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strojnjashka [21]

3 years ago

15

A charge Q is transferred from an initially uncharged plastic ball to an identical ball 24 cm away.The force of attraction is th

en 17 mN. How many electrons were transferred from one ball to the other?

1 answer:

xxMikexx [17]3 years ago

4 0

Answer:

The number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons

Explanation:

Given;

magnitude of the attractive force, F = 17 mN = 0.017 N

distance between the two objects, r = 24 cm = 0.24 m

The attractive force is given by Coulomb's law;

$F = \frac{1}{4\pi \epsilon _0} \times \frac{Q^2}{r^2} = \frac{kQ^2}{r^2} \\\\Q^2 = \frac{Fr^2}{k} \\\\Q = \sqrt{ \frac{Fr^2}{k}} \\\\Q = \sqrt{ \frac{(0.017)(0.24)^2}{9\times 10^9}} \\\\Q = 3.298 \times 10^{-7} \ C$

The charge of 1 electron = 1.602 x 10⁻¹⁹ C

n(1.602 x 10⁻¹⁹ C) = 3.298 x 10⁻⁷

$n = \frac{3.298 \times 10^{-7}}{1.602 \times 10^{-19}} = 2.06 \times 10^{12} \ electrons$

Therefore, the number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons

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White raven [17]

In this case, the PWC will head for the dock. This is because, power needs to be applied in order to maintain steering control. If the throttle is left to idle or if the engine is turn off during operation, all steering control will be lost. In each of this situation, the PWC will continue in the was headed before the engine was turned off or before the throttle was left to idle.

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4 years ago

a teacher drives from school to the mall. then she goes to the doctors office for a physical examination. what is the magnitude

Amanda [17]

The magnitude of the teachers displacement is the difference between

the location of the school and the location of the doctors office

The reason behind the above expression are explained as follows

The known locations of the teacher;

Initial location; The school

Second location; The mall

The eventual location; The doctors office

The required information;

The magnitude of the teachers displacement

The method of obtaining the displacement of the teacher;

Qualitatively, the displacement of the teacher can be obtained by applying the definition of displacement as follows;

The definition of displacement is an object's change in position, given by the difference between its start and finish locations

According to the definition of displacement, therefore, the magnitude of

the teachers displacement is the distance between the school and the

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The magnitude of the displacement of the teacher = The coordinate of the location of the doctors office - The coordinate of the location of the school

Learn more about displacement here;

brainly.com/question/17587058

5 0

3 years ago

Do air bags increase pressure on the body during an accident

victus00 [196]

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7 0

3 years ago

A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com

bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature $=20^{\circ}C$

Final Temperature $=80^{\circ}C$

mass flow rate of cold fluid $\dot{m_c}=1.2 kg/s$

Initial Geothermal water temperature $T_h_i=160^{\circ}C$

Let final Temperature be T

mass flow rate of geothermal water $\dot{m_h}=2 kg/s$

diameter of inner wall $d_i=1.5 cm$

$U_{overall}=640 W/m^2K$

specific heat of water $c=4.18 kJ/kg-K$

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

$\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)$

$2\times (160-T)=1.2\times (80-20)$

$160-T=36$

$T=124^{\circ}C$

As heat exchanger is counter flow therefore

$\Delta T_1=160-80=80^{\circ}C$

$\Delta T_2=124-20=104^{\circ}C$

$LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}$

$LMTD=\frac{80-104}{\ln \frac{80}{104}}$

$LMTD=91.49^{\circ}C$

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

$\dot{m_c}c(80-20)=U\cdot A\cdot$ (LMTD)

$A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2$

$A=\pi DL=5.144$

$L=\frac{5.144}{\pi \times 0.015}$

$L=109.16 m$

6 0

3 years ago

A brick is dropped from a high scaffold. <br> a. How far does the brick fall during this time?

Alecsey [184]

Answer:

a: after 1 seconds it will have fallen 0.2452

after 2 seconds it will have fallen 0.981

after 3 seconds it will have fallen 2.2072

after 4 seconds it will have fallen 3.924

Explanation:

the formula for acceleration due to gravity is (ignoring friction I think)

g = G*M/R^2

earths gravitational constant is about 9.807

g = 9.807*M/R^2

The average weight of a brick is 5 pounds and I'm going to say it's 10 feet off the ground.

g = 9.807*5/10^2. g = 0.4905 so every second the brick will go 0.4905 fps faster. (fps means feet per second.)

after 1 seconds it will have fallen 0.2452

after 2 seconds it will have fallen 0.981

after 3 seconds it will have fallen 2.2072

after 4 seconds it will have fallen 3.924

6 0

3 years ago