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makkiz [27]
2 years ago
11

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

roton and an electron are situated 971 nm from each other and you study the forces that the particles exert on each other. As expected, the predictions of Coulomb's law are well confirmed.
You find that the forces are attractive and the magnitude of each force is:______
Physics
1 answer:
Irina-Kira [14]2 years ago
4 0

Answer:

The magnitude of each force is 2.45 x 10⁻¹⁶ N

Explanation:

The charge of proton, +q = 1.603 x 10⁻¹⁹ C

The charge of electron, -q = 1.603 x 10⁻¹⁹ C

Distance between the two charges, r = 971 nm = 971 x 10⁻⁹ m

Apply Coulomb's law;

F = \frac{kq_1q_2}{r^2}

where;

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

q₁ and q₂ are the charges of proton and electron respectively

F is the magnitude of force between them

Substitute in the given values and solve for F

F = \frac{(8.99*10^9)(1.603*10^{-19})^2}{(971*10^{-9})^2} \\\\F = 2.45*10^{-16} \ N

Therefore, the magnitude of each force is 2.45 x 10⁻¹⁶ N

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2 years ago
A badger is running at a speed of 1 m/s. If the badger moves that was for 2600 seconds, how far will the badger travel?
katen-ka-za [31]

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3 years ago
9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the
mojhsa [17]

Answer:

Moment about SHOULDER  ∑ τ = 3.17 N / m,

Moment respect to ELBOW   Στ= 2.80 N m

Explanation:

For this exercise we can use Newton's second law relationships for rotational motion

         ∑ τ = I α

   

The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.

They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm

Moment about SHOULDER

          ∑ τ = I α

           I = I_forearm + I_sphere

the forearm can be approximated as a fixed bar at one end

            I_forearm = ⅓ m L²

the moment of inertia of the mass in the hand, let's approach as punctual

            I_mass = m L²

we substitute

           ∑ τ = (⅓ m L² + M L²) α

let's calculate

          ∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

           ∑ τ = 3.17 N / m

Moment with respect to ELBOW

In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow

         Στ = I α

         I = I_ forearm + I_mass

         I = ⅓ m (L-0.03)² + M (L-0.03)²

         

let's calculate

        i = ⅓ 2.3 0.47² + 0.5 0.47²

        I = 0.2798 Kg m²

        Στ = 0.2798 10

        Στ= 2.80 N m

3 0
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kiruha [24]

Answer:

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