pizzas have to be ordered, then they should order at least <em>3 pizzas.</em>

4 0

4 years ago

How would the magnetic field lines appear for a bar magnet cut at the midpoint, with the two pieces placed end to end with a spa

Zarrin [17]

Answer:

cutting the magnet in two parts each part has a North and South pole,

Explanation:

In magnetism the magnetic mono-poles are not found, this means that we do not have magnetic charges alone, therefore when cutting the magnet in two parts each part has a North and South pole, the magnetic lines go from the North pole to the South pole, see attached.

The density of the lines is approximately the intensity of the magnetic field.

4 0

3 years ago