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3 years ago

One charge is decreased to one-third of its original value,

Physics

2 answers:

Ann [662]3 years ago

3 0

Answer:

the Third option

Explanation:

Sveta_85 [38]3 years ago

3 0

Answer:

It will decrease to one-sixth the original force.

Explanation:

The electrical force is given by :

$F=k\dfrac{q_1q_2}{r^2}$ ..............(1)

Where

k is the electrostatic constant

r is the distance between charges

One charge is decreased to one-third of its original value, and a second charge is decreased to one-half of its original value.

i.e. $q_1'=\dfrac{q_1}{3}$

and $q_2'=\dfrac{q_2}{2}$

New force is given by :

$F'=k\dfrac{q_1q_2}{6r^2}$

$F'=\dfrac{F}{6}$

So, the new force will decrease to one-sixth the original force. Hence, the correct option is (c).

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Assuming that Bernoulli's equation applies, compute the volume of water ΔV that flows across the exit of the pipe in 1.00 s . In

OLEGan [10]

Answer:

discharge rate (Q) = 0.2005 m^{3} / s

Explanation:

if you read the question you would see that some requirements are missing, by using search engines, you can get the complete question as stated below:

Water flows steadily from an open tank as shown in the figure. (Figure 1) The elevation of point 1 is 10.0m , and the elevation of points 2 and 3 is 2.00 m . The cross-sectional area at point 2 is 4.80x10-2m ; at point 3, where the water is discharged, it is 1.60x10-2m. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe. Part A Assuming that Bernoulli's equation applies, compute the volume of water DeltaV that flows across the exit of the pipe in 1.00 s . In other words, find the discharge rate \Delta V/Delta t. Express your answer numerically in cubic meters per second.

solution:

time = 1 s

elevation of point 1 (z1) = 10 m

elevation of point 2 (z2) = 2 m

elevation of point 3 (z3) = 2 m

cross section area of point 2 = 4.8 x 10^{2} m

cross section area of point 3 = 1.6 x 10^{2} m

acceleration due to gravity (g) = 9.8 m/s^{2}

find the discharge rate at point 3 which is the exit pipe.

discharge rate (Q) = A3 x V3

where A3 is the cross sectional area at point 3 and V3 is the velocity of the fluid and can be gotten by applying Bernoulli's equation below

$\frac{P1}{ρg} + \frac{V1^{2} }{2g} + Z1 = \frac{P3}{ρg} + \frac{V3^{2} }{2g} + Z3$

pressure at point 1 (P1) is the same as pressure at point 3 (P3), and at point 1, the velocity (V1) = 0. therefore the equation now becomes

$\frac{P1}{ρg} + Z1 = \frac{P1}{ρg} + \frac{V3^{2} }{2g} + Z3$

$Z1 = \frac{V3^{2} }{2g} + Z3$

V3 = $\sqrt{2g(Z1-Z3)}$

V3 = $\sqrt{2 x 9.8 x (10 - 3)}$

V3 = 12.53 m/s

discharge rate (Q) = A3 x V3 = 1.6 x 10^{-2} x 12.53

discharge rate (Q) = 0.2005 m^{3} / s

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3 years ago

What is an atomic nucleus

Agata [3.3K]

Answer:

The atomic nucleus is the small, dense region consisting of protons and [[]]s at the center of an atom

Explanation:

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3 years ago

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A certain resistor dissipates 0.5 W when connected to a 3 V potential difference. When connected to a 1 V potential difference,

Stels [109]

Answer:

Explanation:

$Power = IV$

From ohms law we know that

$V= IR\\\\I= \frac{V}{R} \\\\Power= \frac{V}{R}*V\\\\Power= \frac{V^2}{R}$

Given data

P1 = 0.5 Watt

P2 = ?

V1= 3 Volts

V2= 1 Volt

Thus we can solve for the power dissipated as follows

$P1= \frac{V1^2}{R1}\\\\P2= \frac{V2^2}{R2}$

$\frac{P1}{P2} = \frac{V1^2}{V2^2}\\\\ P2=\frac{ V2^2}{ V1^2} *P1\\\\ P2=\frac{ 1^2}{ 3^2} *0.5= 0.055= 0.056 W$

<em>The resistor will dissipate 0.056 Watt</em>

7 0

3 years ago

A mass spectrometer was used in the discovery of the electron. In the velocity selector, the electric and magnetic fields are se

Mama L [17]

Answer:

Explanation:

Radius of dee, r = 8 mm = 0.008 m

Electric field, e = 400 V/m

Magnetic field, B = 4.7 x 10^-4 T

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

(a) Let v is the speed of electrons.

$v = \frac{Bqr}{m}$

$v = \frac{4.7\times 10^{-4}\times 1.6\times 10^{-19}\times 0.008}{9.1 \times 10^{-31}}$

v = 661098.9 = 661099 m/s

(b)

$\frac{e}{m}=\frac{1.6 \times 10^{-19}}{9.1\times 10^{-31}}$

e / m = 1.76 x 10^14 C / kg

K = 0.5 x mv²

K = 0.5 x 9.1 x 10^-31 x 661099 x 661099

K = 1.99 x 10^-19 J

K = 1.24 eV

So, the potential difference is

V = 1.24 V

(d) if the acceleration voltage is doubled

V = 2 x 1.24 = 2.48 V

So, Kinetic energy

K = 2.48 eV

K = 2.48 x 1.6 x 10^-19 = 3.968 x 10^-19 J

Let v is the speed

K = 0.5 x mv²

3.968 x 10^-19 = 0.5 x 9.1 x 10^-31 x v²

v = 933856.5 m/s

Let the new radius is r.

$r=\frac{mv}{Bq}$

$r=\frac{9.1\times 10^{-31}\times 933856.5}{4.7\times 10^{-4}\times 1.6\times 10^{-19}}$

r = 0.0113 m = 1.13 cm

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3 years ago

A speaker vibrates at a frequency of 200 hz. What is it's period ?

Katen [24]

200 Hz = 200 cycles per sec

<span>1 cycle, the period = 1/200 = 0.005 seconds, or 5 milli seconds.</span>

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3 years ago