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3 years ago

The structures shown are primarily involved in

Physics

2 answers:

VLD [36.1K]3 years ago

6 0

Removing waste.
_____

Gennadij [26K]3 years ago

6 0

The correct answer is 3 (removing wastes)
Good luck with future problems m8

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This time particle A starts from rest and accelerates to the right at 65.5 cm/s

FrozenT [24]

Answer:

t = 4 s

Explanation:

As we know that the particle A starts from Rest with constant acceleration

So the distance moved by the particle in given time "t"

$d = v_i t + \frac{1}{2}at^2$

$d = 0 + \frac{1}{2}(65.5)t^2$

$d_1 = 32.75 t^2 cm$

Now we know that B moves with constant speed so in the same time B will move to another distance

$d_2 = 44 \times t$

now we know that B is already 349 cm down the track

so if A and B will meet after time "t"

then in that case

$d_1 = 349 + d_2$

$32.75 t^2 = 349 + 44 t$

on solving above kinematics equation we have

$t = 4 s$

4 0

2 years ago

A block of wood mass 0.60kg is balanced on top of a vertical port 2.0m high. A 10gm bullet is fired horizontally into the block

anzhelika [568]

Answer:

Mass of bullet is m=0.01kg

Mass of the block is M=4kg

Coefficient=0.25,distance=20m

So, let the speed of the block just after the bullet embedded in it be V and v be the speed of bullet before striking the block,

By applying conservation of momentum,

mv=(m+M)V

M+m

Explanation:

please mark me as the brainliest answer and please follow me for more answers to your questions..

7 0

3 years ago

A Porsche challenges a Honda to a 500 m race

zlopas [31]

A Porsche will always win no matter what

8 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:

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

8 0

2 years ago

A football player runs 20 meters North of a football field, and then 15 meters East. The total motion lasted 15 seconds. What wa

vlada-n [284]

Given:
1st run: 20 meters North
2nd run: 15 meters East
time: 15 seconds

Average speed = total distance covered / total time taken
Ave. Speed = (20m + 15m) / 15s
Ave. Speed = 35m / 15s
Ave. Speed = 2 1/3 meters per second

4 0

2 years ago