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2 years ago

Descibe the real-world examples of Newton's third lawthat were idenified in "Applications of Newton's Laws."

Physics

2 answers:

ehidna [41]2 years ago

5 0

Answer:

A horse pulls a cart, a person walks on the ground

Explanation:

ValentinkaMS [17]2 years ago

5 0

Answer:

Know it's late but I couldn't find a good answer when I looked for it so i did it myself.

My answer:

One of the real world examples mentioned in the article was that of a rocket blasting off into space. It was explained that it's an example of Newton's third law because the rocket’s engines push on those gases (action force), and the gases push back (reaction force). This reaction shows an excellent example of Newton's third law.

Sample Response:

When gases escape downward out the tail of a rocket, they push upward on the rocket. When layers of the Earth collide, they force the Earth upward, forming mountains.

Explanation:

hope this helps <3

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A 170 kg astronaut (including space suit) acquires a speed of 2.25 m/s by pushing off with his legs from a 2600 kg space capsule

saw5 [17]

Explanation:

Mass of the astronaut, m₁ = 170 kg

Speed of astronaut, v₁ = 2.25 m/s

mass of space capsule, m₂ = 2600 kg

Let v₂ is the speed of the space capsule. It can be calculated using the conservation of momentum as :

initial momentum = final momentum

Since, initial momentum is zero. So,

$m_1v_1+m_2v_2=0$

$170\ kg\times 2.25\ m/s+2600\ kg\times v_2=0$

$v_2=-0.17\ m/s$

So, the change in speed of the space capsule is 0.17 m/s. Hence, this is the required solution.

8 0

3 years ago

Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram

Troyanec [42]

The student who did the most work is student 2 with 2500 Joules.

Given the following data:

Force 1 = 500 Newton

Distance 1 = 1.2 meter

Force 2 = 500 Newton

Distance 2 = 5 meter

To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

$Work\;done = Force \times distance$

For student 1:

$Work\;done = 500 \times 1.2$

Work done = 600 Joules

For student 2:

$Work\;done = 500 \times 5$

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

Read more: Read more: brainly.com/question/13818347

7 0

2 years ago

Read 2 more answers

Calculate the radius of the orbit of a proton moving at 2.2x10^6 m/s in a magnetic field 0.7 T where v and B are perpendicular.

Juliette [100K]

Answer:

3.28 cm

Explanation:

To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:

r = $\frac{mv}{qB}$