A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.(a) Find the electric force on charge q2.
F12 = ? mN
(b) Find the electric force on q1.
F21 = ? mN
(c) What would your answers for Parts (a) and (b) differ if q2 were -6.0 µC?

Answers

Answer 1

Answer:

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

$F = (kq_1q_2)/(d^2)$

Here

k = Coulomb's Constant

$q_(1,2) =$ Charge of each object

d = Distance

Our values are given as,

$q_1 = 1 \mu C$

$q_2 = 6 \mu C$

d = 1 m

$k = 9*10^9 Nm^2/C^2$

a) The electric force on charge $q_2$ is

$F_(12) = ( (9*10^9 Nm^2/C^2)(1*10^(-6) C)(6*10^(-6) C))/((1 m)^2)$

$F_(12) = 54 mN$

Force is positive i.e. repulsive

b) As the force exerted on $q_2$ will be equal to that act on $q_1$ ,

$F_(21) = F_(12)$

$F_(21) = 54 mN$

Force is positive i.e. repulsive

c) If $q_2 = -6 \mu C$ , a negative sign will be introduced into the expression above i.e.

$F_(12) = ((9*10^9 Nm^2/C^2)(1*10^(-6) C)(-6*10^(-6) C))/((1 m)^(2))$

$F_(12) = F_(21) = -54 mN$

Force is negative i.e. attractive

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Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate from 2.00 to 6.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him.

Answers

Answer:

69.68 N

Explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = $(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

W = $F_(total) .d$

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = $(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

$F_(total) .d =(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

$F_(total) = ((1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i))/(d)$

$F_(total=) ((1)/(2) X 62 X6^(2) -(1)/(2) X 62 X2^(2) )/(25)$

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

$F_(sprinter) = F_(total) + F_(wind) = 39.7 + 30 = 69.68 N$

Answer:

Force exerted by sprinter = 69.68 N

Explanation:

From work energy theorem, we know that, work done is equal to change in kinetic energy.

Thus,

W = ΔK = Kf - Ki = (1/2)m•(v_f)² - (1/2)m•(v_i)² - - - - eq(1)

Now,

Work done is also;

W = Force x Distance = F•d - - - (2)

From the question, we are given ;

v_f = 6 m/s

v_i = 2 m/s

d = 25m

m = 62 kg

Equating equation 1 and 2,we get;

(1/2)m•(v_f)² - (1/2)m•(v_i)² = F•d

Plugging in the relevant values to obtain ;

(1/2)(62)[(6)² - (2)²] = F x 25

31(36 - 4) = 25F

992 = 25F

F = 39.68 N

The force the sprinter exerts backward on the track will be the sum of this force and the headwind force.

Thus,

Force of sprinter = 39.68 + 30 = 69.68N

A screen is placed 60.0 cm from a single slit, which is illuminated with light of wavelength 690 nm. If the distance between the first and third minima in the diffraction pattern is 3.10 mm, what is the width of the slit?

Answers

Answer:

slit width, b = 0.2671 mm

Given:

distance of screen from the slit, x = 60.0 cm

wavelength of light, $\lambda = 690 nm = 690* 10^(- 9) m$

distance between 1st and 3rd minima, t = 3.10 mm = $3.10* 10^(-3) m$

Solution:

Calculation of the distance between 1st and 3rd minima:

$t = ((3 - 1)\lambda x)/(b)$

$3.10* 10^(- 3) = (2* 690* 10^(-9)* 60.0* 10^(-2))/(b)$

b = 0.2671 mm

slit width, b = 0.2671 mm

A pitcher is in 85° of abduction, holding a 1.4 N baseball at point C, 65 cm from the joint axis at point O • The center of gravity of his arm is 25 cm from the joint axis of shoulder abduction at point O • The weight of the arm W is 0.06 of the pitcher’s weight of 100 N • Deltoids muscles are at an angle θ of 15° with respect to the humerus and insert 15 cm from the joint axis at point A • Determine the force applied by the Deltoid muscles and the joint reaction force at the shoulder joint and its orientation β

Answers

I attached a Diagram for this problem.

We star considering the system is in equlibrium, so

Fm makes $90-(\theta+5)$ with vertical

Fm makes 70 with vertical

Applying summatory in X we have,

$\sum F_x = 0$

$W+1.4-Fm cos(70)$

We know that W is equal to

$W= 0.06*100N = 6N$

Substituting,

$Fm cos (70) = W+1.4N$

$Fm cos (70) = 6N + 1.4N$

$Fm = (7.4)/(cos(70))$

$Fm = 21.636N$

For the second part we know that the reaction force Fj on deltoid Muscle is equal to Fm, We can assume also that $\beta = \theta$

) A circular coil of diameter 20. cm, with 16. turns is in a 0.13 Tesla field. (a) Find the total flux through the coil when the field is perpendicular to the coil plane. (b) If the coil is rotated in 10. ms so its plane is parallel to the field, find the average induced emf.

Answers

Answer:

(a) 0.0041 weber

(b) 0.41 volt

Explanation:

diameter of coil, d = 20 cm

radius of coil, r = half of diameter = 10 cm = 0.1 m

magnetic field strength, B = 0.13 tesla

(a)

The angle between the normal of the coil and the magnetic field is 0°.

Magnetic flux, Ф = B x A x Cos 0°

Ф = 0.13 x 3.14 x 0.1 x 0.1 x 1

Ф = 0.0041 Weber

(b)

angle between the magnetic field and the normal of the coil is 90°.

time, t = 10 ms = 0.01 s

final flux = B x A x cos 90° = 0

induced emf = rate of change of magnetic flux

e = (0.0041 - 0) / 0.01

e = 0.41 Volt

Answer:

a) $\phi=0.4084\ T.m^2$

b) $emf=653.44\ V$

Explanation:

Given:

diameter of the coil, $d=20\ cm=0.2\ m$

no. of turns in the coil, $N=16$

magnetic field strength to which the coil is subjected, $B=0.13\ T$

time taken by the coil to rotate from normal the field to parallel, $t=10* 10^(-3)\ s$

The flux through the coil can be given as:

$\phi=BA$

where:

$A=$ area enclosed by the section of the coil

$\phi=0.13* \pi* (0.2^2)/(4)$

$\phi=0.4084\ T.m^2$

When the coil is rotated there is change in flux which lead to an induced emf in the coil according to the Faraday's law:

$emf=N(d\phi)/(t)$

where:

$d\phi=$ change in the flux

here the flux changes from maximum value to zero when the coil becomes parallel to the field lines because then there is no field line intercepting the coil area.

$emf=16* (0.4084)/(0.01)$

$emf=653.44\ V$

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is = 2.93 × 109 W/m2. What is the rms value of (a) the electric field and (b) the magnetic field in the electromagnetic wave emitted by the laser?

Answers

A. The rms value of electric field be "1.05 × 10⁶ N/C".

B. The rms value of magnetic field will be "3.5 × 10⁻³ T".

Magnetic and Electric field

According to the question,

Intensity of the wave, S = 2.93 × 10⁹ W/m²

Free space permittivity, $\epsilon_0$ = 8.86 × 10⁻¹²

Speed of light, c = 3 × 10⁸

A. We know that,

The rms value of electric field,

→ $E_(rms)$ = $\sqrt{(S)/(\epsilon_0 c) }$

By substituting the values,

= $\sqrt{(2.93* 10^9)/((8.85* 10^(-12))(3* 10^8)) }$

= 1.05 × 10⁶ N/C

and,

B. We know that,

The rms value of magnetic field,

→ $B_(rms)$ = $(E_(rms))/(c)$

By substituting the values,

= $(1.05* 10^6)/(3* 10^8)$

= 3.5 × 10⁻³ T

Thus the above response is appropriate.

Find out more information about magnetic field here:

brainly.com/question/25801845

To solve this problem, it is necessary to apply the concepts related to the electric field according to the intensity of the wave, the permittivity constant in free space and the speed of light.

As well as the expression of the rms of the magnetic field as a function of the electric field and the speed of light.

PART A) The expression for the rms of electric field is

$E_(rms) = \sqrt{(S)/(\epsilon_0 c)}$

Where,

S= Intensity of the wave

$\epsilon_0$ = Permitivitty at free space

c = Light speed

Replacing we have that,

$E_(rms) = \sqrt{((2.93*10^9))/((8.85*10^(-12))(3*10^8))}$

$E_(rms) = 1.05*10^6N/C$

The RMS value of electric field is $1.05*10^6N/C$

PART B) The expression for the RMS of magnetic field is,

$B_(rms) = (E_(rms))/(c)\nB_(rms) = (1.05*10^6)/(3*10^8)\nB_(rms) =3.5*10^(-3)T$

The RMS of the magnetic field is $3.5*10^(-3)T$

A power P is required to do work W in a time interval T. What power is required to do work 3W in a time interval 5T? (a) 3P (b) 5P (c) 3P/5 (a) P (e) 5P/3