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MrMuchimi
3 years ago
6

A ball hits a wall. What is true about the magnitude of the force experienced by the ball compared with the force experienced by

the wall?
A. The ball experiences more force than the wall.
B. The ball experiences less force than the wall.
The ball and the wall experience the same force.
D. The ball experiences half the force of the wall.
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Physics
1 answer:
Natalka [10]3 years ago
6 0

Answer:

The ball and the wall experience the same force.

Explanation:

According to the third law of Newton, which states that "for every action, there is an equal and opposite reaction", this means that when an object 1 acts on object 2 with a certain force, object 2 also acts on object 1 with the same magnitude of force but in an opposite direction.

According to this question, a ball hits a wall with a certain force. This means that the wall will react on the ball with the same force magnitude, but in an opposite manner. Hence, the ball and the wall experience the same force.

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A trained eye in the dark for an extended period of time may pick up a light stimulus from a light source, at the lowest radiate
frez [133]

Answer:

    #_photons = 30 photons / s

Explanation:

Let's start by finding the energy of a photon of light, let's use the Planck relation

         E = h f

the speed of light is related to wavelength and frequency

         c = λ f

we substitute

         E = h c /λ

         E₀ = 6.63 10⁻³⁴ 3 10⁸/500 10⁻⁹

         E₀ = 3.978 10⁻¹⁹ J

now let's use a direct proportion rule. If the energy of a photon is Eo, how many fornes has an energy E = 1.2 10⁻¹⁷ J in a second

          #_photons = 1 photon   (E / Eo)

          #_photons = 1  1.2 10⁻¹⁷ /3.978 10⁻¹⁹

          #_photons = 3.0 10¹

          #_photons = 30 photons / s

3 0
2 years ago
______is the rate of change in velocity.
shepuryov [24]

Answer:

Acceleration

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3 years ago
It is known that a shark can travel at a speed of 17 m/s. How far can a shark go in 8 seconds? (please show steps to how you got
Alex777 [14]

Answer:

136 meters.

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4 0
3 years ago
Consider a simple tension member that carries an axial load of P=22.44N. Find the total elongation in the member due to the load
rodikova [14]

Answer:

The total elongation for the tension member is of 0.25mm

Explanation:

Assuming that material is under a linear deformation then the relation between the stress and the specific elongation is given as:

\sigma=E*\epsilon (1)

Where E is the modulus of elasticity, σ the stress and ε the specific deformation. Also, the total longitudinal elongation can be expressed as:

\delta L=L*\epsilon (2)

Here L is the member extension and δL the change total longitudinal elongation.  

Now if the stress is found then the deformation can be calculated by solving the stress-deformation equation (1). The stress applied sigama is computed dividing the axial load P by the cross-sectional area A:

\sigma=P/A  

\sigma=22.44N / 1290 mm^2  

\sigma=0.0174 N/mm^2  

Solving for epsilon and replacing the calculated value for the stress and the value for the modulus of elasticity:

=\sigma=E*\epsilon

\epsilon=\sigma/E

\epsilon=0.0174 \frac{N}{mm^2}/\ 204 \frac{N}{mm^2}

\epsilon=8.53*10^-{5}

Finally introducing the specific deformation and the longitudinal extension in the equation of total elongation (2):

\delta L=3048 mm * 8.53*10^{-5}  

\delta L= 0.25 mm

4 0
2 years ago
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