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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

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1 year ago

A carnival game consists of a two masses on a curved frictionless track, as pictured below. The player pushes the larger object

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Answer:

v₁₀ = 1.90 m / s

Explanation:

In this exercise we are given the maximum height data, with energy we can know how fast the body came out

Final mechanical energy, maximum height

$Em_{f}$ = U = m g h

Initial mechanical energy, in the lower part of the track

Em₀ = K = ½ m v²

Em= $Em_{f}$

½ m v² = m g h

v = √ 2gh

Now we can use the moment to find the speed with which objects collide

The large object has a mass M = 5.41 kg a velocity starts v₁₀, the small object has a mass m = 1.68 kg an initial velocity of zero v₂₀ = 0 and final velocity v

Initial before the crash

p₀ = M v₁₀ + 0

Final after the crash

$p_{f}$ = M v1f + m v

p₀ = $p_{f}$

M v₁₀ = M $v_{1f}$ + m v

As the shock is elastic the kinetic energy is conserved

K₀ = $K_{f}$

½ M v₁₀² = ½ M $v_{1f}$ ² + ½ m v²

Let's write the system of equations

M v₁₀ = M $v_{1f}$ + m v

M v1₁₀² = M $v_{1f}$ ² + m v²

We cleared v1f in the first we replaced in the second

$v_{1f}$ = (M v₁₀ - mv) / M

M v₁₀² = M (M v₁₀ - mv)² / M² + m v²

M v₁₀² = 1 / M (M² v₁₀² - 2mM v v₁₀ + m² v²) +m v²

v₁₀² (M - M) + 2 m v v₁₀ - v² (m2 + m) / M = 0

2 m v₁₀ - v (m + 1) m/ M = 0

v₁₀ = v (m +1) / (2M)

Let's substitute the value of v

v1₁₀= √ (2gh) (m +1) / (2M)

Let's calculate

v₁₀ = √ (2 9.8 3) (1+ 1.68) / (2 5.41)

V₁₀ = 7.668 (2.68) / 10.82

v₁₀ = 1.90 m / s

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3 years ago

The charges Q1=Q and Q2=4Q that are a distance d apart, repel each other with a force of 1.60 N. What would be the force between

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Answer:

50.4 N

Explanation:

Q1 = Q

Q2 = 4 Q

Distance = d

The force is given by

$F = \frac{KQ_{1}Q_{2}}{d^{2}}$

$1.60 = \frac{4KQ^{2}}{d^{2}}$ .... (1)

Now,

Q3 = 2 Q

Q4 = 7 Q

distance = d/3

$F' = \frac{9KQ_{3}Q_{4}}{d^{2}}$

$F' = \frac{126KQ^{2}}{d^{2}}$ .... (2)

Divide equation (2) by equation (1), we get

F' / 1.60 = 126 / 4

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Thus, the force is 50.4 N.

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A ball is thrown upward. what can be said about the system ?

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That you have thrown a ball with kinetic energy upwards at an increasing velocity rate

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2 years ago

Read 2 more answers

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

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3 years ago