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3 years ago

14

Adam and kevin are roommates. Adam can clean their apartment in 45 minutes. Kevin can clean it in 60 minutes. Rounded to the nea

rest minute, how much time will it take if they work together?.

1 answer:

Karo-lina-s [1.5K]3 years ago

5 0

If you add both numbers together it makes 105 but we’re talking time so it’ll be an hour and 45 minutes for them to clean their apartment.

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A 2 kg body is dropped from a height of 3 m. Calculate:

NikAS [45]

Answer:

a) 6.26 m/s

b) 7.67 m/s

Explanation:

The potential energy at height h0 is initially ...

PE0 = mgh0

At height h1, the potential energy is ...

PE1 = mgh1

The difference in potential energy is converted to kinetic energy:

PE0 -PE1 = KE1 = (1/2)m(v1)^2

Solving for v1, we have ...

mg(h0 -h1) = (1/2)m(v1)^2

2g(h0 -h1) = (v1)^2

v1 = √(2g(h0 -h1))

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a) When the body is 1 m high, its speed is ...

v = √(2(9.8)(3 -1)) ≈ 6.26 m/s . . . at 1 m high

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b) When the body is 0 m high, its speed is ...

v = √(2(9.8)(3 -0)) ≈ 7.67 m/s . . . when it reaches the ground

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3 years ago

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to

White raven [17]

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^{-1}$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^{-1}$

speed of the swimmer with respect to water, $v=4\ ft.s^{-1}$

<u>Now the resultant of the two velocities perpendicular to each other:</u>

$v_r=\sqrt{v^2+v_s^2}$

$v_r=\sqrt{4^2+8^2}$

$v_r=8.9442\ ft.s^{-1}$

<u>Now the angle of the resultant velocity form the vertical:</u>

$\tan\beta=\frac{v_s}{v}$

$\tan\beta=\frac{8}{4}$

$\beta=63.43^{\circ}$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d\times \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=\sqrt{d^2-w^2}$

$\Delta s=\sqrt{1118.034^2-500^2}$

$\Delta s=1000\ ft$

2)

On swimming 37° upstream:

<u>The velocity component of stream cancelled by the swimmer:</u>

$v'=v.\cos37$

$v'=4\times \cos37$

$v'=3.1945\ ft.s^{-1}$

<u>Now the net effective speed of stream sweeping the swimmer:</u>

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^{-1}$

<u>The component of swimmer's velocity heading directly towards the opposite bank:</u>

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^{-1}$

<u>Now the angle of the resultant velocity of the swimmer from the normal to the stream</u>:

$\tan\phi=\frac{v_n}{v'_r}$

$\tan\phi=\frac{4.8055}{2.4073}$

$\phi=63.39^{\circ}$

Now let the distance swam in this direction be d'.

$d'\times \cos\phi=w$

$d'=\frac{500}{\cos63.39}$

$d'=1116.344\ ft$

<u>Now the distance swept downstream:</u>

$\Delta s'=\sqrt{d'^2-w^2}$

$\Delta s'=\sqrt{1116.344^2-500^2}$

$\Delta s'=998.11\ ft.s^{-1}$

3)

Time taken in crossing the rive in case 1:

$t=\frac{d}{v_r}$

$t=\frac{1118.034}{8.9442}$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=\frac{d'}{v'_r}$

$t'=\frac{1116.344}{2.4073}$

$t'\approx463.733\ s$

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Longitude lines because latitude lines are the ones that run east to west

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LekaFEV [45]

Answer:

The closest thing to "rub bing two sticks together" is the hand-drill. You will need a fireboard (a small cedar board is good) and a thin, straight stick.

Explanation:

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