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Leya [2.2K]
3 years ago
12

What would the electrostatic force be for two 0.005C charges 50m apart?

Physics
1 answer:
irakobra [83]3 years ago
3 0

Answer:

90 N

Explanation:

The electrostatic force between two charges is given by:

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

In this problem we have

q1 = q2 = 0.005 C

r = 50 m

So the electrostatic force is

F=(9\cdot 10^9 N m^2 C^{-2})\frac{(0.005 C)^2}{(50 m)^2}=90 N

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A solid cylinder of mass M = 45 kg, radius R = 0.44 m and uniform density is pivoted on a frictionless axle coaxial with its sym
user100 [1]

Answer:

w_f = 1.0345 rad/s

Explanation:

Given:

- The mass of the solid cylinder M = 45 kg

- Radius of the cylinder R = 0.44 m

- The mass of the particle m = 3.6 kg

- The initial speed of cylinder w_i = 0 rad/s

- The initial speed of particle V_pi = 3.3 m/s

- Mass moment of inertia of cylinder I_c = 0.5*M*R^2

- Mass moment of inertia of a particle around an axis I_p = mR^2

Find:

- What is the magnitude of its angular velocity after the collision?

Solution:

- Consider the mass and the cylinder as a system. We will apply the conservation of angular momentum on the system.

                                     L_i = L_f

- Initially, the particle is at edge at a distance R from center of cylinder axis with a velocity V_pi = 3.3 m/s contributing to the initial angular momentum of the system by:

                                    L_(p,i) = m*V_pi*R

                                    L_(p,i) = 3.6*3.3*0.44

                                    L_(p,i) = 5.2272 kgm^2 /s

- While the cylinder was initially stationary w_i = 0:

                                    L_(c,i) = I*w_i

                                    L_(c,i) = 0.5*M*R^2*0

                                    L_(c,i) = 0 kgm^2 /s

The initial momentum of the system is L_i:

                                    L_i = L_(p,i) + L_(c,i)

                                    L_i = 5.2272 + 0

                                    L_i = 5.2272 kg-m^2/s

- After, the particle attaches itself to the cylinder, the mass and its distribution around the axis has been disturbed - requires an equivalent Inertia for the entire one body I_equivalent. The final angular momentum of the particle is as follows:

                                   L_(p,f) = I_p*w_f

- Similarly, for the cylinder:

                                   L_(c,f) = I_c*w_f

- Note, the final angular velocity w_f are same for both particle and cylinder. Every particle on a singular incompressible (rigid) body rotates at the same angular velocity around a fixed axis.

                                  L_f = L_(p,f) + L_(c,f)

                                  L_f = I_p*w_f + I_c*w_f

                                  L_f = w_f*(I_p + I_c)

-Where, I_p + I_c is the new inertia for the entire body = I_equivalent that we discussed above. This could have been determined by the superposition principle as long as the axis of rotations are same for individual bodies or parallel axis theorem would have been applied for dissimilar axes.

                                  L_i = L_f

                                  5.2272 = w_f*(I_p + I_c)

                                  w_f =  5.2272/ R^2*(m + 0.5M)

Plug in values:

                                  w_f =  5.2272/ 0.44^2*(3.6 + 0.5*45)

                                  w_f =  5.2272/ 5.05296

                                  w_f = 1.0345 rad/s

5 0
3 years ago
Which is heavier, 1 m3 of steel or 1 m3 of aluminium?​
Strike441 [17]

Answer:

Steel is almost 2.9 times heavier the aluminium.

5 0
2 years ago
Can someone please help me?
svetlana [45]

Answer:

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7 0
3 years ago
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The time constant for RC circuit with the values of R1 and C1 is 5ms. What will be the time constant for a new RC circuit with t
lions [1.4K]

Answer:

d. 25 ms

Explanation:

  • In a RC circuit we call time constant to the product of the resistance times the capacitance, which represents the time when the charge reaches to the 63% of the final value, as follows:

       \tau_{1} = R_{1} *C_{1}  = 5 ms (1)

  • If we have a new circuit with new values for R and C, the time constant will be defined in the same way, as follows:

       \tau_{2} =10* R_{1} *0.5*C_{1}  = 5*(R_{1}* C_{1}) = 5* \tau_{1} = 5* 5 ms = 25 ms (2)

6 0
3 years ago
Could someone please help me with this problem? I’m thinking that they both will have the same force because they have the same
Simora [160]
You would be correct
7 0
4 years ago
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