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jolli1 [7]
3 years ago
12

What are 3 ways a machine serves its purpose?

Physics
2 answers:
gogolik [260]3 years ago
8 0
1 Makes things easier
2 Does (some) work for you
3 ... anyone else?
Alinara [238K]3 years ago
7 0
What type of machines are you talking about?
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In order for combustion to occur what needs to happen
miv72 [106K]
Pretty much combustion is the reaction of oxygen to a compound that contains carbon and hydrogen. That's why it creates CO2 and H2O
5 0
3 years ago
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An object is 45.0 kg here on Earth. What is its mass on Jupiter if the acceleration
puteri [66]

Answer:

B. 45.0 kg

Explanation:

On earth the object has ;

Mass = 45 kg

acceleration = 9.81 m/s²

On Jupiter, acceleration is  24.79 m/s²

The mass of this object on Jupiter will be 45 kg. It will not change. Mass of an object will only change when you remove part of the object from it or add to it another part. The mass is the same on Earth and on Jupiter. However, due to increased acceleration on Jupiter , the weight will change/ increase because;

Weight = mass * acceleration

<u>On Earth </u>

Weight of the object will be : 45 *  9.81 = 441.45 kg

<u>On Jupiter</u>

Weight of the object will be : 45*24.79 =1115.55 kg

8 0
3 years ago
A man stands at the edge of a cliff and throws a rock downward with A speed of 12.0 m s . Sometimes later it strikes the ground
kvasek [131]

B) 48.0 m/s

We can actually start to solve the problem from B for simplicity.

The motion of the rock is a uniformly accelerated motion (free fall), so we can find the final speed using the following suvat equation

v^2 -u^2 = 2as

where

v is the final velocity

u = 12.0 is the initial velocity (positive since we take downward as positive direction)

a=g=9.8 m/s^2 is the acceleration of gravity

s = 110 m is the vertical displacement

Solving for v, we find the final velocity (and so, the speed of the rock at impact):

v = \sqrt{u^2+2as}=\sqrt{12^2+2(9.8)(110)}=48.0 m/s

A) 3.67 s

Now we can find the time of flight of the rock by using the following suvat equation

v=u+at

where

v = 48.0 is the final velocity at the moment of impact

u = 12.0 is the initial velocity

a=g=9.8 m/s^2 is the acceleration of gravity

t is the time it takes for the rock to reach the ground

And solving for t, we find

t=\frac{v-u}{a}=\frac{48.0-12.0}{9.8}=3.67 s

6 0
3 years ago
Magnets are not hammered or rubbed<br> why?​
Nutka1998 [239]
Lose its magnetic properties
8 0
2 years ago
Visualizing the body in the position is significant because all observers have a common point of reference when describing and d
FromTheMoon [43]

Visualizing the body in the anatomical position is significant because all observers have a common point of reference when describing and discussing its region.

<h3>What is the anatomical position?</h3>
  • Anatomical position, also known as conventional anatomical position, refers to the body's position while it is standing erect and looking forward, with each arm hanging on either side of the body and palms facing forward.
  • The legs are parallel, with the feet level on the floor and forward facing.
  • When explaining particular anatomical words and locations in human anatomy and physiology, the anatomical position is a standard point of reference.
  • Visualizing the body in its anatomical location is important because it provides a single point of reference for all observers when describing and discussing its region.

As the description itself says, visualizing the body in its anatomical location is important because it provides a single point of reference for all observers when describing and discussing its region.

Therefore, visualizing the body in the anatomical position is significant because all observers have a common point of reference when describing and discussing its region.

Know more about the anatomical position here:

brainly.com/question/5029605

#SPJ4

The question you are looking for is given here:

Visualizing the body in the _____________ is significant because all observers have a common point of reference when describing and discussing its region.

6 0
2 years ago
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