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Novay_Z [31]
3 years ago
15

Two 1.9 kg masses are 1.1 m apart (center to center) on a frictionless table. Each has + 9.6 μC of charge.

Physics
1 answer:
Anni [7]3 years ago
5 0

Answer:

F= 0.6 N

Explanation:

Fe(electrical force)=k q1q2/r^2

    k=9*10^9\\q1=-9.6 *10^-6 C\\q2= -9.6*10^-6 C\\r= 1.1m

So,    

        F=\frac{9*10^9*-9.6 *10^-6 C* -9.6*10^-6 C}{1.1^2}

         F= 0.6 N

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A spring of force constant 285.0 N/m and unstretched length 0.230 m is stretched by two forces, pulling in opposite directions a
Kaylis [27]

Answer: W = 0.3853 J, e = 0.052 m

Explanation: Given that,

K =285.0N/M , L = 0.230m , F = 15N , e = ?

F = Ke

15 = 285 × e

e = 15÷ 285

e =0.052 m

e + L = 0.052 + 0.230

= 0.282m ( spring new length )

Work needed to stretch the spring

W = 1/2ke2

W = 1/2 × 285 x 0.052 × 0.052

W = 0.3853 J

8 0
3 years ago
Which statement(s) correctly compare the masses of protons, neutrons, and electrons? Check all that apply
umka21 [38]

__Protons and neutrons have similar mass.

__Electrons are smaller than a proton or a neutron.

Explanation:

The three particles involved in this problem are:

  • Proton: it is positively charge, it is found in the nucleus of the atom, and its mass is m_p = 1.67 \cdot 10^{-27} kg
  • Neutron: it has no electric charge, it is also found in the nucleus of the atom, and its mass is approximately equal to that of the proton (just slightly larger)
  • Electron: it has negative electric charge, it orbit around the nucleus of the atom, and its mass is much smaller than that of the proton: m_e = 9.11 \cdot 10^{-31} kg

We can now analyze each of the given statement:

__Protons and neutrons have similar mass.  --> TRUE

__Protons and electrons have similar mass.  --> FALSE, the electron is much lighter

__Neutrons and electrons have similar mass.  --> FALSE, the neutron is much heavier

__Protons are smaller than a neutron or an electron.  --> FALSE, protons are similar to the neutrons

__Neutrons are smaller than a proton or an electron.  --> FALSE, neutrons are similar to the protons

__Electrons are smaller than a proton or a neutron. --> TRUE

Learn more about atoms:

brainly.com/question/2757829

#LearnwithBrainly

3 0
3 years ago
Read 2 more answers
A planet has a period of revolution about the sun equal to T and a mean distance from the sun equal to R. T2 varies directly as
Tamiku [17]

Answer:

T² ∝ R³

Explanation:

Given data,

The period of revolution of the planet around the sun, T

The mean distance of the planet from the sun, R

According to the III law of Kepler, " Law of Periods' states that the square of the orbital period to go around the sun once is directly proportional to the cube of the mean distance between the sun and the planet.

                              T² ∝ R³

                               \frac{T^{2}}{R^{3}} = Constant

From the above equation it is clear that T² varies directly as the R³.

7 0
3 years ago
The mass of an atom is ________________________.
alexandr402 [8]
The mass of an atom comes from the protons and neutrons that is found in the nucleus. The number of protons is the atomic number of an element. To find the number of neutrons, subtract the atomic number from the mass of an atom. For example, sodium’s atomic number is 11. This will tell us that sodium has 11 protons in it. The atomic mass of sodium is 23. So subtract 23 form 11 gives us 12. Therefore, there are 12 neutrons in sodium.
7 0
3 years ago
The 1.0-kg collar slides freely on the fixed circular rod. Calculate the velocity v of the collar as it hits the stop at B if it
soldi70 [24.7K]

Answer:

6.21 m/s

Explanation:

Using work energy equation then

U_{1-2}=T_B- T_A\\58d-mgh=0.5m(v_b^{2}-v_a^{2})

where d is displacement from initial to final position, v is velocity and subscripts a and b are position A and B respectively, m is mass of collar, g is acceleration due to gravity

Substituting 1 Kg for m, 0.4m for h, v_a as 0, 9.81 for g then

58(\sqrt{0.4^{2}+0.3^{2}}-0.1)-(1\times 9.81\times 0.4)=0.5\times 1\times (v_b^{2}-v_a^{2})\\19.276=0.5\times 1v_b^{2}\\v_b=6.209025688 m/s\approx 6.21 m/s

7 0
3 years ago
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