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3 years ago

PLEASE HELP What is it called when the moon passes through the penumbra of Earth’s shadow?(1 point)

Physics

1 answer:

bogdanovich [222]3 years ago

8 0

Partial. a penumbral lunar eclipse is when the moon passes through the earths penumbra

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I need help on both questions

iris [78.8K]

It's not in motion when the line straight and flat . there's #9

3 0

3 years ago

Read 2 more answers

A has the magnitude 14.4 m and is angled 51.6° counterclockwise from the positive direction of the x axis of an xy coordinate sy

Ad libitum [116K]

Answer:

Ã in unit vector notation = 12.26485i + 7.54539j

B in unit vector notation = 16.3516i + 3.11529j

Explanation:

The detailed steps and calculation is shown in the attachment.

7 0

3 years ago

magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,

Mama L [17]

Answer:

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

Explanation:

Given:

height above which the rock is thrown up, $\Delta h=50\ m$

initial velocity of projection, $u=20\ m.s^{-1}$

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

$v=u+g'.t'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0=20-g'.t'$ (-ve sign to indicate that acceleration acts opposite to the velocity)

$t'=\frac{20}{g'}\ s$

The time taken by the rock to reach the top height on the earth:

$v=u+g.t$

$0=20-g.t$

$t=\frac{20}{g} \ s$

Height reached by the rock above the point of throwing on the exoplanet:

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

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0^2=20^2-2\times g'.h'$

$h'=\frac{200}{g'}\ m$

Height reached by the rock above the point of throwing on the earth:

$v^2=u^2+2g.h$

$0^2=20^2-2g.h$

$h=\frac{200}{g}\ m$

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

$(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2$ (during falling it falls below the cliff)

here:

$u=$ initial velocity= 0 $m.s^{-1}$

$\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2$

$t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}$

$t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'} }$

Similarly on earth:

$t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g} }$

Now the required time difference:

$\Delta t=(t'+t_f')-(t+t_f)$

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

3 0

3 years ago

How much space something take up

elixir [45]

Your answer is matter.

5 0

3 years ago

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A 500-kilogram horse runs at 5 meters/second. The horse’s kinetic energy is <br> joules.

kari74 [83]

To solve this, we are going to use the formula for the kinetic energy of an object: $E_{k}= \frac{1}{2} mv^2$
where
$E_{k}$ is the kinetic energy of the object.
$m$ is the mass of the object.
$v$ is the speed of the object.

We know form our problem that the mass of the horse is 500 kilograms, so $m=500$ ; we also know that the speed of the horse is 5 meter/second, so $m=5$ . Lets replace those values in our formula to find $E_{k}$ :
$E_{k}= \frac{1}{2} mv^2$
$E_{k}= \frac{1}{2} (500)(5)^2$
$E_{k}= \frac{1}{2} (500)(25)$
$E_{k}= \frac{12500}{2}$
$E_{k}=6250$ J

We can conclude that the kinetic energy of the horse is 6250 Joules.

7 0

4 years ago

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