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frozen [14]
4 years ago
7

Please help me on 3,4 and 6. Thanks

Physics
1 answer:
rusak2 [61]4 years ago
3 0
3. <span>The second piston will experience the same force as compared with the first. This is because since the </span>pressure is the same everywhere inside the fluid system,<span> the force is proportional to the surface area. We are told that both the first and the second piston have the same surface area, therefore, they will both experience the same force/pressure.


4. </span>The situation is much the same as number 3 above, with the exception that the second piston is twenty times larger than the first. Again, since the pressure is the same everywhere inside the fluid system, the force is proportional to the surface area. We are told that the second piston is 20 times larger than the first, therefore, the larger piston will experience 20 times larger the force of the small one.


6. The answer is TRUE. The <span>hydraulic </span>braking system<span> of most cars makes use of a vacuum servo (or booster), which is located between the </span>brake pedal<span> and the master cylinder piston. </span><span>This vacuum servo amplifies the force applied </span><span>from the </span>brake pedal<span>.</span>

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slava [35]
Both mass and weight would change
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3 years ago
A particle of charge 3.53×10 ​−8 ​​ C experiences a force of magnitude 6.03×10 ​−6 ​​ N when it is placed in a particular point
Cloud [144]
<h2>Electric field at the location of the charge is 169.97 N/C</h2>

Explanation:

Electric field is the ratio of force and charge.

Force, F = 6 x 10⁻⁶ N

Charge, q = 3.53 x 10⁻⁸ C

We have           

       E=\frac{F}{q}\\\\E=\frac{6\times 10^{-6}}{3.53\times 10^{-8}}\\\\E=169.97N/C

Electric field at the location of the charge is 169.97 N/C

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3 years ago
A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC mo
Alla [95]

Answer:

W = 2.74 J

Explanation:

The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.

This is the electrostatic equivalent of the work-energy theorem.

W = \Delta U = U_2 - U_1

where the potential energy is defined as follows

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Let's first calculate the distance 'r' for both positions.

r_1 = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} = \sqrt{(0.170 - 0)^2 + (0 - 0)^2} = 0.170~m\\r_2 = \sqrt{(x_2 - x_0)^2 + (y_2 - y_0)^2} = \sqrt{(0.250 - 0)^2 + (0.250 - 0)^2} = 0.353~m

Now, we can calculate the potential energies for both positions.

U_1 = \frac{kq_1q_2}{r_1^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.170)^2} = -3.57~J\\U_2 = \frac{kq_1q_2}{r_2^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.3530)^2} = -0.829~J

Finally, the total work done on the moving particle can be calculated.

W = U_2 - U_1 = (-0.829) - (-3.57) = 2.74~J

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