What accurately depicts the change in average kinetic energy of the particles undergoes in matter as the temperature of the sample is decreased?

Complete question:

A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa

Answer:

The average pressure exerted on the surface by the block is 9655.17 Pa

Explanation:

Given;

density of the lead, ρ =  1.13 x 10⁴ kg/m³

mass of the lead block, m = 20 kg

surface area of the area of the block, A = 2.03 x 10⁻² m²

Determine the force exerted on the surface by the block due to its weight;

F = mg

F = 20 x 9.8

F = 196 N

Determine the pressure exerted on the surface by the block

P = F / A

where;

P is the pressure

P = 196 / (2.03 x 10⁻²)

P = 9655.17 N/m²

P = 9655.17 Pa

Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa

Answer:

0.85c

Explanation:

Rest mass of Kaon [tex]M_{0K}[/tex] = 494 MeV/c²

Rest mass of proton [tex]M_{0P}[/tex]  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy [tex]E_{0K}[/tex] = 494c² MeV

for the proton, rest energy [tex]E_{0P}[/tex] = 938c² MeV

Recall that the rest energy, and the total energy are related by..

[tex]E[/tex] = γ[tex]E_{0}[/tex]

which can be written in this case as

[tex]E_{K}[/tex] = γ[tex]E_{0K}[/tex] ...... equ 1

where [tex]E[/tex] = total energy of the kaon, and

[tex]E_{0}[/tex] = rest energy of the kaon

γ = relativistic factor = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex]

where [tex]\beta = \frac{v}{c}[/tex]

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

[tex]E_{K}[/tex] = [tex]E_{0P}[/tex] ......equ 2

where [tex]E_{K}[/tex] is the total energy of the kaon, and

[tex]E_{0P}[/tex] is the rest energy of the proton.

From [tex]E_{K}[/tex] = [tex]E_{0P}[/tex] = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex] = 1.89

1.89[tex]\sqrt{1 - \beta ^{2} }[/tex] = 1

squaring both sides, we get

3.57( 1 - [tex]\beta^{2}[/tex]) = 1

3.57 - 3.57[tex]\beta^{2}[/tex] = 1

2.57 = 3.57[tex]\beta^{2}[/tex]

[tex]\beta^{2}[/tex] = 2.57/3.57 = 0.72

[tex]\beta = \sqrt{0.72}[/tex] = 0.85

but, [tex]\beta = \frac{v}{c}[/tex]

v/c = 0.85

v = 0.85c

Answer:

T₂ = 95.56°C

Explanation:

The final resistance of a material after being heated is given by the relation:

R' = R(1 + αΔT)

where,

R' = Final Resistance = 207.4 Ω

R = Initial Resistance = 154.9 Ω

α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹

ΔT = Change in Temperature = ?

Therefore,

207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]

207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT

1.34 - 1 = (0.0045°C⁻¹)ΔT

ΔT = 0.34/0.0045°C⁻¹

ΔT = 75.56°C

but,

ΔT = Final Temperature - Initial Temperature

ΔT = T₂ - T₁ = T₂ - 20°C

T₂ - 20°C = 75.56°C

T₂ = 75.56°C + 20°C

T₂ = 95.56°C

Answer:

q/6Eo

Explanation:

See attached file pls

Answer:

The speed of the electron will be 6x10^8m/s

Explanation:

See attached file

Answer:

magnitude of the resultant of forces is 11.45 N

Explanation:

given data

force F1 = 6N

force F2 = 8N

angle = 240°

solution

we get here resultant force that is express as

F(r) = [tex]\sqrt{F_1^2+F_2^2+2F_1F_2cos\ \theta}[/tex]    ..............1

put here value and we get

F(r) = [tex]\sqrt{6^2+8^2+2\times 6\times 8 \times cos240}[/tex]

F(r) =  11.45 N

so magnitude of the resultant of forces is 11.45 N

Answer:

94.248 g/sec

Explanation:

For solving the total current of the blood passing first we have to solve the cross sectional area which is given below:

[tex]A_1 = \pi R^2\\\\A_1 = \pi (1)^2\\\\A_1 = 3.1416 cm^2[/tex]

And, the velocity of blood pumping is 30 cm^2

Now apply the following formula to solve the total current

[tex]Q = \rho A_1V_1\\\\Q = (1)(3.1416)(30)\\\\[/tex]

Q =  94.248 g/sec

Basically we applied the above formula So, that the total current could come

Answer:

- E = 11.55J

- Q = 0.17C

- E' = (1/4)E

Explanation:

- To calculate the amount of energy stored in the capacitor, you use the following formula:

[tex]E=\frac{1}{2}CV^2[/tex]

C: capacitance = 3800.0*10^-6F

V: potential difference = 78.0V

[tex]E=\frac{1}{2}(3800.0*10^{-6}C)(78.0V)^2=11.55J[/tex]

The energy stored in the capacitor is 11.55J

- If the electrical energy stored in the capacitor is 6.84J, the charge on the capacitor is:

[tex]E=\frac{1}{2}QV\\\\Q=\frac{2E}{V}\\\\Q=\frac{2(6.84J)}{78.0V}=0.17C[/tex]

The charge on the capacitor is 0.17C

- If you take the capacitor as a parallel plate capacitor, you have that the energy stored on the capacitor is:

[tex]E=\frac{1}{2}CV^2=\frac{1}{2}(\frac{\epsilon_oA}{d})V^2=\frac{1}{2}\frac{\epsilon_oAV^2}{d}\\\\[/tex]

A: area of the plates

d: distance between plates

If the distance between plates is increased by a factor of 4, you have:

[tex]E'=\frac{1}{2}\frac{\epsilon_oAV^2}{(4d)}=\frac{1}{4}\frac{\epsilon_oAV^2}{2d}=\frac{1}{4}E[/tex]

Then, the stored energy in the capacitor is decreased by a a factor of (1/4)

Answer:

198.9 x 10^-16

Explanation:

E = hc/ wavelength

E =(6.63 x 10^-34 x 3 x 10^8)/(0.01 x 10^-9)

E = 198.9 x 10^-16

Answer:

The change in potential energy is  [tex]\Delta PE = - 3.8*10^{-16} \ J[/tex]

Explanation:

From the question we are told that

     The  magnitude of the uniform electric field  is  [tex]E = 950 \ N/C[/tex]

      The  distance traveled by the electron is  [tex]x = 2.50 \ m[/tex]

Generally the force on this electron is  mathematically represented as

     [tex]F = qE[/tex]

Where F is the force and  q is the charge on the electron which is  a constant value of  [tex]q = 1.60*10^{-19} \ C[/tex]

    Thus  

      [tex]F = 950 * 1.60 **10^{-19}[/tex]

      [tex]F = 1.52 *10^{-16} \ N[/tex]

Generally the work energy theorem can be mathematically represented as

          [tex]W = \Delta KE[/tex]

Where W is the workdone on the electron by the  Electric field and  [tex]\Delta KE[/tex]  is the change in kinetic energy

Also  workdone on the electron can also  be represented as

        [tex]W = F* x *cos( \theta )[/tex]

Where  [tex]\theta = 0 ^o[/tex] considering that the movement of the electron is along the x-axis  

        So

             [tex]\Delta KE = F * x cos (0)[/tex]

substituting values

         [tex]\Delta KE = 1.52 *10^{-16} * 2.50 cos (0)[/tex]

          [tex]\Delta KE = 3.8*10^{-16} J[/tex]

Now From the law of energy conservation

       [tex]\Delta PE = - \Delta KE[/tex]

Where [tex]\Delta PE[/tex] is the change  in  potential energy  

Thus  

        [tex]\Delta PE = - 3.8*10^{-16} \ J[/tex]

Answer:

C You should deflect the ball back toward your friend.

Explanation:

This is because it would result in a completely inelastic collision, and the final velocity of me would be found using,

with m= mass, V=velocity, i=initial, f=final:

mV(me,i) +mV(ball,i) = [m(me)+m(b)]V(f)

So V(f) would be just the momentum of the ball divided by just MV mass of the ball and it will be higher resulting in inelastic collision

Answer:

A. You should catch the ball.

Explanation:

Catching the ball maximizes your speed by converting most of the momentum of the flying ball into the momentum of you and the ball. Since the ice is smooth, the friction between your feet and the ice is almost negligible, meaning less energy is needed to set your body in motion. Catching the ball means that you and the ball undergoes an inelastic collision, and part of the kinetic energy of the ball is transferred to you, setting you in motion. Deflecting the ball will only give you a relatively small speed compared to catching the ball.

Answer:

The largest  possible distance is [tex]x = 4.720 \ m[/tex]

Explanation:

From the question we are told that

    The distance of  separation is   [tex]d = 4.30 \ m[/tex]

      The  frequency of the tone played by both speakers is [tex]f = 103 \ Hz[/tex]

     The speed of sound is  [tex]v_s = 343 \ m/s[/tex]

The  wavelength of the tone played by the speaker is  mathematically evaluated as

              [tex]\lambda = \frac{v}{f}[/tex]

substituting values

            [tex]\lambda = \frac{343}{103}[/tex]

            [tex]\lambda = 3.33 \ m[/tex]

Let the the position of the observer be O

Given that the line of sight between observer and speaker B is  perpendicular to the distance between A and B then

        The distance between A and the observer is  mathematically evaluated using Pythagoras theorem as follows

               [tex]L = \sqrt{d^2 + x^2}[/tex]

Where x is the distance between the observer and B

  For the observer to observe destructive interference

          [tex]L - x = \frac{\lambda}{2}[/tex]

So  

          [tex]\sqrt{d^2 + x^2} - x = \frac{\lambda}{2}[/tex]

       [tex]\sqrt{d^2 + x^2} = \frac{\lambda}{2} +x[/tex]

        [tex]d^2 + x^2 = [\frac{\lambda}{2} +x]^2[/tex]

         [tex]d^2 + x^2 = [\frac{\lambda^2}{4} +2 * x * \frac{\lambda}{2} + x^2][/tex]

       [tex]d^2 = [\frac{\lambda^2}{4} +2 * x * \frac{\lambda}{2} ][/tex]

substituting values              

       [tex]4.30^2 = [\frac{3.33^2}{4} +2 * x * \frac{3.33}{2} ][/tex]

      [tex]x = 4.720 \ m[/tex]

Answer:  If you are traveling at a speed of 60mph, you will go 220 feet.

Explanation: 60mph is a mile a minute. 5280 feet in a mile, 60 seconds in a minute. Divide to find that is 88 feet per second. Multiply by the number of seconds.

Answer:

The angular displacement is  [tex]\theta = 28.33 \ rad[/tex]

Explanation:

From the question we are told that

     The speed of the driver is  [tex]v =13 \ m/ s[/tex]

     The acceleration of the driver is  [tex]a = 1.02 \ m/s^2[/tex]

      The time taken is [tex]t = 0.70 \ s[/tex]

      The radius of the tire is  [tex]r = 33 cm = 0.33 \ m[/tex]

The distance covered by the car during this  acceleration can be  calculated using the equation of motion as follows

        [tex]s = v*t +\frac{1}{2} * a * t^2[/tex]

Now substituting values  

       [tex]s = 13 * 0.70 +\frac{1}{2} * 1.02 * (0.700)^2[/tex]

      [tex]s = 9.35 \ m[/tex]

Now the angular displacement of the car with respect to the tire movement can be  represented mathematically as

      [tex]\theta = \frac{s}{r}[/tex]

substituting values

      [tex]\theta = \frac{9.35}{0.33}[/tex]

      [tex]\theta = 28.33 \ rad[/tex]

Answer: (a) α = [tex]\frac{3I}{2.\pi.R^{3}}[/tex]

(b) For r≤R: B(r) = μ_0.[tex](\frac{I.r^{2}}{2.\pi.R^{3}})[/tex]

For r≥R: B(r) = μ_0.[tex](\frac{I}{2.\pi.r})[/tex]

Explanation:

(a) The current I enclosed in a straight wire with current density not constant is calculated by:

[tex]I_{c} = \int {J} \, dA[/tex]

where:

dA is the cross section.

In this case, a circular cross section of radius R, so it translates as:

[tex]I_{c} = \int\limits^R_0 {\alpha.r.2.\pi.r } \, dr[/tex]

[tex]I_{c} = 2.\pi.\alpha \int\limits^R_0 {r^{2}} \, dr[/tex]

[tex]I_{c} = 2.\pi.\alpha.\frac{r^{3}}{3}[/tex]

[tex]\alpha = \frac{3I}{2.\pi.R^{3}}[/tex]

For these circunstances, α = [tex]\frac{3I}{2.\pi.R^{3}}[/tex]

(b) Ampere's Law to calculate magnetic field B is given by:

[tex]\int\ {B} \, dl =[/tex] μ_0.[tex]I_{c}[/tex]

(i) First, first find [tex]I_{c}[/tex] for r ≤ R:

[tex]I_{c} = \int\limits^r_0 {\alpha.r.2\pi.r} \, dr[/tex]

[tex]I_{c} = 2.\pi.\frac{3I}{2.\pi.R^{3}} \int\limits^r_0 {r^{2}} \, dr[/tex]

[tex]I_{c} = \frac{I}{R^{3}}\int\limits^r_0 {r^{2}} \, dr[/tex]

[tex]I_{c} = \frac{3I}{R^{3}}\frac{r^{3}}{3}[/tex]

[tex]I_{c} = \frac{I.r^{3}}{R^{3}}[/tex]

Calculating B(r), using Ampere's Law:

[tex]\int\ {B} \, dl =[/tex] μ_0.[tex]I_{c}[/tex]

[tex]B.2.\pi.r = (\frac{Ir^{3}}{R^{3}} )[/tex].μ_0

B(r) = [tex](\frac{Ir^{3}}{R^{3}2.\pi.r})[/tex].μ_0

B(r) = [tex](\frac{Ir^{2}}{2.\pi.R^{3}} )[/tex].μ_0

For r ≤ R, magnetic field is B(r) = [tex](\frac{Ir^{2}}{2.\pi.R^{3}} )[/tex].μ_0

(ii) For r ≥ R:

[tex]I_{c} = \int\limits^R_0 {\alpha.2,\pi.r.r} \, dr[/tex]

So, as calculated before:

[tex]I_{c} = \frac{3I}{R^{3}}\frac{R^{3}}{3}[/tex]

[tex]I_{c} =[/tex] I

Using Ampere:

B.2.π.r = μ_0.I

B(r) = [tex](\frac{I}{2.\pi.r} )[/tex].μ_0

For r ≥ R, magnetic field is; B(r) = [tex](\frac{I}{2.\pi.r} )[/tex].μ_0.

A number, which represents a property, amount, or relationship that does not change under certain situations is constant and further calculations as follows:

The Radius of the cross-section of the wire R

Current passing through the wire I

Current Density [tex]J = \alpha r[/tex]

Constant [tex]\alpha[/tex]

Distance of the point from the center [tex]r[/tex]

For part a)

Consider a circular strip between two concentric circles of radii r and r+dr.

Current passing through the strip [tex]dI =\overrightarrow J \times \overrightarrow{dA}[/tex]

[tex]\to\alpha r (2\pi r dr) cos 0^{\circ}[/tex]

Integration

[tex]\to I =2\pi \alpha \int^R_0 r^2\ dr =2\pi \alpha [r^3]^R_0=2\pi \alpha \frac{r^3}{3}\\\\\to \alpha = \frac{3I}{2\pi R^3}\\\\[/tex]

For part b)

The magnetic field at a point distance [tex]'r'^{(r \ \pounds \ R)}[/tex] from the center is B.

We have the value of the line integral of the magnetic field over a circle of radius ‘r’ given as

[tex]\oint \overrightarrow B \times \overrightarrow{dl} = \mu_0 I\\\\[/tex]

where ‘I’ is the threading current through the circle of radius ‘r’

[tex]\oint B \ dl \cos 0^{\circ} = \mu_0 [2\pi \alpha \frac{r^3}{3}]\\\\ B \int dl = \mu_0 [2\pi \frac{3I}{2\pi R^3} \frac{r^3}{3}]\\\\ B \cdot 2\pi r = \mu_0 I [\frac{r}{R}]^3\\\\ B = \frac{\mu_0}{2\pi} I [\frac{Ir^2}{R^3}]\\\\[/tex]

(ii) Similarly, we can calculate the magnetic field at the point at A distance ‘r’ where

[tex]\to r^3 R\\\\\to \int \overrightarrow{B} \overrightarrow{dl} = \mu_0\ I[/tex] [The threading current is the same]

[tex]\to \beta - 2\pi r = \mu_0 I[/tex] As (I)

[tex]\to \beta =\frac{\mu_o \ I}{2\pi \ r}[/tex]

Find out more about the density here:

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Answer:

q = 2.997*10^-4C

Explanation:

In order to find the required charge that the penny have to have, to acquire an upward acceleration, you take into account that the electric force on the penny must be higher than the weight of the penny.

You use the second Newton law to sum both electrical and gravitational forces:

[tex]F_e-W=ma\\\\qE-mg=ma[/tex]             (1)

Fe: electric force

W: weight of the penny

q: required charge = ?

m: mass of the penny = 3g = 0.003kg

E: magnitude of the electric field = 100N/C

g: gravitational acceleration = 9.8m/s^2

a: acceleration of the penny = 0.19m/s^2

You solve the equation (1) for q, and replace the values of the other parameters:

[tex]q=\frac{ma+mg}{E}=\frac{m(a+g)}{E}\\\\q=\frac{(0.003kg)(0.19m/s^2+9.8m/s^2)}{100N/C}\\\\q=2.997*10^{-4}C[/tex]

It is necessary that the penny has a charge of 2.997*10^-4 C, in order to acquire an upward acceleration of 0.19m/s^2

Answer:

dr = 0.71 ft/min

the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.

Explanation:

Given;

Rate of inflation dV = 140pift^3/min

Radius r = 7 ft

Change in radius = dr

Volume of a spherical balloon is;

V = (4/3)πr^3

The change in volume can be derived by differentiating both sides;

dV = (4πr^2)dr

Making dr the subject of formula;

dr = dV/(4πr^2)

Substituting the given values;

dr = 140π/(4π×7^2)

dr = 0.714285714285 ft/min

dr = 0.71 ft/min

the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.

Answer:

5. The amount of work is the same for both.

Explanation:

Work done is a measure of change in kinetic energy of each egg

For both egg , the initial speed and mass are same , so they have equal initial Kinetic energy

For both egg , the final speed is 0 and mass are same , so they have equal final Kinetic energy which is 0.

So work done is same for both eggs since they have same change in kinetic energy.

Answer:

Explanation:

Given data

mass of  load m= 425 kg

height/distance h=64 m

acceleration a= 1.8 m/s^2

The work done can be calculated using the expression

work done= force* distance

but force= mass *acceleration

hence work done= 425*1.8*64= 48,960 J

work done= 48.96 kJ

Answer:

The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.

Explanation:

The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.

The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery. Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.      

Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.                                  

I hope it helps you!

Answer:

F' = F/16

So, the force of gravity has become 16 times less than initial value.

Explanation:

The force of gravity between two planets, is given by the following formula:

F = Gm₁m₂/r²   ----------- equation 1

where,

F = Force of gravity between two planets

G = Gravitational Constant

m₁ = Mass of one planet

m₂ = Mass of other plant

r = Distance between two planets

Now, if the distance between the planets (r) is 10 times smaller, then Force of gravity will become:

F' = Gm₁m₂/(4r)²

F' = (1/16) (Gm₁m₂/r²)

using equation 1:

F' = F/16

So, the force of gravity has become 16 times less than initial value.

Answer:

The force is  [tex]F = 8*10^{-12} \ N[/tex]

Explanation:

From the question we are told that

     The rate at which ATP molecules are used is [tex]R = 80 ATP/ s[/tex]

       The energy provided by a single ATP is  [tex]E_{ATP} = 0.8 * 10^{-19} J[/tex]

       The velocity of the kinesin is  [tex]v = 800 nm/s = 800*10^{-9} m/s[/tex]

The power provided by the ATP in one second is  mathematically represented as

       [tex]P = E_{ATP} * R[/tex]

substituting values

       [tex]P = 80 * 0.8*10^{-19 }[/tex]

       [tex]P = 6.4 *10^{-18}J/s[/tex]

Now  this power is mathematically represented as

       [tex]P = F * v[/tex]

Where  F  is  the force the kinesin is exerting

  Thus  

          [tex]F = \frac{P}{v}[/tex]

substituting values

            [tex]F = \frac{6.4*0^{-18}}{800 *10^{-9}}[/tex]

           [tex]F = 8*10^{-12} \ N[/tex]

The force exerted by the kinesin  is 8  × 10-12 N.

Let us recall that power is defined as the rate of doing work. Hence, power = Energy/Time.

Since;

Energy  =  0.8 × 10-19 J/molecule

Number ATP molecules transported per second = 80 ATP molecules/s

Power =  0.8 × 10-19 J/molecule × 80 ATP molecules/s

Power = 6.4  × 10-18 J

Again, we know that;

Power = Force × Velocity

Velocity of the ATP molecules =  800 nm/s or 8 × 10-7 m/s

Force = Power/velocity

Force =  6.4  × 10-18 J/ 8 × 10-7 m/s

Force = 8  × 10-12 N

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Answer:

6.44 s

Explanation:

Given:

v₀ = 119 m/s

v = 233 m/s

a = 17.7 m/s²

Find: t

v = at + v₀

(233 m/s) = (17.7 m/s²) t + (119 m/s)

t = 6.44 s

Answer:

a.    σ = 3.82*10^-18C/m^2

b.    d = 2.00m

Explanation:

a. In order to calculate the surface charge density of the sheet, you first calculate the acceleration of the electron on its motion.

You use the following formula:

[tex]v^2=v_o^2-2ad[/tex]                (1)

v: final speed of the electron = 0m/s

vo: initial speed of the electron = 390m/s

a: acceleration of the electron = ?

d: distance traveled by the electron = 2.00m

You solve the equation (1) for a, and replace the values of all parameters:

[tex]a=\frac{v_o^2-v^2}{2d}\\\\a=\frac{(390m/s)^2}{2(2.00m)}=3.8*10^4\frac{m}{s^2}[/tex]

Next, you calculate the electric field that exerts the electric force on the electron, by using the second Newton law, as follow:

[tex]F_e=qE=ma[/tex]               (2)

q: charge of the electron = 1.6*10^-19C

E: electric field of the sheet = ?

m: mass of the electron = 9.1*10^-31kg

You solve the equation (2) for E:

[tex]E=\frac{ma}{q}=\frac{(9.1*10^{-31}kg)(3.8*10^{4}m/s^2)}{1.6*10^{-19}C}\\\\E=2.16*10^{-7}\frac{N}{C}[/tex]

Next, you use the following formula to calculate the surface charge density, by using the value of its electric field:

[tex]E=\frac{\sigma}{2\epsilon_o}[/tex]          (3)

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

σ: surface charge density of the sheet

You solve for σ:

[tex]\sigma=2\epsilon_o E=2(8.85*10^{-12}C^2/Nm^2)(2.16*10^{-7}N/C)\\\\\sigma=3.82*10^{-18}\frac{C}{m^2}[/tex]

The surface charge density of the sheet id 3.82*10^-18C/m^2

b. To calculate the required distance for the electron reaches the sheet, you take into account that the electron acceleration is the same in all places near the sheet, then by the result of the previous point, you can conclude that the electron must be fired from a distance of 2.00m.

Answer:

is this a company name.? java is a computer software right..

Answer:

Towards the west.

Explanation:

The direction of a magnetic field lines is the direction north end of a compass needle points. The magnetic field exert force on positive charge.

Using the magnetic rule,which indicate that in order to find the direction of magnetic force on a moving charge, the thumb of the right hand point in the direction of force, the index finger in the direction of velocity charge and the middle finger in the direction of magnetic field.

According to the right hand rule, the electron moving moving west which is the thumb, the direction of the electron is west which is the middle finger and it is upward

Answer:

V = 2.95*10^-3 V

E = 0 N/C

Explanation:

In order to calculate the electric potential due to the two protons at the midpoint between them, you use the following:

[tex]V=V_1+V_2=k\frac{q}{r}+k\frac{q}{r}=2\frac{kq}{r}[/tex]       (1)

where you have taken into account each contribution to the total electric potential, produced by each proton.

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q: charge of the proton = 1.6*10^{-19}C

r: distance from the point (at the midway distance between the protons) to one proton = 973nm = 973nm/2 = 486.5nm = 486.5*10^-9m

You replace the values of the parameters in the equation (1):

[tex]V=2\frac{(8.98*10^9Nm^2/C^2)(1.6*10^{-19}C)}{486.5*10^{-9}m}\\\\V=2.95*10^{-3}V[/tex]

The electric potential is 2.95*10^-3V

The electric field generated at the midpoint in between the protons is zero, because the electric field generated by each proton has the same magnitude but opposite direction.

E = 0N/C

Answer:

D

Explanation:

The power equation is P= V^2/R

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As a result of the given scenario, power delivered from the battery to combination decreases. The correct option is D.

A resistor is a two-terminal passive electrical component that uses electrical resistance as a circuit element.

Resistors are used in electronic circuits to reduce current flow, adjust signal levels, divide voltages, and bias active elements.

A resistor is a component of an electronic circuit that limits or regulates the flow of electrical current. Resistors can also be used to supply a fixed voltage to an active device such as a transistor.

The current through resistors is the same when they are connected in series. The battery voltage is divided among resistors.

Adding more resistors to a series circuit increases total resistance and thus lowers current. However, in a parallel circuit, adding more resistors in parallel creates more options while decreasing total resistance.

Thus, the correct option is D.

For more details regarding resistors, visit:

https://brainly.com/question/24297401

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Answer:

C

Explanation:

it transforms electrical energy into mechanical energy.

Answer:

2.16 × 10^-16N

Explanation:

The computation of the net force is shown below:

Data provided in the question

Electron mass = 9.11 × 10^-31kg

V_o = 0

V_f =  3.00 × 10^6 m/s

s = 1.90 cm i.e 1.9 × 10^-2

Based on the above information, the force is

As we know that

[tex]Force\ f = ma = \frac{mv^2}{2s}\\\\ = \frac{(9.11\times 10^{-31})(3\times 10^{6})^2}{2(1.9\times 10^{-2})}[/tex]

= 2.16 × 10^-16N

Hence, the last option is correct

Basically we applied the above formula to determine the net force