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siniylev [52]
3 years ago
8

Giving brainliest!

Physics
1 answer:
Drupady [299]3 years ago
7 0

Answer:

  1. An object in space that has no motion has to be shoved to move in the first place because of <u>newton's first law of motion. </u>                          
  2.  A bowling ball will slow down and stop when rolled on a flat surface, like a street because of <u>rolling friction</u> which comes under<u> Newton's first law of motion.</u>                                  
  3. When a 100Kg astronaut throws a 9Kg bowling ball forward in space, she moves backward but slower, this is because of<u> Newton's third law of motion. </u>      

Explanation:

  • <u>This is due to Newton's first law of motion, explaining that an object in space that has no movement has to be moved to move in the first place.</u> There is no external force or gravitational force in space, so the body can stay at rest and be forced to move first.
  • <u>Explaining why a bowling ball, like a street, can slow down and stop when rolled on a flat surface.</u> This, again, is attributed to <u>Newton's first law of motion.</u> <u>Due to rolling friction, it happens.</u> The electrons on the surface of the ground in the atoms push against the electrons that touch the ground in the atoms on the surface of the ball as you roll a ball on the ground. A rolling ball stops when the surface on which it rolls opposes its motion. A ball rolling, due to friction, ceases.
  • <u>Explaining that when a 100 kg astronaut throws a 9 kg bowling ball forward in space, she travels backward but slower</u>, due to <u>Newton's third motion rule , </u>which states that when force is applied, each body exhibits equal and opposite reaction. In this case, the astronaut's mass is greater, so the ball's acceleration will be less than the astronaut 's mass (since the body's mass is inversely proportional to the acceleration). Therefore, because of the reverse direction and slowly because of no gravitational force in space, the girl shifts backward while tossing the ball forward in space.

<u> Newton's Law of Motion -  </u>

<u>FIRST  LAW -</u>  The first law of Newton states that if a body is at rest or moving in a straight line at a constant speed, it will stay at rest or continue to travel at constant speed in a straight line unless it is acted upon by a force.

<u>SECOND LAW -  </u> The second law of Newton is a quantitative explanation of the changes that a force can cause in a body 's motion. It states that the rate of change of a body's momentum in time is proportional to the force exerted on it in both magnitude and direction.  

<u>THIRD LAW -  </u>   The third law of Newton notes that they apply forces to each other when two bodies interact, which are equal in magnitude and opposite in direction. Often known as the law of action and reaction, the third law is. In analyzing static equilibrium problems, where all forces are balanced, this law is essential, but it also applies to bodies in uniform or accelerated motion.  

Hence , all the three explainations consist of newton's first and third  laws of motion.

                                                                         

                                             

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A satellite orbits Earth. The only force on the satellite is the gravitational force exerted by Earth. How does the satellite’s
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Answer:

here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

Explanation:

As we know that gravitational field is defined as the force experienced by the satellite per unit of mass

so we will have

E = \frac{F}{m}

now in order to find the acceleration of the satellite we know by Newton's II law

F = ma

so we will have

a = \frac{F}{m}

so here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

4 0
3 years ago
Read 2 more answers
Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t
earnstyle [38]
1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
B= \frac{\mu_0I}{2 \pi r}
where
\mu_0 is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance 
r=40.0 cm=0.40 m, 
which is the distance at which the other wire is located:
B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N

2) direction of the force: 
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive

A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.
6 0
3 years ago
The next four questions refer to the situation below.
Anna11 [10]

Answer:

 t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in},      t_{out} = \frac{D}{v_s +v_r}

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

         v_{sg 1} = v_{sr} + v_{rg}

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

           v_{sg1} = D / t_{out}

           D = v_{sg1}  t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

        v_{sg 2} =  v_{sr}  - v_{rg}

         v_{sg 2} = D / t_{in}

         D = v_{sg 2}  t_{in}

with the distance is the same we can equalize

           v_{sg1} t_{out} = v_{sg2} t_{in}

          t_{out} =  t_{in}

           t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in}

This must be the answer since the return time is known. If you want to delete this time

            t_{in}= D / v_{sg2}

we substitute

            t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

            t_{out} = \frac{D}{v_s +v_r}

7 0
2 years ago
Belly-flop Bernie dives from atop a tall flagpole into a swimming pool below. His potential energy at the top is 7000 J (relativ
elena55 [62]

Answer:

KE₂ = 6000 J

Explanation:

Given that

Potential energy at top U₁= 7000 J

Potential energy at bottom U₂= 1000 J

The kinetic energy at top ,KE₁= 0 J

Lets take kinetic energy at bottom level =  KE₂

Now from energy conservation

U₁+ KE₁= U₂+ KE₂

Now by putting the values

U₁+ KE₁= U₂+ KE₂

7000+ 0 = 1000+ KE₂

KE₂ = 7000 - 1000 J

KE₂ = 6000 J

Therefore the kinetic energy at bottom is 6000 J.

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3 years ago
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Free_Kalibri [48]

Answer:

the first one is the correct answer

6 0
2 years ago
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