The woman in the first photo lifts a plant 2 m

What will one see during an annular eclipse?

KATRIN_1 [288]

Partial eclipse, Annular eclipse, Total Eclipse and Hybrid Eclipse are the four different types of the eclipses. When the Sun and Moon are exactly in line with the Earth, the annular eclipse occurs. The new moon is invisible from the earth and it is silhouetted against the sun, this can only be seen in annular eclipse. Annular word means ring shaped, we can see a dark ring of fire in annular eclipse. It has five different stages that are first contact, second contact, maximum eclipse, third contact and 4th contact.

7 0

3 years ago

PlzHELP

Naddika [18.5K]

Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion.Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.

To calculate displacement, simply draw a vector from your starting point to your final position and solve for the length of this line. If your starting and ending position are the same, like your circular 5K route, then your displacement is 0. In physics, displacement is represented by Δs.

For me to solve this I would need to know the time, but I can give you a handy displacement calculator I used that helped me.

https://www.easycalculation.com/physics/classical-physics/constant-acc-displacement.php

Hope I helped.

8 0

3 years ago

Which of the following was NOT one of the top 10 reported Stressors for Teens?

olga55 [171]

I think it life changes

4 0

3 years ago

Read 2 more answers

Assapp!!!’!!!!!!!! !!!!!!

dlinn [17]

Answer:

delta r(x) = (delta (r)) * cos(alpha), delta r(y) = (delta(r)) * sin(alpha)

Explanation:

Well it's a simple rule I guess...

6 0

3 years ago

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

posledela

Answer:

a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

$v_{y}$ ² = $v_{oy}$ ² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

y-yo = $v_{oy}² /2 g
 y- 0 = 10.0²/2 9.8
 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [tex]v_{y}$ = $v_{oy}$ - g t

t = $v_{oy}$ / g

t = 10 / 9.8

t = 1.02 s

b) the maximum height

y- 44.0 = $v_{y}$ ² / 2 g

y - 44.0 = 5.1

y = 5.1 +44.0

y = 49.1 m

The time is the same because it does not depend on the initial height

t = 1.02 s

7 0

3 years ago