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2 years ago

7

Does anyone know the answers to these ?

2 answers:

geniusboy [140]2 years ago

8 0

Answer:

thirty degrease west to north

Explanation:

OverLord2011 [107]2 years ago

3 0

A.........................

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Why do adults make bigger splashes when they jump into swimming pools than small children?

Semenov [28]

Answer:

it bc the adults are bigger then us kid so when they dip in the pool it makes bigger splashes

Explanation:

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3 years ago

Ja-Anna, has a mass of 45 kg, climbed 3m up a ladder to rescue her cat from a tree. how much work against gravity did she do?

lara [203]

45*3*2=20kg in total

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3 years ago

If you were lost in space how would you get back to the space station explain scientifically

iris [78.8K]

You would use a object called The big dipper to find the closest space station.

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3 years ago

A child is riding a merry-go-round that is turning at 7.18 rpm. If the child is standing 4.65 m from the center of the merry-go-

iVinArrow [24]

Answer:

B) 3.50 m/s

Explanation:

The linear velocity in a circular motion is defined as:

$v=\omega r(1)$

The angular frequency ( $\omega$ ) is defined as 2π times the frequency and r is the radius, that is, the distance from the center of the circular motion.

$\omega=2\pi f(2)$

Replacing (2) in (1):

$v=2\pi fr$

We have to convert the frequency to Hz:

$7.18rpm*\frac{1Hz}{60rpm}=0.12Hz$

Finally, we calculate how fast is the child moving:

$v=2\pi(0.12Hz)(4.65m)\\v=3.5\frac{m}{s}$

3 0

3 years ago

A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

<u>Now the horizontal component of velocity:</u>

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

<u>The vertical component of the velocity:</u>

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

<u>Time taken by the projectile to travel the distance of 30 m:</u>

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

$t=\frac{30}{28.191}$

$t=1.0642\ s$

<u>Vertical position of the projectile at this time:</u>

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

$h=5.3701\ m$

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

<u>We observe that </u> $t>t'$ <u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

8 0

3 years ago