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3 years ago

IF YOU ANSWER ALL THE QUESTIONS THEN I WILL GIVE YOU BRAINEST!!! 20 POINTS!!!!

Physics

1 answer:

Firdavs [7]3 years ago

5 0

Answer:

1.) A, 2.) D, 3.) C

Explanation:

Hope this Helps! :D

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What is the term for the process by which a portion of a glacier breaks off and falls into the water

zhenek [66]

The term for the process by which a portion of a glacier breaks off and falls into the water is called calving.

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3 years ago

i throw a tennis ball straight down from the roof of a building 400 meters tall.id it leaves my hand at 10 m/s, how fast will it

Sidana [21]

consider the motion of the tennis ball in downward direction

Y = vertical displacement = 400 m

a = acceleration = acceleration due to gravity = 9.8 m/s²

v₀ = initial velocity of the ball at the top of building = 10 m/s

v = final velocity of the ball when it hits the ground = ?

using the kinematics equation

v² = v²₀ + 2 a Y

inserting the values

v² = 10² + 2 (9.8) (400)

v = 89.11 m/s

6 0

3 years ago

Kon'nichiwa~ please help me with this question!!

andrezito [222]

Let's check the relationship

$\\ \rm\hookrightarrow g=\dfrac{GM}{r^2}$

$\\ \rm\hookrightarrow g\propto G$

Raindrops will fall faster . .
Also walking on ground would become more difficult as g increases.

Option C is wrong by now .Let's check D once

$\\ \rm\hookrightarrow T\propto \dfrac{1}{\sqrt{g}}$

So time period of simple pendulum would decrease.

4 0

2 years ago

Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could

Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

$\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}$

Where:

$d=17.91 mm =17.91(10)^{-3} m$ is the diameter of a dime

$D$ is the diameter of the Sun

$x_{sun-pinhole}=150,000,000 km=1.5(10)^{11} m$ is the distance between the Sun and the pinhole

$x_{sunball-pinhole}=100 d=1.791 m$ is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding $D$ :

$D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}$

$D=\frac{17.91(10)^{-3} m}{1.791 m} 1.5(10)^{11} m$

$D=1.5(10)^{9} m$ This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

$1 AU=149,597,870,700 m$

So, we have to divide this distance between $D$ in order to find how many suns could it fit in this distance:

$\frac{149,597,870,700 m}{1.5(10)^{9} m}=99.73 suns \approx 100 suns$

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3 years ago

Most atoms in the Sun exist as: A)solids B)liquids C)plasma

-BARSIC- [3]

U did it the way i asked. nice lol. its plasma btw

5 0

3 years ago

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