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3 years ago

10

If you were looking up at the sky during the day and the Sun appeared to be blacked out by an object passing in front of it, you

would most likely be witnessing

1 answer:

vaieri [72.5K]3 years ago

4 0

You would most likely be witnessing an eclipse

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Four point charges have the same magnitude of 2.4×10^−12C and are fixed to the corners of a square that is 4.0 cm on a side. Thr

Gwar [14]

Answer:

7.2N/C

Explanation:

Pls see attached file

6 0

3 years ago

What accurately describes what happens when water vapor condenses into dew in terms of energy

Andreyy89

The awnser is condensation

5 0

3 years ago

A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the

irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

$x=V_{o} cos \theta t$ (1)

$y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2}$ (2)

Where:

$x$ is the horizontal distance between the cannon and the ball

$V_{o}=23 m/s$ is the cannonball initial velocity

$\theta=0\°$ since the cannonball was shoot horizontally

$t$ is the time

$y=0$ is the final height of the cannonball

$y_{o}=1.96 m$ is the initial height of the cannonball

$g=9.8 m/s^{2}$ is the acceleration due gravity

Isolating $t$ from (2):

$t=\sqrt{-\frac{2(y-y_{o})}{g}}$ (3)

$t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}}$ (4)

$t=0.63 s$ (5)

Substituting (5) in (1):

$x=(23 m/s) cos(0\°) 0.63 s$ (6)

Finally:

$x=14.49 m$

5 0

3 years ago

A substance that accelerates any chemical reaction but is not consumed in the reaction is a _____.

Mandarinka [93]

A. Catalyst. ---------------

8 0

3 years ago

Read 2 more answers

A man throws a ball straight up to his friend on a balcony who catches it at its highest point. The ball was thrown with an init

Ede4ka [16]

Answer:

The maximum height reached by the ball is 16.35 m.

Explanation:

Given;

initial velocity of the ball, u = 17.9 m/s

the final velocity of the ball at the maximum height, v = 0

The maximum height reached by the ball is given by;

v² = u² + 2gh

During upward motion, gravity is negative

v² = u² + 2(-g)h

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u² / 2g

h = (17.9)² / (2 x 9.8)

h = 16.35 m

Ttherefore, the maximum height reached by the ball is 16.35 m.

3 0

3 years ago