A positive electric charge is moved at a constant speed between two locations in an electric field, with no work done by or against the field at any time during the motion. This situation can occur only if the

Answers

Answer 1

Answer:

The options are

A) charge is moved in the direction of the field

(B) charge is moved opposite to the direction of the field

(C) charge is moved perpendicular to an equipotential line

(D) charge is moved along an equipotential line

(E) electric field is uniform

The answer is (D) charge is moved along an equipotential line

For all the criteria in the question to be met it means that the motion is perpendicular to the field and the force is perpendicular to the displacement of the charge respectively. In this scenario,

Change in velocity is zero. Work can be calculated by multiplying charge and change in velocity which when calculated will give zero.


Related Questions

A truck on a straight road starts from rest, accelerating at 2.00m/s^2 until it reaches a speed of 20.0m/s. Then the truck travels for 20.0s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00s.
(a) How long is the truck in motion?
(b) What is the average velocity of the truck for the motion described?

Answers

Answer:

Explanation:

a )

Time to reach the speed of 20 m/s with an acceleration of 2 m/s² can be calculated as follows .

v = u + a t

20 = 0 + 2 t

t = 20 /2 = 10 s .

Total time = 10 s + 20 s + 5 s = 35 s .

b) Average velocity = Total distance travelled / total time

Distance travelled in first 10 s

S₁ = ut + 1/2 a t²

= 0 + .5 x 2 x 10²

= 100 m

Distance travelled in next 20 s

S₂= 20s x 20 m/s  = 400 m

Distance travelled in last 5 s .

deceleration in last 5 s

v = u + at

0 = 20 m/s + a x 5

a = - 4 m/s²

v² = u² - 2 a s

0 = (20 m/s)² - 2 x 4 m/s² x s

s = 50 m

S₃ = 50 m

Total distance = S₁ + S₂ + S₃

= 100 m + 400 m + 50 m

= 550 m .

Average velocity = 550 m / 35 s

= 15.71 m /s .

3. How can a generator that otherwise produces AC
current be modified to produce DC current?

Answers

The change requires a commutator. ... The brushes attached to each half of the commutator are arranged so that at the moment the direction of the current in the coil reverses, they slip from one half of the commutator to the other.

heat travel through vacuum by
a. conduction. b.convention
c. radiation. d. both a&b​

Answers

Answer:

Option C

Explanation:

C. Radiation.....

Answer:

heat travel through vacuum by radiation

What are two things that are different about this refugee crisis compared to other
refugee events in the past?

Answers

Answer:

The current humanitarian crisis is unprecedented with an appalling and unacceptable human cost. The number of

refugees is unparalleled in recent times. The diversity of nationalities, motives for migration and individual profiles

also creates a huge challenge for asylum systems and welcoming communities in main European destination

countries. Moreover, given the complexity of its main driving forces, there is unfortunately little hope that the

situation will improve significantly in the near future.

This issue of Migration Policy Debates looks at the most recent developments in the humanitarian migration crisis

and what makes this crisis different from previous ones.

_______________________________________________________

Choose the smallest item from the list below.
1 glass of water
1 droplet of water
1 atom of oxygen
1 molecule of water

Answers

Answer

One molecule of water

One molecule of water

write the full form of MBBS pliz now​

Answers

Answer:

The full form of MBBS is Bachelor of Medicine, Bachelor of Surgery

pls mark me brainliest :))

The full form of MBBS in India is 'Bachelor of Medicine, Bachelor of Surgery'. However, MBBS is an abbreviation of Medicinae Baccalaureus Baccalaureus Chirurgiae, which is the term used for this course in Latin.

What is the difference between human caused Climate Change and natural
climate cycles?

Answers

Human caused climate change is climate change caused by human made creations such as cars, trucks, factories, and plastic in oceans these all are causing the ozone layer to deteriorate. And natural climate change cycles is any change occurring to the planet's climate either permanently or lasting for long periods of time.

While running, a person dissipates about 0.6 J of mechanical energy per step per kilogram of body mass. If a 60-kg person runs with a power of 70 Watts during a race, how fast is the person running

Answers

Complete question is: While running, a person dissipates about 0.60 J of mechanical energy per step per kilogram of body mass. If a 60-kg person develops a power of 70 W during a race, how fast is the person running? (Assume a running step is 1.5 m long).

Answer: The person running at a speed of 2.91 m/s.

Explanation:

Given: Mass of runner = 60 kg

Runner dissipates = 0.6 J/kg per step

Average power = 70 W

1 step = 1.50 m

Energy dissipated by the runner is as follows.

[tex]\Delta E_{step} = 0.60 \times 60\\= 36 J[/tex]

Formula used to calculate the value of one step 'S' is as follows.

[tex]\frac{S}{\Delta t} = \frac{P_{avg}}{\Delta E_{step}} = \frac{70}{36}\\= 1.94\\S = 1.94 \Delta t[/tex]

It is known that average velocity is equal to the total distance divided by time interval.

So, total distance for the given situation is as follows.

[tex]d = S \times 1.5[/tex]

Hence, speed of the person is calculated as follows.

[tex]v_{avg} = \frac{d}{\Delta t}\\= \frac{S \times 1.5}{\Delta t}\\= \frac{1.94 \Delta t \times 1.5}{\Delta t}\\= 2.91 m/s[/tex]

Thus, we can conclude that the person running at a speed of 2.91 m/s.

An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 43.8 N, the spring is stretched by 15.5 cm from its equilibrium length. Calculate the additional work required by F to stretch the spring by an additional 10.4 cm from that position. (in J)

Answers

A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant k is

43.8 N = k (0.155 m)   ==>   k = (43.8 N) / (0.155 m) ≈ 283 N/m

The total work done on the spring to stretch it to 15.5 cm from equilibrium is

1/2 (283 N/m) (0.155 m)² ≈ 3.39 J

The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be

1/2 (283 N/m) (0.259 m)² ≈ 9.48 J

Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.

An electric motor consumes 8.40 kJ of electrical energy in 1.00 min. Part A If one-third of this energy goes into heat and other forms of internal energy of the motor, with the rest going to the motor output, how much torque will this engine develop if you run it at 2900 rpmrpm

Answers

Answer:

The torque is 0.31 Nm.

Explanation:

Electrical energy, E = 8400 J

time, t = 1 min

Angular speed, w = 2900 rpm = 303.53 rad/s

efficiency = 2/3 of input power

The toque is given by  

[tex]P =\tau w\\\\\frac{2}{3}\times \frac{E}{t}=\tau w\\\\\frac{2}{3}\times \frac{8400}{60}=\tau \times 303.53\\\\\tau =0.31 Nm[/tex]

Hệ số giản nở nhiệt là gì?

Answers

Answer:

Hâ ÇÒ ÑÁ ÚIT LÃ

Explanation:

MÃÑJTÎWJWJ

A tennis player swings at a ball at a constant speed, taking 0.50 s to rotate her arms and racquet from horizontal to vertical. What acceleration is felt by a small bug at the tip of her racquet if it is 1.3 m from her shoulder

Answers

Answer:

the acceleration of the small bug is 12.83 m/s²

Explanation:

Given;

time of motion, t = 0.5 s

radius of the circular path created by his arm, r  = 1.3 m

if he rotates his arm from horizontal to vertical, the angular displacement = 90⁰

The centripetal acceleration of the ball is calculated as;

[tex]a_c = \omega^2 r\\\\a_c = (\frac{\theta}{t} )^2 r\\\\[/tex]

[tex]a_c = (\frac{90}{360} \times\frac{ 2\pi }{t} )^2r\\\\a_c = (\frac{\pi}{2t} )^2 r\\\\a_c = \frac{\pi^2r}{4t^2} = \frac{\pi^2 \times1.3 }{4\times 0.5^2} = 12.83 \ m/s^2[/tex]

Therefore, the acceleration of the small bug is 12.83 m/s²

At what frequency would an inductor of inductance 0.8H have a reactance of 12000^?​

Answers

Explanation:

The inductive reactance [tex]X_L[/tex] is given by

[tex]X_L = \omega L = 2 \pi fL[/tex]

Solving for f, we get

[tex]f = \dfrac{X_L}{2 \pi L} = \dfrac{12000\:\text{ ohms}}{2\pi (0.8\:H)}[/tex]

[tex]\:\:\:\:\:\:\:= 2387.3\:\text{Hz}[/tex]

Calculate how fast the ball would be moving at the instant it leaves the projectile launcher of the spring is compressed by 3.75 cm. Use a value of k = 500 N/m for the spring constant, 10 g for the mass of the ball, and 75 g for the effective mass of the ball holder. Show your work.

Answers

Answer:

V = 8.34m/s

Explanation:

Given that

1/2ke^2 = 1/2mv^2 ......1

Where e = 3.75cm = (3.75/100)m

e = 0.0375m

K = 500 N/m

m = 10g = 10/1000

= 0.01kg

Substitute the values into equation 1

0.5×500×(0.0375)^2 = 0.5×0.01×v^2

250×0.001395 = 0.005v^2

0.348 = 0.005v^2

v^2 = 0.348/0.005

v^2 = 69.6

V = √69.6

V = 8.34m/s

The ball launches at the speed of V = 8.34m/s

Q2 A source of frequency 500 Hz emits waves of
wavelength 0.2m. How long does it take the waves to
travel 400m?

Answers

Answer:

4 secs

Explanation:

The first step is to calculate the velocity

V= frequency × wavelength

= 500× 0.2

= 100

Therefore the time can be calculated as follows

= distance/velocity

= 400/100

= 4 secs

At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At what distance from the line will the field strength to be 2000 N/C?
a. D/√2.
b. √2D.
c. D/2.
d. 2D.
e. D/4.

Answers

Answer:

The correct option is (a).

Explanation:

We know that, the E is inversely proportional to the distance as follows :

[tex]E=\dfrac{k}{d^2}[/tex]

We can write it as follows :

[tex]\dfrac{E_1}{E_2}=(\dfrac{d_2}{d_1})^2[/tex]

Put all the values,

[tex]\dfrac{1000}{2000}=(\dfrac{d_2}{d})^2\\\\\sqrt{\dfrac{1000}{2000}}=(\dfrac{d_2}{D})\\\\0.7071=\dfrac{d_2}{d}\\\\d_1=0.7071D\\\\d_1=\dfrac{D}{\sqrt2}[/tex]

So, the correct option is (a).

You need to produce a set of cylindrical copper wire 3.5 m long that will have a

resistance of 0.125 Ω each. What will be the mass of each of these wires?

(ρ = 1.72X10-8 Ωm, density of copper = 8.9X103 kg/m3)​

Answers

Solution :

We know, resistance is given by :

[tex]R = \dfrac{\rho l}{A}[/tex]

[tex]A = \dfrac{\rho l }{R}\\\\A = \dfrac{1.72\times 10^{-8} \times 3.5 }{0.125}\\\\A = 4.816 \times 10^{-7} \ m^2[/tex]

Now, we know mass of wire is given by :

[tex]Mass = Density \times Volume\\\\\M = 8.9 \times 10^3 \times 4.816 \times 10^{-7} \times 3.5 kg\\\\M = 0.01500\ kg\\\\M = 15.00\ gram[/tex]

Hence, this is the required solution.

The mass of the wire will be "15.00 g".

Given:

Length of wire, l = 3.5 mResistance, R = 0.125 Ω

The resistance will be:

→ [tex]R = \frac{\rho l}{A}[/tex]

or,

→ [tex]A = \frac{\rho l}{R}[/tex]

By substituting the values, we get

      [tex]= \frac{1.72\times 10^{-8}\times 3.5}{0.125}[/tex]

      [tex]= 4.816\times 10^{-7} \ m^2[/tex]

hence,

The mass will be:

→ [tex]Mass = Density\times Volume[/tex]

             [tex]= 8.9\times 10^3\times 4.816\times 10^{-7}\times 3.5[/tex]

             [tex]= 0.01500 \ kg[/tex]

             [tex]= 15.00 \ g[/tex]

Thus the above answer is right.  

Learn more about mass here:

https://brainly.com/question/17108656

What does this circle graph tell you about water on Earth? (2 points)

a pie graph with a big blue section covering seventy one percent and small gray section covering twenty nine percent with a key indicating that blue is water and gray is land
Fresh water covers 71 percent of Earth's surface.
Oceans covers 71 percent of Earth's surface.
Salt water covers 71 percent of Earth's surface.
Water covers 71 percent of Earth's surface.

Answers

Answer:

ocean covers 71 percent of the earth

Answer:

the ocean  covers 71 percent of Earth's surface.

Explanation:

(A) State the relation between acceleration and momentum (10 marks).

(B) Two small steel balls A and B have mass 0.6 kg and 0.2 kg respectively. They are moving towards each other in opposite directions on a smooth horizontal table when they collide directly. Immediately before the collision, the speed of A is 8 m/s and the speed of B is 2 m/s. Immediately after the collision, the direction of motion of A is unchanged and the speed of B is twice the speed of A. Find : a. The speed of A immediately after the collision (10 marks), b. The magnitude of the impulse exerted on B in the collision (10 marks).

Answers

Answer:

momentum is directly proportional to acceleration

If v = 5.00 meters/second and makes an angle of 60° with the negative direction of the y–axis, calculate all possible values of vx.

Answers

Vx = + 4.33 m/s. Hope this helps

Question 24 of 33 Which of the following is an example of uniform circular motion? A. A car speeding up as it goes around a curve O B. A car slowing down as it goes around a curve 23 C. A car maintaining constant speed as it goes around a curve D. A car traveling along a straight road​

Answers

Answer:

Pluto revolves around sun with constant speed in a circular orbit, so it is an example of uniform circular motion.

I need help solving my homework problem , can somebody help me please . This class is physics two . My question is “ where the potential difference across ab is 49.5 V “ ?

Answers

Explanation:

a) [tex]Q_{Total} = C_{eq}V = (3.5\:\mu \text{F})(49.5V) = 1.73×10^{-4}\:\text{C}[/tex]

b) The individual capacitors in series circuit carry the same amount of charge as the [tex]Q_{Total}[/tex] so

[tex]Q_{10} = Q_{Total} = 1.73×10^{-4}\:\text{C}[/tex]

c) Likewise, [tex]C_9[/tex] will carry the same amount of charge as [tex]C_{10}[/tex]

An electric generator has an 18-cmcm-diameter, 120-turn coil that rotates at 60 HzHz in a uniform magnetic field that is perpendicular to the rotation axis. Part A What magnetic field strength is needed to generate a peak voltage of 330 VV

Answers

Explanation:

OK ok nosepo [tex]2825.55[/tex]

The magnetic field strength is needed to generate a peak voltage is 0.287 T.

The given parameters;

diameter of the generator, d = 18 cm = 0.18number of turn, N = 120 turnfrequency of the coil, f = 60 Hzmaximum voltage in the coil emf = 330 V;

The maximum voltage in the coil is calculated as follows;

[tex]E_{max} = NAB\omega \\\\[/tex]

where;

N is the number of turnsA is the area of the coilB is the magnetic field strengthω is angular speed

The magnetic field strength is needed to generate a peak voltage is calculated as;

[tex]B = \frac{E_{max} }{NA \omega } \\\\B = \frac{E_{max} }{N (\frac{\pi d^2}{4} ) \times 2\pi f}\\\\ B = \frac{330}{120 \times (\frac{\pi \times 0.18^2}{4} ) \times 2\pi \times 60} \\\\B = 0.287 \ T[/tex]

Thus, the magnetic field strength is needed to generate a peak voltage is 0.287 T.

Learn more here:https://brainly.com/question/22784792

The total power input to a pumped storage power station is 600 MW
The useful power output is 540 MW calculate the efficiency of this pumped storage power station.
Calculate how much power is wasted by the pumped storage power station.

Answers

Answer:

60MW wasted

Explanation:

600-540

=60MW

Un gas a una temperatura de 38°C, tienen un volumen de 21 Lts Litros. ¿Qué volumen tendrá si la temperatura sube a 67°C?

Answers

English Translation :

A gas at a temperature of 38 °C, have a volume of 21 Lts Liters. What volume will it have if the temperature rises to 67°C?

Solution :

Since, there is no information about pressure, let us assume it is constant.

So, by ideal gas equation at constant pressure :

[tex]V_1 T_2 = V_2 T_1[/tex]

Putting given volume and temperature ( in Kelvin ) in above equation, we get :

[tex]21 \times ( 67 + 273 ) = V_2 \times ( 38 + 273 )\\\\V_2 = \dfrac{21 \times (67+273) }{(38+273)}\\\\V_2 = 22.96 \ L[/tex]

Hence, this is the required solution.

An object of mass m 1 moving with speed v collides with another object of mass m 2 at rest and stick to it. Find the impulse to the second object.

Answers

Answer:

(m1+m2)vo=m1v+m2×0

⟹vo=m1+m2m1v

The impulse imparted to second object is equal to change is momentum is -

J=m2vo=m1+m2m2m1v

A satellite has a mass of 6463 kg and is in a circular orbit 4.82 × 105 m above the surface of a planet. The period of the orbit is 2.0 hours. The radius of the planet is 4.29 × 106 m. What would be the true weight of the satellite if it were at rest on the planet’s surface?

Answers

Answer:

The weight of the planet is 29083.5 N .

Explanation:

mass of satellite, m = 6463 kg

height of orbit, h = 4.82 x 10^5 m

period, T = 2 h

radius of planet, R = 4.29 x 10^6 m

Let the acceleration due to gravity at the planet is g.

[tex]T = 2\pi\sqrt\frac{(R+h)^3}{gR^2}\\\\2\times 3600 = 2\times3.14\sqrt\frac{(4.29+0.482)^3\times10^{18}}{g\times 4.29\times 4.29\times 10^{12} }\\\\24.2 g =108.67\\\\g = 4.5 m/s^2[/tex]

The weight of the satellite at the surface of the planet is

W = m g = 6463 x 4.5 = 29083.5 N

potential Energy is associated with position
True or false​

Answers

Answer:

yes.

Explanation:

potential Energy is associated with position

Answer:

true

Explanation:

the energy possessed by a body by virtue of its position or configuration is called potential energy.Everything at a height from the earth surface posseses potential energy.

A singly charged ion (q=−1.6×10−19) makes 7.0 rev in a 45 mT magnetic field in 1.29 ms. The mass of the ion in kg is

Answers

Answer:

[tex]m=1.47\times 10^{-24}\ Kg[/tex]

Explanation:

Given that,

Charge, [tex]q=1.6\times 10^{-19}\ C[/tex]

Revolution = 7 rev

magnetic field, B = 45 mT

Time, t = 1.29 ms

We need to find the mass of the ion. Let m be the mass. The formula for the mass in terms of time period is given by :

[tex]m=\dfrac{qBT}{2\pi}\\\\m=\dfrac{1.6\times 10^{-19}\times 45\times 10^{-3}\times 1.29\times 10^{-3}}{2\pi}\\\\m=1.47\times 10^{-24}\ Kg[/tex]

So, the mass of the ion is equal to [tex]1.47\times 10^{-24}\ Kg[/tex].

g to generate electricity, solar panels typically absorb visible light. How many photons of light with a frequency of 5045x10 14 hz does a solar panel absorb to create 360 kj

Answers

Answer:

the number of photons absorbed by the solar panel is 1.08 x 10²¹

Explanation:

Given;

frequency of each photon absorbed, f = 5045 x 10¹⁴ Hz

energy to be created by the solar panel, E = 360 kJ = 360,000 J

The energy of each photon absorbed is calculated as;

[tex]E_{photon} = hf\\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\E_{photon} = (6.626 \times 10^{-34} )(5045 \times 10^{14})\\\\E_{photon} = 3.343 \times 10^{-16} \ J[/tex]

let the number of photons absorbed = n

[tex]n(E_{photon}) = 360,000 \ J\\\\n = \frac{360,000 \ J}{3.343 \times 10^{-16} \ J} \\\\n = 1.08 \times 10^{21} \ photons[/tex]

Therefore, the number of photons absorbed by the solar panel is 1.08 x 10²¹

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