PHYSICS COLLEGE

Two charged particles attract each other with a force of magnitude F acting on each. If the charge of one is doubled and the distance separating the particles is also doubled, the force acting on each of the two particles has magnitude (a) F/2,
(b) F/4,
(c) F,
(d) 2F,
(e) 4F,
(f) None of the above.

Answers

Answer 1

Answer:

F/2

Explanation:

In the first case, the two charges are Q1 and Q2 and the distance between them is r. K is the Coulomb's constant

Hence;

F= KQ1Q2/r^2 ------(1)

Where the charge on Q1 is doubled and the distance separating the charges is also doubled;

F= K2Q1 Q2/(2r)^2

F2= 2KQ1Q2/4r^2 ----(2)

F2= F/2

Comparing (1) and (2)

The magnitude of force acting on each of the two particles is;

F= F/2

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Question Part Points Submissions Used A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.30 m/s in 0.812 s. (a) What is the magnitude of the linear impulse experienced by a 62.0-kg passenger in the car during the time the car accelerates? kg · m/s (b) What is the magnitude of the average total force experienced by a 62.0-kg passenger in the car during the time the car accelerates? N

Answers

(a) 328.6 kg m/s

The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

$I=\Delta p = m\Delta v$

where

m = 62.0 kg is the mass of the passenger

$\Delta v$ is the change in velocity of the car (and the passenger), which is

$\Delta v = 5.30 m/s - 0 = 5.30 m/s$

So, the linear impulse experienced by the passenger is

$I=(62.0 kg)(5.30 m/s)=328.6 kg m/s$

(b) 404.7 N

The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

$I=F \Delta t$

where in this case

$I=328.6 kg m/s$ is the linear impulse

$\Delta t = 0.812 s$ is the time during which the force is applied

Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

$F=(I)/(\Delta t)=(328.6 kg m/s)/(0.812 s)=404.7 N$

The temperature within a thin plate with thermal conductivity of 10 W/m/K depends on position as given by the following expression: TT=(100 K)????????−xx2/????????xx 2cos�yy/????????yy�+300 K Where, Lx = 1 m, and Ly = 2 m. At the point (0.4 m, 1 m), find: a. The magnitude of the heat flux b. The direction of the heat flux

Answers

Answer:

Heat flux = (598.3î + 204.3j) W/m²

a) Magnitude of the heat flux = 632.22 W/m²

b) Direction of the heat flux = 18.85°

Explanation:

- The correct question is the first image attached to this solution.

- The solution to this question is contained on the second and third images attached to this solution respectively.

Hope this Helps!!!

We often refer to the electricity at a typical household outlet as being 120 V. In fact, the voltage of this AC source varies; the 120 V is __________. We often refer to the electricity at a typical household outlet as being 120 V. In fact, the voltage of this AC source varies; the 120 V is __________. the minimum value of the voltage the peak value of the voltage the average value of the voltage the rms value of the voltage

Answers

Explanation:

We often refer to the electricity at a typical household outlet as being 120 V. In fact, the voltage of this AC source varies; the 120 V is "the rms value of the voltage".

The rms value of voltage is given by :

$V_(rms)=(V_(pk))/(√(2))$

Where

$v_(pk)$ is the peak value of voltage

So, the correct option is (d). " rms value of voltage".

A train station bell gives off a fundamental tone of 505 Hz as the train approaches the station at a speed of 27.6 m/s. If the speed of sound in air on that day is 339 m/s, what will be the apparent frequency of the bell to an observer riding the train

Answers

Answer:

Apparent frequency of the bell to the observer is 546.12 Hz

Explanation:

The frequency of train bell (frequency of source) = 505 Hz

The speed of train (observer) = 27.6 m/s

The speed of sound in the air is (velocity of sound) = 339 m/s

The apparent frequency of the bell to the observer is calculated as follows:

Apparent frequency of bell to the observer.

$= \text{frequency of source} * (Observer + velocity \ of \ sound )/( velocity \ of \ sound ) \n= 505 * (27.6 + 339)/(339) \n= 546.12 Hz$

A 0.47 kg block of wood hangs from the ceiling by a string, and a 0.070kg wad of putty is thrown straight upward, striking the bottom of the block with a speed of 5.60 m/s. The wad of putty sticks to the block. (Answer on previous exams) How high does the putty-block system rise above the original position of the block Is the kinetic energy of the system conserved during the collision Is the mechanical energy of the system conserved during the collision Is the mechanical energy conserved after the collision

Answers

Answer:

The height is $h =0.0269 \ m$

The kinetic energy during collision is not conserved

The Mechanical energy during the collision is not conserved

The mechanical energy after the collision is not conserved

Explanation:

From the question we are told that

The mass of the block is $m_b = 0.47\ kg$

The mass of the wad of putty is $m_p = 0.070 \ kg$

The speed o the wad of putty is $v_p = 5.60 \ m/s$

The law of momentum conservation can be mathematically represented as

$p_i = p_f$

Where $p_i$ is the initial momentum which is mathematically represented as

$p_i =m_p * v_p$

While $p_f$ is the initial momentum which is mathematically represented as

$p_f = (m_b + m_p)v_f$

Where $v_f$ s the final velocity

$m_p v_p = (m_p + m_b) * v_f$

Making $v_f$ the subject

$v_f = (m_p v_p)/(m_b +m_p)$

substituting values

$v_f = ((0.070)*(5.60))/(0.47 + 0.070)$

$v_f = 0.726 \ m/s$

According to the law of energy conservation

$KE = PE$

Where KE is the kinetic energy of the system which is mathematically represented as

$KE = (1)/(2) (m_p + m_b)v_f^2$

And PE is the potential energy of the system which is mathematically represented as

$PE = (m_p +m_b) gh$

$(1)/(2) (m_p + m_b)v_f^2 = (m_p +m_b) gh$

Making h the subject of the formula

$h = (v_f^2)/(2g)$

substituting values

$h = ((0.726 )^2 )/(2 * 9.8)$

$h =0.0269 \ m$

Now the kinetic energy is conserved during collision because the system change it height during which implies some of the kinetic energy was converted to potential energy during collision

The the mechanical energy of the system during the collision is conserved because this energy consists of the kinetic and the potential energy.

Now after the collision the mechanical energy is not conserved because the external force like air resistance has reduced the mechanical energy of that system

A 13.0-Ω resistor, 13.5-mH inductor, and 50.0-µF capacitor are connected in series to a 55.0-V (rms) source having variable frequency. If the operating frequency is twice the resonance frequency, find the energy delivered to the circuit during one period.