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cestrela7 [59]
3 years ago
9

Two point charges of equal magnitude (and opposite sign) are 7.5 cm apart. At the midpoint of the line connecting them, their co

mbined electric field has a magnitude of 45 N/C. What is the magnitude of each charge?
Physics
1 answer:
Shkiper50 [21]3 years ago
7 0

Comment

The only reason you can do this is that the charges are the same. If they were not, the problem would not be possible.

Equation

The field equation is, in its simplest form,

E = kq/r^2

So each of the charges are pulling / pushing in the same direction. The equation becomes.

kq/r^2 - (-kq/r^2) = Field magnitude in N/C

Givens

  • K = 9 * 10^9 N m^2 / c^2
  • E = 45 N/C
  • r = 7.5/2 = 3.75 cm * ( 1 m / 100 cm) = 0.0375 m
  • Find Q

Solution

k*q/0.0375 ^2 - (-kq/0.0375^2) = 45 N/C           Combine

2*k*q / 0.0375^2 = 45 N/C                                  Divide by 2

kq /(0.0375^2) = 22.5 N/C                                   Multiply by 0.0375^2

kq = 22.5 * 0.0375 ^2                                           Find d^2

kq = 22.5 * 0.001406                                            Combine

kq = 0.03164 N/C * m^2                                        Divide by k

q = 0.03164 N * m^2 /C / 9*10^9 N m^2 / c^2

q = 2.84760 * 10 ^8 C

I've left the cancellation of the units for you. Notice that only 1 C is left and it is in the numerator as it should be.


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Answer:

Q_T=63313.5\ J

Explanation:

Given:

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  • initial temperature of steam vapour, T_v=100^{\circ}C
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  • mass of steam, m=25\ g
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<em>Assuming that no heat is lost in the surrounding.</em>

<u>We know:</u>

Q=m.c.\Delta T

<u>Now the total heat given by the steam to form water at the given conditions:</u>

Q_T=Q_{Lv}+Q_w ..............................(1)

where:

Q_{Lv}= latent heat given out by vapour to form water of 100°C

Q_w= heat given by water of 100°C to come at 34°C.

putting respective values in eq. (1)

Q_T=m(L+c.\Delta T)

Q_T=25(2256+4.19\times 66)

Q_T=63313.5\ J

is the heat transferred to the skin.

4 0
3 years ago
5. Three children are at work. One is pushing a chair, another is pushing the sofa and the third one is pushing an empty trolley
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Explanation:

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Which example best demonstrates how unbalanced forces change the speed of an object's motion?
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A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.
fgiga [73]

The vertical velocity of the projectile upon returning to its original is 17. 74 m/s

<h3>How to determine the vertical velocity</h3>

Using the formula:

Vertical velocity component , Vy = V * sin(α)

Where

V = initial velocity = 36. 6 m/s

α = angle of projectile = 29°

Substitute into the formula

Vy = 36. 6 * sin ( 29°)

Vy = 36. 6 * 0. 4848

Vy = 17. 74 m/s

Thus, the vertical velocity of the projectile upon returning to its original is 17. 74 m/s

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8 0
2 years ago
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
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