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Kitty [74]
3 years ago
7

Based on Newton’s 3rd law if you were to push on the wall with a force of 100 N, how much force would the wall push back towards

you?
Physics
1 answer:
rusak2 [61]3 years ago
3 0
The answer is 100N. Look up the definition of Newton's third law.
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A falling object accelerates from -10.0 m/s to -30.0 m/s. how much time does it take?
Zepler [3.9K]

Answer:

2.04 s

Explanation:

v = at + v₀

(-30.0 m/s) = (-9.8 m/s²) t + (-10.0 m/s)

t = 2.04 s

8 0
3 years ago
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What do all elements have in common?
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The common<span> feature is that the atoms of </span>all elements<span> consist of electrons, protons, and neutrons. Hope this helps!</span>
7 0
3 years ago
A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm
Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

  • In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

       F = - k * \Delta x (1)

  • where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
  • Solving for k:

       k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)

b)

  • Assuming no friction present, total mechanical energy mus keep constant.
  • When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

        U = \frac{1}{2}* k* (\Delta x)^{2} (3)

  • Replacing k and Δx by their values, we get:

       U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)

c)

  • When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
  • We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
  • So, we can write the following expression:

        \frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2}  = \frac{1}{2}*k*\Delta x^{2}   (5)

  • Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

        \frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2}  = 4J   (6)

⇒      \frac{1}{2}* 200N/m* \Delta x_{1} ^{2}   = 4J  - 3.6 J = 0.4 J (7)

⇒     \Delta x_{1}   = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)

6 0
2 years ago
All please..............
rodikova [14]
4 water has a low heat capacity and a high vaporization temperate and coasts have low temps
3 0
3 years ago
14. If the spring constant of a simple harmonic oscillator is doubled, by what factor will the mass of the system need to change
klio [65]

Lets se

And

\\ \rm\Rrightarrow T=2\pi\sqrt{\dfrac{m}{k}}

\\ \rm\Rrightarrow \sqrt{k}T=2\pi\sqrt{m}

So

\\ \rm\Rrightarrow k\propto m

If spring constant is doubled mass must be doubled

8 0
2 years ago
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