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Harlamova29_29 [7]
2 years ago
12

Benjamin, a transport technician, was punched by a patient in the

Physics
1 answer:
Masja [62]2 years ago
6 0
Yes....................
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A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found
kati45 [8]

Answer:

1)  \rm q_2 is<u> positive.</u>

<u></u>

2) \rm q_2=4.56\times 10^{-10}\ C.

Explanation:

<h2><u>Part 1:</u></h2>

<u></u>

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., \rm q_2 is <u>positive.</u>

<u></u>

<h2><u>Part 2:</u></h2>

<u></u>

<u>Given:</u>

  • Mass of the balloon, m = 0.00275 kg.
  • Charge on the balloon, \rm q_1 = -3.50\times 10^{-8}\ C.
  • Distance between the rod and the balloon, d = 0.0640 m.
  • Acceleration due to gravity, \rm g = 9.81\ m/s^2.

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, \rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.

The magnitude of the electrostatic force on the balloon due to the rod is given by

\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.

\rm \dfrac{1}{4\pi \epsilon_o} is the Coulomb's constant.

For the elecric force and the weight to be balanced,

\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.

3 0
3 years ago
What are the applications of pascal's principle​
Murrr4er [49]

Explanation:

  • The applications are, hydraulic lift- to transmit equal pressure throughout a fluid.
  • Hydraulic jack- used in the braking system of cars.
  • use of a straw- to suck fluids, which goes because of air pressure.
<h3>The question simply asks, where pressure can be applied. There are many others, such as <em><u>l</u></em><em><u>i</u></em><em><u>f</u></em><em><u>t</u></em><em><u> </u></em><em><u>p</u></em><em><u>u</u></em><em><u>m</u></em><em><u>p</u></em><em><u>.</u></em></h3>
5 0
3 years ago
⚠Heeeeeelelepelepleplepelepeplepe⚠
snow_tiger [21]

Answer:

It's B, anything about a circle is Stationary

4 0
3 years ago
supose you have two wires of equal length made from same material. how is it possible for the wires to have different resistance
Ivenika [448]
I'm not sure but I had this question on a benchmark I think its the density of the wire  you need to  find the density or the mass I'm not sure but i do remember this question 
6 0
3 years ago
Water flows into a horizontal, cylindrical pipe at 1.4 m/s. the pipe then narrows until its diameter is halved. what is the pres
inna [77]

According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by

\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )

Here,  v_{1} is the velocity of water through wide ends of cylindrical pipe and v_{2} is the velocity of water through narrow ends of cylindrical pipe.

Given, v_{1} =1.4 m/s

Now from equation continuity,

v_{1} A_{1} = v_{2} A_{2}.

Here, A_{1} and A_{2} are cross- sectional areas of wide and narrow ends of cylindrical pipe.

As pipe is circular, so

v_{1} \pi r^2_{1} = v_{2} \pi r^2_{2}.

At the second point, the diameter is halved, which means the radius is also halved. Therefore,

v_{1} r^2_{1} = v_{2}(\frac{1}{2} r_{1})^2 \\\\ v_{2} = 4 v_{1}

v_{2} = 4 \times 1.4 = 5.6 m/s

Substituting these values  with the density of water is 1000 \ kg/m^3 in pressure difference formula we get.

\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )=\frac{1}{2}\times 1000 kg/m^3(5.6^2-1.4^2)\\\\ \Delta P = 14700\ Pa

3 0
3 years ago
Read 2 more answers
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