What are the sign and magnitude in coulombs of a point charge that produces a potential of -3.50 V at a distance of 2.00 mm

While moving over a separate electric field line, however, has a non-zero value. If the test charge is transferred in a direction perpendicular to E, no work is done and the electric potential does not change. A smooth equipotential surface characterizes every such path.

Generally, The equation of the potential difference is mathematically given as

[tex]V = Ed cos \theta.[/tex]

Where

[tex]\theta=Angle[/tex]

E=electric field

d= the varing distance

When the charge to be tested is transferred via an electric field line. Since =0, the maximum possible velocity (Vmax) is calculated.

In conclusion, Since no work is done when the test charge is shifted in a direction perpendicular to E, the electric potential does not shift in this case. Any such route is a smooth equipotential surface.

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Answer:

final velocity=u+at 15+2×5=25

Answer:

The speed registered by the speedometer depends on the rotations of the axle to which it is connected -

Larger diameter tires will contribute to a larger distance traveled in one rotation of the axle and hence a larger speed and the  speed shown on the speedometer will be less than the actual speed.

Since acceleration is constant, the average and instantaneous accelerations are the same, so that

[tex]a = a_{\rm ave} = \dfrac{\Delta v}{\Delta t} = -\dfrac{v_i}{10.0\,\rm s}[/tex]

By the same token, we have the kinematic relation

[tex]v^2 - {v_i}^2 = 2a\Delta x[/tex]

where [tex]v[/tex] is final speed, [tex]v_i[/tex] is initial speed, [tex]a[/tex] is acceleration, and [tex]\Delta x[/tex] is displacement.

Substitute everything you know and solve for [tex]v_i[/tex] :

[tex]0^2 - {v_i}^2 = 2\left(-\dfrac{v_i}{10.0\,\rm s}\right)(75.0\,\mathrm m)[/tex]

[tex]\implies {v_i}^2 - \left(15.0\dfrac{\rm m}{\rm s}\right) v_i = 0[/tex]

[tex]\implies v_i \left(v_i - 15.0\dfrac{\rm m}{\rm s}\right) = 0[/tex]

[tex]\implies v_i = \boxed{15.0\dfrac{\rm m}{\rm s}}[/tex]

A

A 0 = –(45 kg • v) + (20 kg • 3 m/s)

Not moving, means momentum is zero. Thus, momentum after = momentum before (& that momentum is zero)

V = 4/3m/s

(-45 x 4/3) + (20x3) = 0

Solving to find v :

0. = –(45 kg • v) + (20 kg • 3 m/s)

= - (45 kg • v) + 60

-60

-60 = - 45 x v

÷ — 45

4/3 = v

(4/3 = 1.33333333333)

Hope this helps!

The velocity for the entire trip is 0.4 m/s as It takes her 500 seconds to make the round trip and 60 kg • m/s2 = (45 kg • v) + (20 kg • 3 m/s).

Path 1 = 400 m В B. A Path 2 = 200 m Path 3 = 300 m. Thus, option C is correct.

A particle's settling velocity known as the rate at which is travels through a still fluid. The specific gravity of the particles, their size, and their shape all have an impact on settling velocity.

A particle in still air will gravitationally settle and reach its terminal velocity fairly quickly. A particle's terminal velocity in a still fluid is referred to as the settling velocity (also known as the "sedimentation velocity").

Understanding variations in the hydraulic regime and interactions between sediment and fluid in the surf zone depends heavily on the particle settling velocity at the foreshore region. In contrast to sedimentation, which is the end product of the settling process, settling is the movement of suspended particles through the liquid.

Therefore, Thus, option C is correct.

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The will dog catch up with the rabbit in 6 minutes assuming both their velocities remain constant during the chase.

The time that the dog will catch up with the rabbit is given as follows:

Let the distance covered by the rabbit be x.

Distance covered by dog = x + 30

The time taken will be the same T

Equating both times.

(x + 30)/10 = x/5

x = 30 m

Solving for T in equation (ii);

T = 30/5 = 6 minutes

In conclusion, time is obtained as a ratio of distance and speed.

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Answer:

measured at constant temperature and pressure

Hope this helps :)

The frequency (in Hz) heard by an observer in the moving car as he approaches the police car is 552 Hz

The Doppler Effect equation says

f = f0 c+v / c

Given

f0 = 500 Hz

c = 343 m/s

v = 36 m/s

by plugging the given data, we get

500 (343+36) / 343  = 552 Hz

The frequency (in Hz) heard by an observer in the moving car as he approaches the police car is 552 Hz

Doppler effect is the apparent difference between the frequency at which sound or light waves leave a source and that at which they reach an observer, caused by the relative motion of the observer and the wave source.

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The difference in the mass of carbon dioxide in 500 kg of air in 2013 compared to 1800 is 0.06 Kg

Mass = percent × mass of air

Mass of CO₂ = 0.028% × 500

Mass of CO₂ = 0.14 Kg

Mass = percent × mass of air

Mass of CO₂ = 0.040% × 500

Mass of CO₂ = 0.2 Kg

Difference = mass in 2013 - mass in 1800

Difference = 0.2 - 0.14

Difference = 0.06 Kg

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Answer:

Point A

Explanation:

Because it reaches maximum height

(A) The angle of this in polar coordinates is 30⁰.

(B) The North/South and East/West components of its displacement is 250 m and 433 m respectively.

60° N of W lies in the fourth quadrant;

θ = 90 - 60 = 30⁰

in polar coordinate = (r,θ) = (500 m, 30⁰)

Dy = D sinθ

Dy = 500 m x sin(30) = 250 m

Dx = D cosθ

Dx = 500 m x cos(30) = 433 m

Thus, the angle of this in polar coordinates is 30⁰.

The North/South and East/West components of its displacement is 250 m and 433 m respectively.

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When the mass of Earth doubles, acceleration due to gravity doubles as well.

Apply Newton's second law of motion;

F = mg --- (1)

where;

F = GmM/R²  --- (2)

where;

Solve (1) and (2)

mg =  GmM/R²

g = GM/R²

when the mass of Earth doubles, acceleration due to gravity becomes;

g' = G(2M)/R²

g' = 2(GM/R²)

g' = 2g

Thus, when the mass of Earth doubles, acceleration due to gravity doubles as well.

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a) The motion along the vertical direction and the motion along the horizontal direction.

b) The object remains in the air for a time period of 2usin(θ)/g.

Any object that is thrown in the air when gravity is acting on it is called a projectile. The motion of this projectile is called projectile motion.

When the projectile is thrown in the air at some angle θ, then there are two independent motions taking place at the same time. First is the component of motion along the vertical direction along which gravity acts. Second is the component of motion along the horizontal direction along which the object moves with a constant velocity. No force acts along the horizontal direction. The horizontal motion does not affect the vertical motion and the converse is also true. So these are independent of each other.

The time of flight is the time during which a projectile remains in the air. This time of flight is calculated using the formula,

T=2usin(θ)/g

where T is the time of flight, u is the initial velocity and g is the acceleration due to gravity.

Hence, the object remains in the air for a time period of 2usin(θ)/g.

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The time taken for the mass to reach the bottom of the inclined plane after it is released from rest is 0.78 s.

The height of the inclined plane is calculated as follows;

h = L(sin 37)

where;

h = 5 x sin(37)

h = 3.01 m

t = √(2h/g)

t = √(2 x 3.01 / 9.8)

t = 0.78 s

Thus, the time taken for the mass to reach the bottom of the inclined plane after it is released from rest is 0.78 s.

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Answer:

Newton's corpuscular theory stated that light consisted of particles that travelled in straight lines. Huygens argued that if light were made of particles, when light beams crossed, the particles would collide and cancel each other. He proposed that light was a wave.

Class 1 (Explosives) is the class of hazards that is characterized by thermal and mechanical hazards in the form of blast pressure waves, shrapnel and fragmentation, and incendiary thermal effects.

There are different classes of Hazards

Class 1 - Explosives

Class 2 - Gases

Class 3 - Flammable liquids

Class 4 - Flammable solids

Class 5 - Oxidizers

Class 6 - Toxic materials

Class 7 - Radioactive materials

Class 8 - Corrosive materials

Class 9 - Miscellaneous dangerous goods

Any substance or item, including a gadget, that is intended to function by explosion or that, through a chemical reaction inside itself, is capable of functioning similarly even if not intended to function by explosion, falls within the category of explosive materials (class 1).

Hence, Class 1 (Explosives) is the class of hazards that is characterized by thermal and mechanical hazards.

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50 V to dissipate a power of 10 W

A resistor dissipates 2.00 W when the rms voltage of the emf is 10.0 V.

[tex]$\mathrm{V}_{r m s}=\mathrm{P}$[/tex]

10 v provides 2 W of power whereas V provides 10 W of power.

10 V = 2 W

x V=10 W

By using cross multiplication

2 x = 100

x = 50 V

So, 50 V to dissipate a power of 10 W

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The magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.

Given values:

Length of non-conducting rod, l = 1.20 m

Charge on positive rod, +Q = +2.50 μC = +2.50 × 10⁻⁶ C

Charge on negative rod, -Q = -2.50 μC = -2.50 × 10⁻⁶ C

Distance from point P of each rod, x = 60 cm = 0.60 m

Calculation of Net electric force exerted on point P:

Consider an electron released at point P, then the net electric force exerted will be given as:

F = e. E_net       - ( 1 )

Step 1:

The net electric field value is given as:

E_net  = E₁ cos Φ + E₂ cos Φ      

           = 2E₁ cos Φ                  -( 2 )

where, E₁ & E₂ are electric fields due to positive and negative rod                

            respectively.

            Φ is phase angle

Step 2:

The electric field due to positive rod is given as:

E₁ = k (λ/r)             - ( 3 )

where, k is Coulomb's force constant

            λ is linear charge density

            r is distance between point P and half of the rod.

Now, the linear charge density is given as:

λ = Charge/length = Q/x

The value of r is given as:

r = √x²-a²

where, x is length of rod

           a is half length of rod

Applying values in above equation, we get:

r = √x²-(x/2)²

r = √(1.20 m)²-(1.20/2)²

  = √1.08

  = 1.04 m

Substituting all the determined values in equation 3 we get:

E₁ = k (λ/r)

   = k [(Q/x)/r]

   = k [ Q/xr ]

   = (9×10⁹ Nm²/C²) [ |+2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]

   = 1.803×10⁴ N/C

Step 3:

Similarly, the electric field due to negative rod is given as:    

E₂ = k [ Q/xr ]

    = (9×10⁹ Nm²/C²) [ |-2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]

    = 1.803×10⁴ N/C

Step 4:

Consider equation 2:

E_net  = 2E₁ cos Φ

From the figure we get the phase angle as:

Φ = tan⁻¹ (0.60 m/0.60 m)

   = tan⁻¹ ( 1 )

   = π/4  

Now, the electric field produced due to each rod is equal and mutually perpendicular. Thus, the net electric field after applying values can be calculated as:

E_net = 2(1.803×10⁴ N/C) cos π/4

          = 2(1.803×10⁴ N/C) (0.5)

          = 18030 N/C

Step 5:

Consider equation 1 :

F = e. E_net

where, e is charge on an electron

Applying values in above equation we get:

F = (1.6 × 10⁻¹⁹ C)(18030 N/C)

  = 2.885 × 10⁻¹⁵ N

Therefore, the magnitude of the net force that the rods exert after an electron is released at point P is  2.885 × 10⁻¹⁵ N.

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The frequency of the other wave is 613 Hz or 607 Hz.

The difference between the frequencies of two waves is called the beat frequency.

Here, one wave has a frequency 610 Hz and the beat frequency is 3 beats per second.

Which has a higher frequency is not mentioned. Therefore, there are two possibilities.

Δf = | 610 - 613 | = 3

or

Δf = | 610 - 607 | = 3

Therefore, the frequency of the other wave is 613 Hz or 607 Hz.

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Answer:

(OC) a nucleus of protons and neutrons surrounded by a cloud of electrons - modern description of an atom - where the nucleus may be hundreds of times smaller in diameter than the cloud of electrons

D. The efficiency of the ramp is 54.6 %.

V.R = distance moved by effort/distance moved by load = L/h

V.R = (6.12)/(1.53) = 4

M.A = Load/Effort

M.A = (451 x 9.8)/(2025)

M.A = 2.183

E = (M.A / V.R) x 100%

E = (2.183 / 4) x 100%

E = 54.6 %

Thus, the efficiency of the ramp is 54.6 %.

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The work that is done when twice the load is lifted twice the distance is

four times as much

The net work performed by forces acting on an object equals the change in kinetic energy, according to the work-energy theorem.

when an item slows down, the net work applied to it decreases, its change in kinetic energy is negative, and its ultimate kinetic energy is less than its starting kinetic energy. When an item accelerates, positive net work is done on it. All the forces acting on an item must be taken into consideration when determining the net work. You will obtain an incorrect result if you exclude any forces that affect an item or if you add any forces that do not affect it.

Hence The work that is done when twice the load is lifted twice the distance is four times as much

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The values of the density, volume and masses are:

Density is defined as the ratio of mass and volume of  a substance.

The given data is as follows:

Mass of empty density bottle = 43.2 g

Mass of density bottle full of water = 66.4 g

Mass of density bottle with some sand = 67.5 g

Mass of density bottle with the sand

filled up with water = 82.3 g:

(a) mass of water that completely filled the bottle = 66.4 - 43.2

mass of water that completely filled the bottle = 23.2 g

(b) volume of water that completely filled the bottle = mass/density

density of water = 1 g/cm³

volume of water that completely filled the bottle = 23.3 g/1 g/cm³

volume of water that completely filled the bottle = 23.2 cm³

(c) volume of the density bottle = 23.2 cm³

(d) mass of sand = 67.5 g - 43.2 g = 24.4 g

(e) mass of water that filled the space above the sand = 82.3 - 67.5 = 14.8 g

(f) volume of sand = volume of bottle - volume of water

volume of water = 14.8 cm³

volume of sand = 23.2 cm³ - 14.8 cm³

volume of sand = 8.4 cm³

(g) density of sand = 24.4 g/8.4 cm³ = 2.9 g/cm³

Therefore, the density of sand is obtained by taking the ratio of the mass of the sand and the volume if the sand.

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Answer: A positively charged insulating rod is brought close to an object that is suspended by a string. If the object is repelled away from the rod, we can conclude that the object is positively charged.

Explanation: To find the answer, we need to know more about the Electric charges and its properties.

Thus, from the above, we can conclude that, when a positively charged insulating rod is brought close to an object that is suspended by a string. If the object is repelled away from the rod, we can conclude that the object is positively charged.

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Answer:

positively charged

Explanation:

when a positively charged object is being repelled, that means the charge should be the same for both.

Answer:

Soln:Here

Given,

Distance(s)=5m

Time taken(t)=20min=2400seconds

Speed(s)=?

We know that,

Speed(s)=Distance/timetaken

=5m/2400sec

580m/s

Thus the average speed of his bicycle is 580 m/s..

Thank you...

The speed of tsunami is a.0.32 km. 

Steps involved  :

The equation s = 356d models the maximum speed that a tsunami can move at. It reads as follows: s = 200 km/h d =?

Let's now change s to s in the equation to determine d: s = 356√d 200 = 356√d √d = 200 ÷ 356 √d = 0.562 Let's square the equation now by squaring both sides: (√d)² = (0.562) ² d = (0.562)² = 0.316 ≈ 0.32

As a result, 0.32 km is roughly the depth (d) of water for a tsunami moving at 200 km/h.

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Answer:

In pair production, after the loss of Kinetic energy, the angular separation between the two photons is 180°.

Explanation:

Therefore, the angle of separation between the two photons is 180°.

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The time required to reach the velocity is 15.9s.

The acceleration of the plane is 5.67m/s²

Acceleration is solved by the given formula.

Acceleration = velocity ÷ Time

Finding Time

Time = distance ÷ speed

Time = 1435m ÷ 90.1m/s = 15.9s

Finding Acceleration

Acceleration = velocity ÷ time

Acceleration = 90.1m/s ÷ 15.9s = 5.67m/s²