n a location where the train tracks run parallel to a road, a high-speed train traveling at 60 m/s passes a car traveling at 30 m/s in the opposite direction. How long does it take for the train to be 180 m away from the car? Group of answer choices

Answer:

More current will be loss through the metal wire strands if the force on them was repulsive, and more stress will be induced on the wire strands due to internal and external flexing.

Explanation:

A wire bundle is made up of wire strands bunched together to increase flexibility that is not always possible in a single solid metal wire conductor. In the strands of wire carrying a high voltage power, each strand carries a certain amount of current, and the current through the strands all travel in the same direction. It is know that for two conductors or wire, separated by a certain distance, that carries current flowing through them in the same direction, an attractive force is produced on these wires, one on the other. This effect is due to the magnetic induction of a current carrying conductor. The forces between these strands of the high voltage wire bundle, pulls the wire strands closer, creating more bond between these wire strands and reducing internal flex induced stresses.

If the case was the opposite, and the wires opposed themselves, the effect would be that a lot of cost will be expended in holding these wire strands together. Also, stress within the strands due to the repulsion, will couple with external stress from the flexing of the wire, resulting in the weakening of the material.

The biggest problem will be that more current will be lost in the wire due to increased surface area caused by the repulsive forces opening spaces between the strand. This loss is a s a result of the 'skin effect' in wire transmission, in which current tends to flow close to the surface of the metal wire. The skin effect generates power loss as heat through the exposed surface area.

Answer:

f = 2200 Hz

Explanation:

It is given that,

Speed of a wave is 242 m/s

The distance between crests is 0.11 m

We need to find the frequency of the wave. The distance between crests is called wavelength of a wave. So,

[tex]v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{242}{0.11}\\\\f=2200\ Hz[/tex]

So, the frequency of the wave is 2200 Hz.

Answer:2200 hz

Explanation:

Answer:

Explanation:

Given that

a metal rod 0.4 m long

mass of 2 kg

lightweight 3.0-m-long cord

a) I=mr^2, where m = 2kg and r = 3.2m

b) I=1/3mr^2, where m = 2kg and r = 3.4

c) I=1/12mr^2, where m = 2kg and r = 1.7m

d) I=mr^2, where m= 2kg and r = 3.4m

[tex]I=\int\limits^{3.4}_3 \lambda dx \,x^2\\\\=\lambda\frac{x^3}{3} |^{3.4}_{3}[/tex]

[tex]I=\frac{M}{L}[\frac{3.4^3}{3}-\frac{3^3}{3} ] \\\\I=\frac{2}{(0.4\times3)}[\frac{3.4^3}{3}-\frac{3^3}{3} ]\\\\I=\frac{2}{1.2}(39.304-9)\\\\I=1.67\times12.304\\\\I=20.51[/tex]

[tex]a)I_1=mr^2\\\\I_1=2\times3.2^2\\\\I_1=20.48kgm^2\\\\b)I_2=\frac{1}{3} mr^2\\\\=\frac{1}{3} \times2\times3.4^2\\\\I_2=7.71kgm^2\\\\c)I_3=\frac{1}{12} mr^2\\\\I_3=\frac{1}{12} \times2\times1.7^2\\\\I_3=0.461kg/m^2\\\\d)I_4=mr^2\\\\I_4=2\times3.462\\\\I_4=23.12kg/m^2[/tex]

Sun's energy comes from the nuclear fusion taking place inside. In nuclear fusion two light nuclei fuses together to form a heavy nuclei with the release of greater amount of energy.

Nuclear fusion is the process of combining two light nuclei to form a heavy nuclei. In this nuclear process, tremendous energy is released. This is the source of heat and light in stars.

On the other hand, nuclear fission is the process of breaking of a heavy nuclei into two lighter nuclei. Fission also produces massive energy. But in comparison, more energy is produced by nuclear fusion.

Nuclear fission is used in nuclear power generators. The light energy and  heat energy comes form the nuclear fusion of hydrogens to form helium nuclei. Hence, option D is correct.

Find more on nuclear fusion:

https://brainly.com/question/12701636

#SPJ2

Answer:

E = V/5 x10⁻³

Explanation:

if the potential difference is V

then electric field E is given by

E = V/d

d = 0.5cm = 5 x 10⁻³m

E = V/5 x10⁻³

Answer:

the coefficient of volume expansion of the glass is [tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]

Explanation:

Given that:

Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³

temperature of the glass flask and mercury= 1.00° C

After heat is applied ; the final temperature = 52.00° C

Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C

Volume of the mercury overflow = 8.50 cm^3 = 8.50 ×  10⁻⁶ m³

the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K

The increase in the volume of the mercury =  10⁻³ m³ ×  51.00 × 1.80 × 10⁻⁴

The increase in the volume of the mercury = [tex]9.18*10^{-6} \ m^3[/tex]

Increase in volume of the glass =  10⁻³ × 51.00 × [tex]\beta _{glass}[/tex]

Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask

the mercury overflow = [tex](9.18*10^{-6} - 51.00* \beta_{glass}*10^{-3})\ m^3[/tex]

[tex]8.50*10^{-6} = (9.18*10^{-6} -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]-6.8*10^{-7} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]6.8*10^{-7} = ( 51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}= ( \beta_{glass} )[/tex]

[tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]

Thus; the coefficient of volume expansion of the glass is [tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]

Answer:

-18896.49 V/m

Explanation:

Distance between the two plates = 10 cm = 10 x [tex]10^{-2}[/tex] m = 0.1 m

Also, one of the plates is taken as zero volt.

a. The potential strength between the zero volt plate, and 7.05 cm (0.0705 m) away is 393 V

b. The potential strength between the other plate, and 2.95 cm (0.0295 m) away is 393 V

Potential field strength = -dV/dx

where dV is voltage difference between these points,

dx is the difference in distance between these points

For the first case above,

potential field strength = -393/0.0705 = -5574.46 V/m

For the second case ,

potential field strength = -393/0.0295 = -13322.03 V/m

Magnitude of the field strength across the plates will be

-5574.46 + (-13322.03) = -5574.46 + 13322.03 = -18896.49 V/m

Answer:

(a) σ = 3.41*10⁻7C/m^2

(b) E = 38,530.1 N/C

Explanation:

(a) In order to calculate the resulting surface charge density, you use the following formula:

[tex]\sigma=\frac{Q}{S}[/tex]     (1)

σ: surface charge density

Q: charge of the satellite = 3.1 µC = 3.1*10^-6C

S: surface area of the satellite

The satellite has a spherical form, then, the area of the surface is given by:

[tex]S=4\pi r^2[/tex]     (2)

r: radius of the satellite = d/2 = 1.7m/2 = 0.85m

You replace the equation (2) into the equation (1) and solve for the surface charge density:

[tex]\sigma=\frac{3.1*10^{-6}C}{4\pi (0.85m)^2}=3.41*10^{-7}\frac{C}{m^2}[/tex]

The surface charge density acquired by the satellite on one orbit is 3.41*10⁻7C/m^2

(b) The electric field just outside the surface is calculate d by using the following formula:

[tex]E=k\frac{Q}{R^2}[/tex]      (3)

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

R: radius of the satellite = 0.85m

[tex]E=(8.98*10^9Nm^2/C^2)\frac{3.1*10^{-6}C}{(0.85m)^2}=38530.1\frac{N}{C}[/tex]

The magnitude of the electric field just outside the sphere is 38,530.1 N/C

Answer:

the acceleration [tex]a^{\to} = (0.0159 \ \ m/s^2 )i[/tex]

Explanation:

Given that:

the initial speed v₁ = 0 m/s i.e starting from rest ; since the car accelerates at a distance Δx = 6 miles in order to teach that final speed v₂ of 63.15 km/h.

So;  the acceleration for the first 6 miles can be calculated by using the formula:

v₂² = v₁² + 2a (Δx)

Making acceleration  a the subject of the formula in the above expression ; we have:

v₂² - v₁² = 2a (Δx)

[tex]a = \dfrac{v_2^2 - v_1^2 }{2 \Delta x}[/tex]

[tex]a = \dfrac{(63.15 \ km/s)^2 - (0 \ m/s)^2 }{2 (6 \ miles)}[/tex]

[tex]a = \dfrac{(17.54 \ m/s)^2 - (0 \ m/s)^2 }{2 (9.65*10^3 \ m)}[/tex]

[tex]a =0.0159 \ m/s^2[/tex]

Thus;

Assume the car moves in the +x direction;

the acceleration [tex]a^{\to} = (0.0159 \ \ m/s^2 )i[/tex]

There is only one possible state: constant uniform motion. That means constant speed in a straight line.

(If the constant speed happens to be zero, this description also covers the case where the object isn't moving. That special case is called "at rest".)

Answer:

Hope this helps

Answer:

Hooking up the bulb to the battery in a series arrangement will draw the least amount of current.

Explanation:

In this case now, the bulb will serve as the load on the battery (resistance).

For the current to last longer, the least amount of energy must be drawn.

The least amount of energy will be drawn when the arrangement provides the maximum resistance possible.

Let us take the resistance of each bulb as 'R'

If we arrange the bulbs in series, then, the total resistance will be

Rt = R + R = 2R

at a EMF of V from the battery, current I through the battery will be

I = V/2R

If we arrange the bulbs in parallel, then , the total resistance will be

1/Rt = 1/R + 1/R

1/Rt = 2/R

therefore

Rt = R/2

at an EMF of V from the battery, the current I that will be drawn through the battery will be

I = 2V/R

we see that arranging the bulbs in parallel draws 4 times the current compared to arranging the bulb in series

From the above, we see that arranging the bulbs in series provides the maximum resistance, which means a lesser amount of current is drawn from the battery

Answer:

0.48 cm

Explanation:

given data

wavelength = 480 nm

wavelength = 560 nm

slit spacing = 0.040 mm

distance between double slits and the screen  = 1.2 m

solution

we know that  (1 nm= [tex]10^{-9}[/tex] m)

we wil take here equation of equations of interference that is

ym = R × (m λ)/d    ..........................1

here m = 2 R  i.e distance of screen and slit

 so put here value and we get

separation between the second-order bright fringes = 0.48 cm

Answer:

Option B:

The normal force

Explanation:

The normal force does no work as the box slides down the ramp.

Work can only be done when the force succeeds in moving the object in the direction of the force.

All the other forces involved have a component that is moving the box in their direction.

However, the normal force does not, as it points downwards into the ramp. Since the normal force is pointing into the ramp, and the box is sliding down the ramp, we can say that no work is being done by the normal force because the box is not moving in its direction (which would have been the box moving into the ramp)

Answer: No, we can have a displacement equal to 0 while the distance traveled is different than zero.

Explanation:

Ok, let's write the definitions:

Displacement: The displacement is equal to the difference between the final position and the initial position.

Distance traveled: Total distance that you moved.

So, for example, if at t = 0s, you are in your house, then you go to the store, and then you return to your house, we have:

The displacement is equal to zero, because the initial position is your house and the final position is also your house, so the displacement is zero.

But the distance traveled is not zero, because you went from you traveled the distance from your house to the store two times.

So no, we can have a displacement equal to zero, but a distance traveled different than zero.

Answer:

By convention a negative torque leads to clockwise rotation and a positive torque leads to counterclockwise rotation.

here weight of the child =21kgx9.8m/s2 = 205.8N

the torque exerted by the child Tc = - (1.8)(205.8) = -370.44N-m ,negative sign is inserted because this torque is clockwise and is therefore negative by convention.

torque exerted by adult Ta = 3(151) = 453N , counterclockwise torque.

net torque Tnet = -370.44+453 =82.56N , which is positive means counterclockwise rotation.

b) Ta = 2.5x151 = 377.5N-m

Tnet = -370.44+377.5 = 7.06N-m , positive ,counterclockwise rotation.

c)Ta = 2x151 = 302N-m

Tnet = -370.44+302 = -68.44N-m, negative,clockwise rotation.

Answer:

Explanation:

each grid corresponding  0.1s⁻¹.

0.2grid unit = 0.2×0.1 =0.02s⁻¹

distance of the star from telescope

d = 1/p

d= 1/0.02= 50 par sec

1par sec = 3.26 light year

1 light year = 9.5×10¹²km

3.26ly=3.084×10¹³km

d= 50×3.084×10¹³=1.55×10¹⁵km

Answer:

the two elements are resistor R and inductor L

answers in two significant figures

R = 9.6Ω

L = 54mH

Explanation:

Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

[tex]U_{g,1} = U_{g,2} + W_{dis}[/tex]

Where:

[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.

[tex]W_{dis}[/tex] - Dissipated work due to friction, measured in joules.

Gravitational potential energy is described by the following formula:

[tex]U = m \cdot g \cdot y[/tex]

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]g[/tex] - Gravitational constant, measured in meters per square second.

[tex]y[/tex] - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

[tex]W_{dis} = f \cdot \Delta s[/tex]

Where:

[tex]f[/tex] - Friction force, measured in newtons.

[tex]\Delta s[/tex] - Travelled distance, measured in meters.

Now, the energy equation is expanded and frictional force is cleared:

[tex]m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s[/tex]

[tex]f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}[/tex]

If [tex]m = 1000\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 40\,m[/tex], [tex]y_{2} = 25\,m[/tex] and [tex]\Delta s = 400\,m[/tex], then:

[tex]f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}[/tex]

[tex]f = 367.763\,N[/tex]

The magnitude of the frictional force between the car and the track is 367.763 N.

Answer:

A) 12752.55 J

B) 12507.75 J

C) 12262.95 J

Explanation:

We are given;

Mass;m = 85 kg

Vertical distance; d = 15 m

From change in kinetic energy, the work done by the applied force to pull the spelunker is given by;

Change in kinetic energy = Wa + Wg

Where Wg = -mgd

A) In the first stage;the the work done is given by;

Wa = mgd + ½m(v_f)² - ½m(v_i)²

Since initially stationary, v_i = 0

So, we have;

Wa = mgd + ½m(v_f)²

v_f = 2.4 m/s

So,

Wa = (85 × 9.81 × 15) + ((1/2) × 85 × 2.4²)

Wa = 12752.55 J

B) for the second stage, there is a constant speed of 2.4 m/s

So, v_f = v_i

Thus; Wa = mgd

Wa = (85 × 9.81 × 15)

Wa = 12507.75 J

C) he finally decelerated to zero.

So v_f = 0 while v_i is 2.4 m/s

Thus;

Wa = mgd - ½m(v_i)²

Wa = (85 × 9.81 × 15) - (½ × 85 × 2.4²)

Wa = 12507.75 - 244.8

Wa = 12262.95 J

Answer:

τ = 1679.68Nm

Explanation:

In order to calculate the required torque you first take into account the following formula:

[tex]\tau=I\alpha[/tex]           (1)

τ: torque

I: moment of inertia of the merry-go-round

α: angular acceleration

Next, you use the following formulas for the calculation of the angular acceleration and the moment of inertia:

[tex]\omega=\omega_o+\alpha t[/tex]         (2)

[tex]I=\frac{1}{2}MR^2[/tex]           (3)       (it is considered that the merry-go-round is a disk)

w: final angular speed = 3.1 rad/s

wo: initial angular speed = 0 rad/s

M: mass of the merry-go-round = 432 kg

R: radius of the merry-go-round = 2.3m

You solve the equation (2) for α. Furthermore you calculate the moment of inertia:

[tex]\alpha=\frac{\omega}{t}=\frac{3.1rad/s}{2.1s}=1.47\frac{rad}{s^2}\\\\I=\frac{1}{2}(432kg)(2.3)^2=1142.64kg\frac{m}{s}[/tex]

Finally, you replace the values of the moment of inertia and angular acceleration in the equation (1):

[tex]\tau=(1142.64kgm/s)(1.47rad/s^2)=1679.68Nm[/tex]

The required torque is 1679.68Nm

Answer:

d > a > b > c

Explanation:

Given that

a) radius = 3r and drift speed = 1v

b) radius = 4r and drift speed = 0.5v

c) radius = 1r and drift speed = 5v

d) radius = 2r and  drift speed = 2.5v

Based on the above information, the ranking of the wires for reducing the electron current is

As we can see that the radius i.e to be less and the drift speed that is highest so it should be rank one

And, According to that, other options are ranked

Therefore, the ranking would be d > a > b > c

Answer:

c. The astronaut does not need to worry: the charge will remain on the outside surface.

Explanation:

The astronaut need not worry because according to Gauss's law of electrostatic, a hollow charged surface will have a net zero charge on the inside. This is the case of a Gauss surface, and all the charges stay on the surface of the metal chamber. This same principle explains why passengers are safe from electrostatic charges, in an enclosed aircraft, high up in the atmosphere; all the charges stay on the surface of the aircraft.

Answer:

[tex]\mathbf{1.51\times10^{-15}N}[/tex]

Explanation:

The computation of the weight of the paper in newtons is shown below:

On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.

Therefore the paper charge is

[tex]q_{paper}=-4.1\times10^{-15}C[/tex]

Now the distance from the charge is

[tex]r=1cm=0.01m[/tex]

Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.

[tex]mg=\frac{k_{e}q_{1}q_{2}}{r^{2}}[/tex]

[tex]\Rightarrow W=mg[/tex]

[tex]=\frac{9\times10^{9}\times(4.1\times10^{-15})^{2}}{0.01^{2}}[/tex]

[tex]=\mathbf{1.51\times10^{-15}N}[/tex]

Answer:

a) correct answer is b higher , b) correct answer is b higher , c) correct answer is b faster , d)  traveling wave , e)

Explanation:

A traveling wave is described by the expression

            y = A sin (kx - wt)

where k is the wave vector and w is the angular velocity

let's examine every situation presented

a) a new faster pulse is generated

A faster pulse should have a higher angular velocity

equal speed is related to the period and frequency

            w = 2π f = 2π / T

therefore in this case the period must decrease so that the angular velocity increases

the correct answer is c narrower

b) Generate a pulse, but move your hand more.

Moving the hand increases the amplitude (A) of the pulse

the correct answer is b higher

c) generates a pulse but the force is tightened

Set means that more tension force is applied to the string, so the velicate changes

       v = √ (T /μ)

the correct answer is b faster

d) move your hand back and forth

in this case you would see a pulse series whose sum corresponds to a traveling wave

e) Advance a frame the movie

in this case the wave will be displaced a whole period to the right

the correct answer is b

f) move your hand faster

the waves will have a maximum fast, so they are closer

answer C

g) decrease wave frequency

Since the speed of the wave is a constant m ak, decreasing the frequency must increase the wavelength to keep the velocity constant.

the correct answer is b increases its wavelength

Answer:

[tex]F = -1.107*10^{-8} N\\|F| = 1.107*10^{-8} N[/tex]

Explanation:

[tex]q_1 = 4.5 * 10^{-9} C[/tex]

[tex]q_2 = -2.8 * 10^{-9} C[/tex]

The distance separating the two charges, r = 3.2 m

According to Coulomb's law of electrostatic attraction, the electrostatic force between the two charges can be given by the formula:

[tex]F = \frac{kq_{1} q_{2} }{r^2}[/tex]

Where [tex]k = 9.0 * 10^9 Nm^2/C^2[/tex]

[tex]F = \frac{9*10^9 * 4.5*10^{-9} * (-2.8*10^{-9}}{3.2^2} \\F = \frac{-113.4*10^{-9}}{10.24}\\F = -11.07 *10^{-9}\\F = -1.107*10^{-8}N[/tex]

Answer:

F = 1.1074 × [tex]10^{-8}[/tex] N

Explanation:

An electrostatic force is either a force or attraction or repulsion between two charges. It can be determined by:

F = [tex]\frac{kq_{1}q_{2} }{r^{2} }[/tex]

where: F is the force, k is a constant, [tex]q_{1}[/tex] is the first charge, [tex]q_{2}[/tex] is the second charge and r the distance between the charges.

Given that: k = 9 × [tex]10^{9}[/tex] N[tex]m^{2}[/tex][tex]C^{-2}[/tex], [tex]q_{1}[/tex] = 4.5 × [tex]10^{-9}[/tex]C, [tex]q_{2}[/tex] = -2.8 × [tex]10^{-9}[/tex]C and r = 3.2 m.

Then,

F = [tex]\frac{9*10^{9}*4.5*10^{-9}*2.8*10^{-9} }{3.2^{2} }[/tex]

 = [tex]\frac{1.134*10^{-7} }{10.24}[/tex]

 = 1.1074 × [tex]10^{-8}[/tex]

The electrostatic force exerted is 1.1074 × [tex]10^{-8}[/tex] N, and it is a force of attraction.

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic  fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is [tex]F_{ma}=BqV---(i)[/tex]

The given formula to find the electric force is [tex]F_{el}=qE---(ii)[/tex]

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

[tex]Bqv=qE\\\\v=\frac{E}{B}[/tex]

[tex]v=\frac{187500}{0.125} \\\\v=15\times10^5m/s[/tex]

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

[tex]F_{ce}=\frac{mv^2}{r}---(iii)[/tex]

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

[tex]Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br} \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg[/tex]

Therefore,  the particle's charge-to-mass ratio is [tex]958.1\times10^5C/kg[/tex]

b)

To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

Answer:

a.     F = 2.32*10^-18 N

b.     The force F is 2.59*10^11 times the weight of the electron

Explanation:

a. In order to calculate the magnitude of the force exerted on the electron you first calculate the acceleration of the electron, by using the following formula:

[tex]v^2=v_o^2+2ax[/tex]         (1)

v: final speed of the electron = 6.60*10^5 m/s

vo: initial speed of the electron = 4.00*10^5 m/s

a: acceleration of the electron = ?

x: distance traveled by the electron = 5.40cm = 0.054m

you solve the equation (2) for a and replace the values of the parameters:

[tex]a=\frac{v^2-v_o^2}{2x}=\frac{(6.60*10^5m/s)^2-(4.00*10^5m/s)^2}{2(0.054m)}\\\\a=2.55*10^{12}\frac{m}{s^2}[/tex]

Next, you use the second Newton law to calculate the force:

[tex]F=ma[/tex]

m: mass of the electron = 9.11*10^-31kg

[tex]F=(9.11*10^{-31}kg)(2.55*10^{12}m/s^2)=2.32*10^{-18}N[/tex]

The magnitude of the force exerted on the electron is 2.32*10^-18 N

b. The weight of the electron is given by:

[tex]F_g=mg=(9.11*10^{-31}kg)(9.8m/s^2)=8.92*10^{-30}N[/tex]

The quotient between the weight of the electron and the force F is:

[tex]\frac{F}{F_g}=\frac{2.32*10^{-18}N}{8.92*10^{-30}N}=2.59*10^{11}[/tex]

The force F is 2.59*10^11 times the weight of the electron

Answer:

-1486 KJ

Explanation:

The work done by an electric field on a charged body is:

W = ΔV * q

where ΔV = change in voltage

q = total charge

The total charge of Avogadro's number of electrons is:

6.0221409 * 10^(23) * -1.6023 * 10^(-19) = -9.65 * 10^(4)

The change in voltage, ΔV, is:

9.20 - (6.90) = 15.4

Therefore, the work done is:

W = -9.65 * 10^(4) * 15.4 = -1.486 * 10^6 J = -1486 KJ

The negative sign means that the motion of the electrons is opposite the electrostatic force.

Answer:

497.6 N

Explanation:

From the question,

The net force on the skydiver = weight of the skydiver- the resistive force of air

F' = W-F...................... Equation 1

Where W = weight of the skydiver, F = resistive force of air.

But,

W = mg................ Equation 2

Where m = mass of the skydiver, g = acceleration due to gravity.

Substitute equation 2 into equation 1

F' = mg-F............ Equation 3

Given: m = 87 kg, F = 355 N, g = 9.8 m/s²

Substitute these values into equation 3

F' = 87(9.8)-355

F' = 852.6-355

F' = 487.6 N

Answer:

498 N down

Explanation: