Two parallel plates 0.800 cm apart are equally and oppositely charged. An electron is released from rest at the surface of the negative plate and simultaneously a proton is released from rest at the surface of the positive plate.
How far from the negative plate is the point at which the electron and proton pass each other?

Express your answer with the appropriate units.

Answer:

the result is the quantization of __Energy__ of the particle

Explanation:

Answer:

power dissipated in the target is 100 W

Explanation:

given data

electrons = 1 mev = [tex]10^{6}[/tex] eV

1 eV = 1.6 × [tex]10^{-19}[/tex] J

current =  100 microampere = 100 × [tex]10^{-6}[/tex] A

solution

when energy of beam strike with 1 MeV so energy of electron is

E = e × v   ...................1

e is charge of electron and v is voltage

so put here value and we get voltage

v = 1 ÷ 1.6 × [tex]10^{-19}[/tex]

v =  [tex]10^{6}[/tex] volt

so power dissipated in target

P = voltage × current   ..............2

put here value

P =  [tex]10^{6}[/tex]  × 100 × [tex]10^{-6}[/tex]

P = 100 W

so power dissipated in the target is 100 W

Answer:

Explanation:

When a body is launched in air and allowed to fall freely under the influence of gravity, the motion experienced by the body is known as a projectile motion. The body is launched at a particular velocity and at an angle theta to the horizontal. The velocity of the body ca be resolved towards the horizontal component and the vertical component.

Along the horizontal Ux = Ucos(theta)

Along the vertical Uy = Ucos(theta)

Ux and Uy are the velocities of the body along the horizontal and vertical components respectively.

This means that the only factor connecting horizontal and vertical components of projectile motion is its velocity since we are able to calculate the velocity of the body along both components irrespective of its initial velocity.

Answer:

(a) V = 65.625 Volts

(b) V = 131.25 Volts

(c) V = 131.25 Volts

Explanation:

Recall that:

1) in a metal sphere the charges distribute uniformly around the surface, and the electric field inside the sphere is zero, and the potential is constant equal to:

[tex]V=k\frac{Q}{R}[/tex]

2) the electric potential outside of a charged metal sphere is the same as that of a charge of the same value located at the sphere's center:

[tex]V=k\frac{Q}{r}[/tex]

where k is the Coulomb constant ( [tex]9\,\,10^9\,\,\frac{N\,m^2}{C^2}[/tex] ), Q is the total charge of the sphere, R is the sphere's radius (0.24 m), and r is the distance at which the potential is calculated measured from the sphere's center.

Then, at a distance of:

(a) 48 cm = 0.48 m, the electric potential is:

[tex]V=k\frac{Q}{r}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.48} =65.625\,\,V[/tex]

(b) 24 cm = 0.24 m, - notice we are exactly at the sphere's surface - the electric potential is:

[tex]V=k\frac{Q}{r}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.24} =131.25\,\,V[/tex]

(c) 12 cm (notice we are inside the sphere, and therefore the potential is constant and the same as we calculated for the sphere's surface:

[tex]V=k\frac{Q}{R}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.24} =131.25\,\,V[/tex]

Answer:

c) a difference in electric potential

Explanation:

my insta: priscillamarquezz

Answer:

20,000 kg m/s

Explanation:

Given:

v₀ = 50 m/s

a = -5 m/s²

t = 2 s

Find: v

v = at + v₀

v = (-5 m/s²) (2 s) + (50 m/s)

v = 40 m/s

p = mv

p = (500 kg) (40 m/s)

p = 20,000 kg m/s

Answer:

Explanation:

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Answer:

The  length is  [tex]D = 5 \ cm[/tex]

Explanation:

From the question we are told  that

     The  length of the  hand is  [tex]l = 10.0 \ cm[/tex]

      The  angle at the hand is  held is  [tex]\theta = 30 ^o[/tex]

Generally resolving the length the length of the hand to it vertical component we obtain that the length of the shadow on the vertical wall is mathematically evaluated as

             [tex]D = l * sin(\theta )[/tex]

substituting values

             [tex]D = 10 * sin (30)[/tex]

             [tex]D = 5 \ cm[/tex]

Answer:

The two small charged spheres are now 4.382 cm apart

Explanation:

Given;

distance between the two small charged sphere, r = 7.59 cm

The force on each of the charged sphere can be calculated by applying Coulomb's law;

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

where;

F is the force on each sphere

q₁ and q₂ are the charges of the spheres

r is the distance between the spheres

[tex]F = \frac{kq_1q_2}{r^2} \\\\kq_1q_2 = Fr^2 \ \ (keep \ kq_1q_2 \ constant)\\\\F_1r_1^2 = F_2r_2^2\\\\r_2^2 = \frac{F_1r_1^2}{F_2} \\\\r_2 = \sqrt{\frac{F_1r_1^2}{F_2}} \\\\r_2 = r_1\sqrt{\frac{F_1}{F_2}}\\\\(r_1 = 7.59 \ cm, \ F_2 = 3F_1)\\\\r_2 = 7.59cm\sqrt{\frac{F_1}{3F_1}}\\\\r_2 = 7.59cm\sqrt{\frac{1}{3}}\\\\r_2 = 7.59cm *0.5773\\\\r_2 = 4.382 \ cm[/tex]

Therefore, the two small charged spheres are now 4.382 cm apart.

Explanation:

Mass and energy are closely related. Due to mass–energy equivalence, any object that has mass when stationary (called rest mass) also has an equivalent amount of energy whose form is called rest energy, and any additional energy (of any form) acquired by the object above that rest energy will increase the object's total mass just as it increases its total energy. For example, after heating an object, its increase in energy could be measured as a small increase in mass, with a sensitive enough scale.

Answer:

The atomic number

Explanation:

Answer:

  E = 1/9  E₀

Explanation:

In this exercise we are told that the electric field is Eo when the diameter of the balloon is D, the expression

we are asked to shorten the electric field when the diameter is 3D with the same eclectic charge

For this we can use the gauss law to find the field in the new diameter, for this we create a Gaussian surface in the form of a sphere

                Ф = ∫ E. dA = [tex]q_{int}[/tex] /ε₀

In this case the lines of the electric field and the radii of the sphere are parallel, therefore the scalar product is reduced to the algebraic product and the charge inside the sphere is the initial charge Q

               A = 4π r²

               E 4π r² = Q /ε₀

               E = 1 /4πε₀     Q / r²

the value of the indicated distance is 3 times the initial diamete

                 r  = 3 D / 2

we substitute

              E = 1/4 πε₀ Q (2/ 3D)²

for the initial conditions

              E₀ = 1 / 4πε₀ Q  (2/D)²

subtitled in the equation above

         E = 1/9  E₀

Answer:

1 yr = 24 * 3600 * 365 = 3.2 * 10E7 sec

C = 6 R = 1.5 * 10E8 * 6 = 9 * 10E8 km     circumference of orbit

v = C / t = 9 * 10E8 km / 3 * 10E7  sec = 30 km / sec = 18 mi/sec

Answer:

T= 2p√m/k

Explanation:

This is because the period of oscillation of the mass of spring system is directly proportional to the square root of the mass and it is inversely proportional to the square root of the spring constant.

The period of a mass on a spring is given by the equation

T=2π√m/k.

Where T is the period,

M is mass

K is spring constant.

An increase in mass in a spring increases the period of oscillation and decrease in mass decrease period of oscillation.

When there is the relationship between the Period (T) caused by the oscillation of the mass should be considered as the T= 2p√m/k.

The mass of the spring system with respect to period of oscillation should be directly proportional to the square root of the mass and it is inversely proportional to the square root of the spring constant.

So the following equation should be considered

T=2π√m/k.

Here,

T is the period,

M is mass

K is spring constant.

An increase in mass in a spring rises the period of oscillation and reduce in mass decrease period of oscillation.

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Answer:

E = 2750 J at h = 5 m

Explanation:

The gravitational potential energy is given by :

[tex]E=mgh[/tex]

In this case, m is the mass of swimmer is constant at every heights. So,

At h = 1 m, E = 550 J

[tex]550=m\times 10\times 1\\\\m=55\ kg[/tex]

So, at h = 5 m, gravitational potential energy is given by :

[tex]E=55\times 10\times 5\\\\E=2750\ J[/tex]

So, the correct option is (B).

Answer:

The horizontal force acting on m2 is F + 9.8m1

Explanation:

Given;

Block m1 on left of block m2

Make a sketch of this problem;

                         F →→→→→→→→→→→-------m1--------m2

Apply Newton's second law of motion;

F = ma

where;

m is the total mass of the body

a is the acceleration of the body

The horizontal force acting on block m2 is the force applied to block m1 and force due to weight of block m1

F₂ = F + W1

F₂ = F + m1g

F₂ = F + 9.8m1

Therefore, the horizontal force acting on m2 is F + 9.8m1

The force acting on the block of mass m₂ is  [tex]\frac{m_2F}{m_1+m_2}[/tex]

Given that there are two blocks of mass m₁ and m₂.

m₁ is on the left of block m₂. They are in contact with each other.

A force F is applied on m₁ to the right.

According to Newton's laws of motion:

The equation of motion of the blocks can be written as:

F = (m₁ + m₂)a

here, a is the acceleration.

so, acceleration:

a = F / (m₁ + m₂)

Now, the force acting on the block of mass m₂ is:

f = m₂a

[tex]f = \frac{m_2F}{m_1+m_2}[/tex]

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Answer:

I hope it is correct ✌️

Answer:

C. R^2

Explanation:

A cyclotron is a particle accelerator which employs the use of electric and magnetic fields for its functioning. It consists of two D shaped region called dees and the magnetic field present in the dee is responsible for making sure the charges follow the half-circle and then to a gap in between the dees.

R is denoted as the radius of the final orbit then the final particle energy is proportional to the radius of the two dees. This however translates to the energy being proportional to R^2.

Answer:

a) ±7.08×10⁻⁷ C/m²

b) 1.77×10⁻⁷ C

Explanation:

For a conductor,

σ = ±Eε₀,

where σ is the charge density,

E is the electric field,

and ε₀ is the permittivity of space.

a)

σ = ±Eε₀

σ = ±(8×10⁴ N/C) (8.85×10⁻¹² F/m)

σ = ±7.08×10⁻⁷ C/m²

b)

σ = q/A

7.08×10⁻⁷ C/m² = q / (0.5 m)²

q = 1.77×10⁻⁷ C

Answer:

Explanation:

Mass ( m ) = 2 kg

Speed of the object (v) = 6 metre per second

Kinetic energy =?

Now,

We have,

Kinetic Energy = [tex] \frac{1}{2} \times m \times {v}^{2} [/tex]

Plugging the values,

[tex] = \frac{1}{2} \times 2 \times {(6)}^{2} [/tex]

Reduce the numbers with Greatest Common Factor 2

[tex] = {(6)}^{2} [/tex]

Calculate

[tex] = 36 \: joule[/tex]

Hope this helps...

Good luck on your assignment...

The Kinetic energy of the object will be "36 joules".

The excess energy of moving can be observed as that of the movement of an object, component, as well as the group of components. There would never be a negative (-) amount of kinetic energy.

According to the question,

Mass of object, m = 2 kg

Speed of object, v = 6 m/s

As we know the formula,

→ Kinetic energy (K.E),

= [tex]\frac{1}{2}[/tex] × m × v²

By substituting the values, we get

= [tex]\frac{1}{2}[/tex] × 2 × (6)²

=  [tex]\frac{1}{2}[/tex] × 2 × 36

= 36 joule

Thus the above answer is appropriate.

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Answer:

8 ft/s

Explanation:

This is a straight forward question without much ado.

It is given from the question that she walks with a speed of 8 ft/s

The correct answer is B. thrill and adventure seeking, experience seeking, disinhibition, and susceptibility to boredom

Explanation:

Marvin Zuckerman was an important American Psychologists mainly known for his research about personality and the creation of a model to study this aspect of human psychology. This model purposes five factors define personality, these are the thrill and adventure-seeking that involves seeking for adventures and danger; experience seeking that implies a strong interest in participating in new activities; disinhibition that implies being open and extrovert; and susceptibility to boredom that implies avoiding boredom or repetition. Thus, option B correctly describes the characteristics used in Zuckerman's test.

Answer:

D is determined by distance between the telescopes.

Explanation:

Answer:

  I = 2.18 10⁻⁴ W / m²

Explanation:

The two-slit interference pattern is described by the expression for constructive interference.

             d sin θ = m λ

If we also want to know the distribution of intensities we must perform the su of the electric field of the two waves, and find the intensity as the square of the velvet field, obtaining the expression

              I = I_max cos² ((π d /λ L) y)

where d is the separation of the slits, λ  the wavelength, L the distance to the screen e and the separation of the interference line with respect to the central maximum

let's reduce the magnitudes to the SI system

λ  = 583 nm = 583 10⁻⁹ m

L = 75.0 cm = 75.0 10⁻² m

d = 0.640 mm = 0.640 10⁻³ m

y = 0.900 mm = 0.900 10⁻³ m

let's calculate the intensity of this line

        I = 5 10⁻⁴ cos² ((π 0.640 10⁻³ /583 10⁻⁹ 0.75 10⁻²) 0.900 10⁻³)

        I = 5 10⁻⁴ cos2 (413.84)

         I = 5 10⁻⁴ 0.435

        I = 2.18 10⁻⁴ W / m²

Answer:

5000N

Explanation:

According to Newton's second law of motion, the net force (∑F) acting on a body is the product of the mass (m) of the body and the acceleration (a) of the body caused by the force. i.e

∑F = m x a             -------------(i)

From the question, the net force is the combined effect of the thrust (F) and the friction force (Fₓ). i.e

∑F = F + Fₓ             -------------(ii)

Where;

Fₓ = -200N       [negative sign because the friction force opposes motion]

Combine equations(i) and (ii) together to get;

F + Fₓ = m x a

F = ma - Fₓ         -------------(iii)

Where;

m = mass of car = 1200kg

a = acceleration of the car = 4m/s²

Now substitute the values of m, a and Fₓ into equation (iii) as follows;

F = (1200 x 4) - (-200)

F = 4800 + 200

F = 5000N

Therefore, the force the thrust is providing is 5000N

Answer:

A.  Z = 185.87Ω

B.  I  =  0.16A

C.  V = 1mV

D.  VL = 68.8V

E.  Ф = 30.59°

Explanation:

A. The impedance of a RL circuit is given by the following formula:

[tex]Z=\sqrt{R^2+\omega^2L^2}[/tex]       (1)

R: resistance of the circuit = 160-Ω

w: angular frequency = 220 rad/s

L: inductance of the circuit = 0.430H

You replace in the equation (1):

[tex]Z=\sqrt{(160\Omega)^2+(220rad/s)^2(0.430H)^2}=185.87\Omega[/tex]

The impedance of the circuit is 185.87Ω

B. The current amplitude is:

[tex]I=\frac{V}{Z}[/tex]                     (2)

V: voltage amplitude = 30.0V

[tex]I=\frac{30.0V}{185.87\Omega}=0.16A[/tex]

The current amplitude is 0.16A

C. The current I is the same for each component of the circuit. Then, the voltage in the resistor is:

[tex]V=\frac{I}{R}=\frac{0.16A}{160\Omega}=1*10^{-3}V=1mV[/tex]            (3)

D. The voltage across the inductor is:

[tex]V_L=L\frac{dI}{dt}=L\frac{d(Icos(\omega t))}{dt}=-LIsin(\omega t)\\\\V_L=-(0.430H)(160\Omega)sin(220 t)=68.8sin(220t)\\\\V_L_{max}=68.8V[/tex]

E. The phase difference is given by:

[tex]\phi=tan^{-1}(\frac{\omega L}{R})=tan^{-1}(\frac{(220rad/s)(0.430H)}{160\Omega})\\\\\phi=30.59\°[/tex]

Answer:

The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.

Explanation:

Let suppose that lion and Thomson's gazelle are running at constant speed before and after collision and that collision is entirely inelastic. Given the absence of external force, the Principle of Momentum Conservation is applied such that:

[tex]\vec p_{L} + \vec p_{G} = \vec p_{F}[/tex]

Where:

[tex]\vec p_{L}[/tex] - Linear momentum of the lion, measured in kilograms-meters per second.

[tex]\vec p_{G}[/tex] - Linear momentum of the Thomson's gazelle, measured in kilograms-meters per second.

[tex]\vec p_{F}[/tex] - Linear momentum of the lion-Thomson's gazelle, measured in kilograms-meters per second.

After using the definition of momentum, the system is expanded:

[tex]m_{L}\cdot \vec v_{L} + m_{G}\cdot \vec v_{G} = (m_{L} + m_{G})\cdot \vec v_{F}[/tex]

Vectorially speaking, the final velocity of the lion-gazelle system is:

[tex]\vec v_{F} = \frac{m_{L}}{m_{L}+m_{G}}\cdot \vec v_{L} + \frac{m_{G}}{m_{L}+m_{G}}\cdot \vec v_{G}[/tex]

Where:

[tex]m_{L}[/tex], [tex]m_{G}[/tex] - Masses of the lion and the Thomson's gazelle, respectively. Measured in kilograms.

[tex]\vec v_{L}[/tex], [tex]\vec v_{G}[/tex], [tex]\vec v_{F}[/tex] - Velocities of the lion, Thomson's gazelle and the lion-gazelle system. respectively. Measured in meters per second.

If [tex]m_{L} = 177\,kg[/tex], [tex]m_{G} = 32\,kg[/tex], [tex]\vec v_{L} = 81.8\cdot j\,\left[\frac{km}{h} \right][/tex] and [tex]\vec v_{G} = 59.0\cdot i\,\left[\frac{km}{h} \right][/tex], the final velocity of the lion-gazelle system is:

[tex]\vec v_{F} = \frac{177\,kg}{177\,kg+32\,kg}\cdot \left(81.8\cdot j\right)\,\left[\frac{km}{h} \right] + \frac{32\,kg}{177\,kg+32\,kg}\cdot \left(59.0\cdot i\right)\,\left[\frac{km}{h} \right][/tex]

[tex]\vec v_{F} = 9.033\cdot i + 69.276\cdot j\,\left[\frac{km}{h} \right][/tex]

The speed of the system is the magnitude of the velocity vector, which can be found by means of the Pythagorean theorem:

[tex]\|\vec v_{F}\| = \sqrt{\left(9.033\frac{km}{h} \right)^{2}+\left(69.276\frac{km}{h} \right)^{2}}[/tex]

[tex]\|\vec v_{F}\| \approx 69.862\,\frac{km}{h}[/tex]

The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.

A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12t² + 35 t + 1

Answer:

Velocity = 131 m/s

Speed = 131 m/s

Explanation:

Equation of motion, s = f(t) = 12t² + 35 t + 1

To get velocity of the particle, let us find the first derivative of s

v (t) = ds/dt = 24t + 35

At t = 4

v(4) = 24(4) + 35

v(4) = 131 m/s

Speed is the magnitude of velocity. Since the velocity is already positive, speed is also 131 m/s

Answer:

(a) force is 1066.56N

Explanation:

(a) MV²/R

Answer:

2000 N

Explanation:

20 km/h = 5.56 m/s

100 km/h = 27.78 m/s

F = ma

F = m Δv/Δt

F = (200 kg + 80 kg) (27.78 m/s − 5.56 m/s) / (3 s)

F = 2074 N

Rounded to one significant figure, the force is 2000 N.

Answer:

F = umg where u is coefficient of dynamic friction

Explanation:

F = 0.4 x 80 x 9.81 = 313.92 N