A positively charged rod is held near a neutral conducting sphere as illustrated below. A positively charged particle is moved from point A to point B. Thee electrostatic work done on the positively charged particle during the motio

Answers

Answer 1

Answer:

The movement of a positively charged particle from point A to point B. the motion-induced electrostatic work done on the positively charged particle.

Whether positively or negatively charged, an object that is neutral will interact with it in a pleasing way. Both positively charged and neutral items attract one another, as do negatively charged and neutral objects. These electrons gather on the further surface of sphere B, depleting the electron supply in sphere A. Therefore, sphere A (which is closer to the rod) obtains a positive charge and sphere B acquires a negative charge when the two spheres separate in the presence of the rod. The change in the particle's electrostatic potential energy in the external field equals the work done by the external force. When a charge is pushed from point A to point B, its potential energy changes, representing the labor of an outside force.

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What would the position of arrows on a target need to be to illustrate measurements that are neither accurate nor precise

Answers

Answer:

The position of the arrows will not be on the target i.e. outside the bull's eye, neither will they be close to one another (widely scattered).

Explanation:

Accuracy refers to the closeness of a measurement to an actual or accepted value while precision refers to the closeness of measurements to one another.

Using archery as an illustration of precision and accuracy, measurements (arrows) that are neither accurate not precise are those arrows that will be far away or outside the bull's eye region (target) of the board and also far apart from one another.

In a nutshell, the arrows will be distant from the bull's eye or target (not accurate) and also distant from one another (not precise).

Two narrow slits separated by 1.5 mm are illuminated by 514 nm light. Find the distance between adjacent bright fringes on a screen 5.0 m from the slits. Express your answer in meters using two significant figures.

Answers

The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M

Given data :

separation between slits ( d ) = 1.5 x 10⁻³ m

wavelength of light ( λ ) = 514 * 10⁻⁹ m

Distance from narrow slit ( D ) = 5.0 m

Determine the distance between the adjacent bright fringes

we apply the formula below

w = D * λ / d ---- ( 1 )

where : w = distance between adjacent bright fringes

Back to equation ( 1 )

w = ( 5 * 514 * 10⁻⁹ ) / 1.5 x 10⁻³

= 1.7 * 10⁻³ M

Hence we can conclude that The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M

Learn more about bright fringes calculations : brainly.com/question/4449144

Answer:

$1.7* 10^(-3)$ m

Explanation:

d = separation between the two narrow slits = 1.5 mm = 1.5 x 10⁻³ m

λ = wavelength of the light = 514 nm = 514 x 10⁻⁹ m

D = Distance of the screen from the narrow slits = 5.0 m

w = Distance between the adjacent bright fringes on the screen

Distance between the adjacent bright fringes on the screen is given as

$w = (D\lambda )/(d)$

$w = ((5.0)(514* 10^(-9)) )/(1.5* 10^(-3))$

$w = 1.7* 10^(-3)$ m

A rigid tank initially contains 3kg of carbon dioxide (CO2) at a pressure of 3bar.The tank is connected by a valve to a frictionless piston-cylinder assembly located vertically above, initially containing 0.5 m^3 of CO2. The piston area is 0.1 m^2. Initially the pressure of the CO2 in the piston-cylinder assembly is 2 bar. The ambient pressure and temperature are 1 bar and 290 K. Although the valveis closed, a small leak allows CO2 to flow slowly into the cylinder from the tank. Owing to heat transfer, the temperature of the CO2 throughout the tank and the piston-cylinder assembly stays constant at 290K. You can assume ideal gas behavior for CO2.Determine the following:

a. The total amount of energy transfer by work (kJ)
b. The total amount of energy transfer by heat (kJ)

Answers

Answer:

Part a: The total amount of energy transfer by the work done is 54.81 kJ.

Part b: The total amount of energy transfer by the heat is 54.81 kJ

Explanation:

Mass of Carbon Dioxide is given as m1=3 kg

Pressure is given as P1=3 bar =300 kPA

Volume is given as V1=0.5 m^3

Pressure in tank 2 is given as P2=2 bar=200 kPa

T=290 K

Now the Molecular weight of $CO_2$ is given as

M=44 kg/kmol

the gas constant is given as

$R=\frac{\bar{R}}{M}\nR=(8.314)/(44)\nR=0.189 kJ/kg.K$

Volume of the tank is given as

$V=(mRT)/(P_1)\nV=(3 * 0.189 * 290)/(300 )\nV=0.5481 m^3$

Final mass is given as

$m_2=(P_2V)/(RT)\nm_2=(200* 0.5481)/(0.189* 290)\nm_2=2 kg$

Mass of the CO2 moved to the cylinder

$m=m_1-m_3\nm=3-2=1 kg$

The initial mass in the cylinder is given as

$m_((cyl)_1)=(P_((cyl)_1)V_1)/(RT)\nm_((cyl)_1)=(200* 0.5)/(0.189 * 290)\nm_((cyl)_1)=1.82 kg$

The mass after the process is

$m_((cyl)_2)=m_((cyl)_1)+m\nm_((cyl)_2)=1.82+1\nm_((cyl)_2)=2.82\n$

Now the volume 2 of the cylinder is given as

$V_((cyl)_2)=(m_((cyl)_2)RT)/(P_2)\nm_((cyl)_2)=(2.82* 0.189* 290)/(200)\nm_((cyl)_1)=0.774 m^3$

Part a:

So the Work done is given as

$W=P(V_2-V_1)\nW=200(0.774-0.5)\nW=54.81 kJ$

The total amount of energy transfer by the work done is 54.81 kJ.

Part b:

The total energy transfer by heat is given as

$Q=\Delta U+W\nQ=0+W\nQ=54.81 kJ$

As the temperature is constant thus change in internal energy is 0.

The total amount of energy transfer by the heat is 54.81 kJ

Suppose a fast-pitch softball player does a windmill pitch, moving her hand through a circular arc with her arm straight. She releases the ball at a speed of 25.5 m/s (about 57.0 mph ). Just before the ball leaves her hand, the ball's radial acceleration is 1060 m/s2 . What is the length of her arm from the pivot point at her shoulder

Answers

Answer:

61.3 cm

Explanation:

Radial acceleration of the object in circular motion is given by formula

$a = (v^2)/(R)\n$

Given:

$a = 1060 m/s^2\nv = 25.5 m/s$

Plugging in the values in the formula

$1060 = (25.5^2)/(R)\nR = 0.613 m$

so length of his arm is 61.3 cm

A quarterback claims that he can throw the football a horizontal distance of 167 m. Furthermore, he claims that he can do this by launching the ball at the relatively low angle of 33.1 ° above the horizontal. To evaluate this claim, determine the speed with which this quarterback must throw the ball. Assume that the ball is launched and caught at the same vertical level and that air resistance can be ignored. For comparison a baseball pitcher who can accurately throw a fastball at 45 m/s (100 mph) would be considered exceptional.

Answers

Answer:u=42.29 m/s

Explanation:

Given

Horizontal distance=167 m

launch angle $=33.1^(\circ)$

Let u be the initial speed of ball

Range $=(u^2\sin 2\theta )/(g)$

$167=(u^2\sin (66.2))/(9.8)$

$u^2=1788.71$

$u=√(1788.71)$

$u=42.29 m/s$

A baseball player standing on a platform throws a baseball out over a level playing field. The ball is released from a point 3.50 m above the field with an initial speed of 14.3 m/s at an upward angle of 27.0 degrees above the horizontal. What horizontal distance will the ball travel before hitting the ground?