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pychu [463]
3 years ago
12

How does newton 3 laws relate to an egg drop

Physics
2 answers:
crimeas [40]3 years ago
7 0
Newton's third law, which states that every action has an equal and opposite reaction, can be related to an egg drop because the forces pushing the egg up and down must follow the law.
puteri [66]3 years ago
7 0
The egg's drop is related to newton 3 laws. This because he has the 3 laws of motion.
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The more domains that are aligned in a magnet ___________.
I am Lyosha [343]
Hey, Sissy05pooh!
The more domains that are aligned in a magnet it is stronger, and the magnetic field is larger.
6 0
3 years ago
Must physics always be explained by math?
cluponka [151]

Answer:

Because Physics is mainly composited of math

Explanation:

If you really think about it the only way any science works is off maths, whether that be measuring the amount of NaOH to put into a solution or determening the amount of sugar you want in a cake it all relys on maths

5 0
3 years ago
Which equation represents the law of conservation of energy in a closed system?
Irina-Kira [14]

Answer:

KE + PE = KE + PE

Explanation:

In a closed system, the mechanical energy of the system is constant.

Mechanical energy is given by the sum of kinetic energy and potential energy; mathematically:

U = KE + PE

where

KE is the kinetic energy

PE is the potential energy

This means that if we consider two situations, one at the beginning and one at the end, the value of U will not change if the system is closed; this means that the sum KE + PE will remain the same, so we can write:

KE + PE = KE + PE

7 0
4 years ago
Read 2 more answers
1) A spring, which has a spring constant k=7.50 N/m, has been stretched 0.40 m from ts equilibrium position . What the potential
cupoosta [38]
<h3>Answer:</h3>

\displaystyle U_s = 0.6 \ J

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Physics</u>

<u>Energy</u>

Elastic Potential Energy: \displaystyle U_s = \frac{1}{2} k \triangle x^2

  • U is energy (in J)
  • k is spring constant (in N/m)
  • Δx is displacement from equilibrium (in m)
<h3>Explanation:</h3>

<u>Step 1: Define</u>

k  = 7.50 N/m

Δx = 0.40 m

<u>Step 2: Find Potential Energy</u>

  1. Substitute in variables [Elastic Potential Energy]:                                        \displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.40 \ m)^2
  2. Evaluate exponents:                                                                                      \displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.16 \ m^2)
  3. Multiply:                                                                                                           \displaystyle U_s = (3.75 \ N/m) (0.16 \ m^2)
  4. Multiply:                                                                                                           \displaystyle U_s = 0.6 \ J
6 0
3 years ago
There are two identical, positively charged conducting spheres fixed in space. The spheres are 44.0 cm apart (center to center)
Aneli [31]

Answer:

q₁ =± 1.30 10⁻⁶ C  and   q₂ = ± 1.28 10⁻⁶ C

Explanation:

We will solve this problem with Coulomb's law

    F = K q₁q₂ / r²

Where the Coulomb constant is value 8.99 10⁹ N m² / C²

Let's apply this equation to our problem

Case 1

    F1 = k q₁ q₂ / r₁²

Where r₁ = 0.440 m and F1 = 0.0765 N

Case 2

The charges are the same

    F2 = k q q / r₂²

With r₂ = r₁ = 0.440 m, the spheres are fixed and the force is F2 = 0.100 N

When the spheres are joined with the wire, the charge is distributed, distributed and matched in the two spheres

    q₁ + q₂ = 2 q

Let's replace

    F2 = k ½ (q₁ + q₂) / r²

Let's write the two equations and solve the system of equations

    F1 = k q₁ q₂ / r²

    F2 = ½ k (q₁ + q₂) / r²    

    F1 r² / k = (q₁ q₂)

    F2 r² / k = (q₁ + q₂)/2

    q₁ = 2F2 r² / k - q₂

We substitute in the other equation

    F1 r² / k = (2F2 r² / k - q₂) q₂

    0 = -F1 r² / k + (2F2 r² / k) q₂ - q₂²

Let's solve the second degree equation

    F1 r² / K = 0.0765 0.440² / 8.99 10⁹

    F1 r² / K = 1.65 10⁻¹²

   (2F2 r² / k) =2  0.10 0.44² / 8.99 10⁹

    (2F2 r2 / k) = 4.30 10⁻¹²

    q₂² - 4.30 10⁻¹² q₂ + 1.65 10⁻¹² = 0

    q₂ = ½ {4.30 10⁻¹² ± √ [(4.30 10⁻¹²)² - 4 1.65 10⁻¹²]}

    q₂ = ½ {4.30 10⁻¹² ± 2,569 10⁻⁶}

    q₂ = ± 1.2845 10⁻⁶ C

Now we calculate q1

    F1 = k q₁ q₂ / r²

    q₁ = F1 r² / (k q₂)

    q₁ = 0.0765 0.440² / (8.99 10⁹ 1.2845 10⁻⁶)

    q₁ = 1.30 10⁻⁶ C

3 0
3 years ago
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