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3 years ago

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A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i

ncredible intensity. The rain is falling vertically, and it accumulates in the boat at the rate of 100 kg/hr. What is the speed of the boat after time 0.500 hour has passed? Assume that the water resistance is negligible.

1 answer:

bija089 [108]3 years ago

4 0

Answer:

$v_o\approx0.7059\ m.s^{-1}$

Explanation:

Given:

mass of the boat, $m=120\ kg$

uniform speed of the boat, $v=1\ m.s^{-1}$

rate of accumulation of water mass in the boat, $\dot m=100\ kg.hr^{-1}$

time of observation, $t=0.5\ hr$

The mass of the boat after the observed time:

$m_o=m+\dot m\times t$

$m_o=120+100\times 0.5$

$m_o=170\ kg$

<u>Now using the conservation of momentum:</u>

$m.v=m_o.v_o$

$120\times 1=170\times v_o$

$v_o\approx0.7059\ m.s^{-1}$

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A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.

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Answer: $f_{r}$ = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

$W_{x}$ = horizontal component of the weight = mgsinФ

$W_{y}$ = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - $f_{r}$ = ma

where m = mass of object=5kg

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we need to first get the acceleration before the frictional force which is gotten by using the equation below

$v^{2} = u^{2} + 2aS$

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

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$2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}$

we slot in a into the equation below to get frictional force

mgsinФ - $f_{r}$ = ma

3 * 9.8 * sin 37 - $f_{r}$ = 3* 0.4

17.9633 - $f_{r}$ = 1.2

$f_{r}$ = 17.9633 - 1.2

$f_{r}$ = 16.49N

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