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3 years ago

During a total lunar eclipse, the moon

2 answers:

8 0

A, C, and D all happen at different stages
of a total lunar eclipse.

I'll describe the stages of the eclipse, but before I do, I just need
to clarify: The Earth doesn't have an umbra or a penumbra, but
its shadow does.

-- the eclipse begins when the first edge of the moon
   moves into the penumbra of Earth's shadow; ( C )
   this part of the moon grows steadily.

-- After a while, the first edge of the moon begins to move
   into the umbra of Earth's shadow ( A ), and gets very dark.

-- The total phase of the eclipse begins when the ENTIRE
    moon is in the umbra of Earth's shadow.

Then everything happens in reverse.

-- Eventually, the leading edge of the moon moves out
   of the shadow's umbra, into the penumbra. This part
   steadily grows.

-- After a while, none of the moon is in the umbra, and
   the whole thing is in the penumbra. The moon is
   fully illuminated, but not quite as bright as it should be.

-- Soon, the leading edge of the moon leaves the penumbra
of Earth's shadow, and gets brighter. This portion of the moon
    steadily grows, until ...

-- the moon completely leaves the penumbra, all of it is as bright
as it's supposed to be. The eclipse is completely over. ( B )

==> The whole process lasts several hours.

==> Everybody on the night side of the Earth sees the same thing
   at the same time. It doesn't matter WHERE you are on the night
   side ... if you can see the moon in the sky, you see the present
   phase of the eclipse.

==> The lunar eclipse can only happen at the Full Moon. In fact, the
   mid-point of the total phase is the exact moment of Full Moon.

lidiya [134]3 years ago

5 0

The answer would be A) Moves into Earth's umbra. Hope this helps!!

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In the long jump, an athlete launches herself at an angle above the ground and lands at the same height, trying to travel the gr

NikAS [45]

A) 2.64t

B) 2.64h

C) 2.64D

Explanation:

The motion of the athlete is equivalent to the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion (constant acceleration) along the vertical direction

The time of flight of a projectile can be found from the equations of motion, and it is found to be

$t=\frac{2u sin \theta}{g}$

where

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the time of flight is $t$ .

When she is on Mars, the acceleration due to gravity is:

$g'=0.379 g$

where g is the acceleration due to gravity on Earth. Therefore, the time of flight on Mars will be:

$t'=\frac{2usin \theta}{g'}=\frac{2u sin \theta}{0.379g}=\frac{1}{0.379}t=2.64t$

The maximum height reached by a projectile can be also found using the equations of motion, and it is given by

$h=\frac{u^2 sin^2\theta}{2g}$

where

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the maximum height is $h$ .

When she is on Mars, the acceleration due to gravity is:

$g'=0.379 g$

where g is the acceleration due to gravity on Earth. So, the maximum height reached on Mars will be:

$h'=\frac{u^2 sin^2\theta}{2g'}=\frac{u^2 sin^2\theta}{(0.379)2g}=\frac{1}{0.379}h=2.64h$

The horizontal distance covered by a projectile is also found from the equations of motion, and it is given by

$D=\frac{u^2 sin(2\theta)}{g}$

where:

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the horizontal distance covered is D.

When she is on Mars, the acceleration due to gravity is:

$g'=0.379 g$

where g is the acceleration due to gravity on Earth. Therefore, the horizontal distance reached on Mars will be:

$D'=\frac{u^2 sin(2\theta)}{g'}=\frac{u^2 sin(2\theta)}{(0.379)g}=\frac{1}{0.379}D=2.64D$

7 0

3 years ago

In the design of a supermarket, there are to be several ramps connecting different parts of the store. Customers will have to pu

tatuchka [14]

Answer:

3.90 degrees

Explanation:

Let g= 9.81 m/s2. The gravity of the 30kg grocery cart is

W = mg = 30*9.81 = 294.3 N

This gravity is split into 2 components on the ramp, 1 parallel and the other perpendicular to the ramp.

We can calculate the parallel one since it's the one that affects the force required to push up

F = WsinΘ

Since customer would not complain if the force is no more than 20N

F = 20

$294.3sin\theta = 20$

$sin\theta = 20/294.3 = 0.068$

$\theta = sin^{-1}0.068 = 0.068 rad = 0.068*180/\pi \approx 3.90^0$

So the ramp cannot be larger than 3.9 degrees

6 0

3 years ago

The maximum value of magnetic in an electric field 3.2 *10^4

Svetradugi [14.3K]

Answer:

the answer is 12 because if your magnetic value and Electric field is 3.2 the answer will be 12

6 0

2 years ago

I dropped an apple (mass 0.1kg) from the window because i'm weird. (15m above the ground). How fast was it going when it hit the

Olegator [25]

Answer:

I think the answer is 1 m per second.

5 0

3 years ago

Superman does an exhibition run at a track meet. When he runs the 200 m

Ber [7]

Answer:

6.32s

Explanation:

Given parameters:

Length of track and distance covered = 200m

Acceleration = 10m/s²

Unknown:

Time taken to cover the track = ?

Solution:

To solve this problem, we apply one of the motion equations as shown below:

S = ut + $\frac{1}{2}$ at²

S is the distance covered

t is the time taken

a the acceleration

u is the initial velocity

The initial velocity of Superman is 0;

So;

S = $\frac{1}{2}$ at²

200 = $\frac{1}{2}$ x 10 x t²

200 = 5t²

t² = 40

t = 6.32s

5 0

3 years ago