A small, dense ball is launched from ground level at an angle of 50° above the horizontal. The ballâs initial speed is 22 m/s, and it lands on a hard, level surface at the same height from which it was launched. The ball then bounces and reaches a height of 75% its peak height it achieved at launch. Assume air resistance is negligible.(a) Find the maximum height reached by the ball during its first parabolic arc. (b) Find the distance between the launch point to where the ball lands the first time. (c) Find the distance between the launch point to where the ball lands the second time?

Answers

Answer 1

Answer:

a) Hmax =   10.86 m

b) Rx = 48.64 m

c) Rx¹ = 36.48m

Explanation:

Given that Ф = 50°

v₀ = 22 m/s

a)

hmax = v₀²sin²Ф / 2g

hmax = 22² × sin²50  / 2 × 9.8

hmax =   14.49 m

Hmax = 75% of hmax

Hmax = 0.75 × 14.49

Hmax = 10.86 m

the maximum height reached by the ball during its first parabolic arc is 10.86 m

 

b)

Rx = v₀²sin2Ф / g

Rx = 22² × sin 100  / 9.8

Rx = 48.64 m

the distance between the launch point to where the ball lands the first time is 48.64 m

c)

Rx¹ = 75% 0f Rx

Rx¹ = 0.75 × 48.64

Rx¹ = 36.48m

the distance between the launch point to where the ball lands the second time is 36.48m  


Related Questions

With the LED illuminated, flip the direction of the LED on the breadboard (notch on top now), and then flip it back to its original position (notch on bottom). What does your observation tell you about diodes (and LEDs)?

Answers

Explanation:

The full form of LED is Light Emitting Diode. It is a semi conductor source of light which when electrons passes or flows through them, it emits light. In the semi conductor, the electrons recombine with the electron holes to release high amount of energy. This energy is called photon.

When the LED direction is flipped in the breadboard and back to the original position, the LEDs are forced to be grounded. They will not be able to function without being grounded.

While the block hovers in place, is the density of the block (top left) or the density of the liquid (bottom center) greater?

Answers

Answer:

for the body to float, the density of the body must be less than or equal to the density of the liquid.

Explanation:

For a block to float in a liquid, the thrust of the liquid must be greater than or equal to the weight of the block.

Weight is

        W = mg

let's use the concept of density

        ρ_body = m / V

        m = ρ_body V

        W = ρ_body V g

The thrust of the body is given by Archimedes' law

        B = ρ_liquid g V_liquid

 

as the body floats the submerged volume of the liquid is less than or equal to the volume of the block

       ρ_body V g = ρ_liquid g V_liquid

     

       ρ_body = ρ liquid Vliquido / V_body

As we can see, for the body to float, the density of the body must be less than or equal to the density of the liquid.

Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 6.00 m away from the slits.
Which laser has its first maximum closer to the central maximum?
What is the distance Image for Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas las between the first maxima (on the same side of the central maximum) of the two patterns?
Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas las= ______ m
What is the distance Deltay_max-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?
Deltay_max-min = ______ m

Answers

Answer:

A)   Therefore laser1 has the maximum closest to the central maximum

B) Δₓ = 0.8

Explanation:

A) The expression for the constructive interference of a double slit is

           d sin θ  = m λ

let's use trigonometry to find the angle

          tan θ = y / L

in interference phenomena the angles are small

         tan θ = sin θ / cos θ = sin θ

         sin θ = y / L

we subjugate

         d y / L = m λ

         y = m λ L / d

let's apply this equation for each case

a) Lares 1 has a wavelength λ₁ = d / 20, the screen is at L = 6.00 m

they ask us for the first axiom m = 1,

let's calculate

           y₁ = 1 (d / 20) 6.00 / d

           y₁ = 0.3

Laser 2, λ₂ = d / 15

            λ₂ = 1 (d / 15) 6.00 / d

           λ₂ = 0.4

Therefore laser1 has the maximum closest to the central maximum

b) let's find the distance of each requested value

second maximum m = 2 of laser 1

            yi '= 2 (d / 20) 6 / d

            y1 '= 0.6

3rd minimum of laser 2

the expression for destructive interference is

               d sinθ = (m + 1/2) lam

               y = (m ) λ L / d

in this case m = 3

let's calculate

              y2 '= (3+0.5) (d / 15) 6 / d

              y2 '=21/15

They ask us for the dalt of these interference

            Δₓ = y3 -y2'          

           Δₓ = 21/15 - 0.6

           Δₓ = 0.8

Moving mirror M2 of a Michelson interferometer a distance of 90 μm causes 470 bright-dark-bright fringe shifts.What is the wavelength of the light?

Answers

Answer:

Wavelength, [tex]\lambda=382.9\ nm[/tex]

Explanation:

It is given that,

Distance moved by mirror in Michelson interferometer is 90 μm

Number of bright fringe shift = 470

We need to find the wavelength of the light.

For Michelson interferometer experiment,

[tex]2d=m\lambda[/tex]

here, [tex]\lambda[/tex] is the wavelength of the light

[tex]\lambda=\dfrac{2d}{m}\\\\\lambda=\dfrac{2\times 90\times 10^{-6}}{470}\\\\\lambda=3.829\times 10^{-7}\ m\\\\\lambda=382.9\ nm[/tex]

So, the wavelength of the light is 382.9 nm.

The wavelength of the light should be 382.9nm.

Calculation of the wavelength of the light:

Since

Distance moved by the mirror in Michelson interferometer is 90 μm

And, Number of bright fringe shift = 470

So

we know that

2d = m*wavelength

wavelength = 2*90*10^-6/470

= 3.829*10^-7m

= 382.9nm

hence, The wavelength of the light should be 382.9nm.

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Cousin Throckmorton is playing with the clothesline. One end of the clothesline is attached to a vertical post. Throcky holds the other end loosely in his hand, so that the speed of waves on the clothesline is a relatively slow 0.700 m/s . He finds several frequencies at which he can oscillate his end of the clothesline so that a light clothespin 40.0 cm from the post doesn't move. What are these frequencies?

Answers

Answer:

The  frequencies are  [tex]f_n = n (0.875 )[/tex]

Explanation:

From the question we are told that

   The speed of the wave is  [tex]v = 0.700 \ m/s[/tex]

   The  length of vibrating  clothesline is  [tex]L = 40.0 \ cm = 0.4 \ m[/tex]

Generally the fundamental frequency is  mathematically represented as

        [tex]f = \frac{v}{2 L }[/tex]

=>     [tex]f = \frac{ 0.700 }{2 * 0.4 }[/tex]

=>     [tex]f = 0.875 \ Hz[/tex]

Now  this other frequencies of vibration experience by the clotheslines are know as harmonics and they are obtained by integer multiple of  the fundamental frequency

So  

   The  frequencies are mathematically represented as

       [tex]f_n = n * f[/tex]

=>     [tex]f_n = n (0.875 )[/tex]

Where  n  =  1, 2, 3 ....

       

You are looking down on a N = 17 turn coil in a magnetic field B = 0.5 T which points directly down into the screen. If the diameter of the coil d = 3.8 cm, and the field goes to zero in t = 0.24 seconds, what would be the magnitude of the voltage (in Volts) and direction of the induced current? Indicate the direction of the current by the sign in front of your voltage: counterclockwise is positive, clockwise is negative.

Answers

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The  voltage is [tex]\epsilon = 0.40163 \ V[/tex]

The  direction of the induced current is clockwise

Explanation:

From the question we are told that

   The number of turns is  N  = 17

     The  magnetic field is  [tex]B_2 = 0.5 \ T[/tex]

     The diameter is  [tex]d = 3.8 \ cm = 0.038 \ m[/tex]

      The  time interval is [tex]dt = 0.24 \ s[/tex]

The induce emf is mathematically represented as

       [tex]\epsilon = - N \frac{d\phi}{dt}[/tex]

       [tex]\epsilon = - N \frac{d ( B_2 - B_1 ) A }{dt}[/tex]

Here [tex]B_1[/tex] is the magnetic field experienced by the coil before entering the magnetic field given in the question  i.e  [tex]B_1 = 0[/tex]

Here the negative sign show that the induced voltage is moving in a direction opposite to the change magnetic flux

  The  area is mathematically represented as

      [tex]A = \pi \frac{d^2}{4}[/tex]

=>  [tex]A = 3.142 * \frac{ 0.038^2 }{4}[/tex]

=>   [tex]A = 0.01134 \ m^2[/tex]

Hence

     [tex]\epsilon = - 17 * \frac{ 0.5 * 0.01134 }{ 0.24}[/tex]

     [tex]\epsilon = 0.40163 \ V[/tex]

The  direction of the induced current is the same as that of induced voltage

    Thus the direction is clockwise

A 26.5-mW laser beam of diameter 1.88 mm is reflected at normal incidence by a perfectly reflecting mirror. Calculate the radiation pressure on the mirror.

Answers

Answer:

Explanation:

Area= pier^2

=( 1.88/2)^2= 0.0000000314m^2

Intensity= 0.0265/0.000000314

= 7962W/m^2

Pressure= 2*7962)/345= 5.13*10^-5pa

Imagine you derive the following expression by analyzing the physics of a particular system: M= (mv2r)(mGr2). Simplify the expression for M using the techniques mentioned above.

Answers

Answer:

The simplified expression is [tex]M = \frac{v^2 r}{G}[/tex]

Explanation:

From the question we are told that  

     [tex]M = \frac{ \frac{m v^2}{r} }{\frac{ mG}{r^2 } }[/tex]

So simplifying we have

    [tex]M = \frac{m v^2}{r} * \frac{r^2 }{ mG }[/tex]

    [tex]M = \frac{v^2 r}{G}[/tex]

Thus the simplified formula is [tex]M = \frac{v^2 r}{G}[/tex]

A cellist tunes the C string of her instrument to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.590 m long and has a mass of 15.0 g.
(a) Calculate the wavelength corresponding to this fundamental frequency?
(b) Calculate the wave speed.
(c) With what tension must the musician stretch the string?

Answers

Answer:

a

[tex]\lambda = 1.18 \ m[/tex]

b

[tex]v = 77.172 \ m/s[/tex]

c

[tex]T = 151.41 \ N[/tex]

Explanation:

From the question we are told that

   The frequency is  [tex]f = 65.4 \ Hz[/tex]

   The  length of the vibrating string is  [tex]L = 0.590 \ m[/tex]

   The  mass is  [tex]m = 15.0 \ g = 0.015 \ kg[/tex]

Generally the wavelength is mathematically represented as

           [tex]\lambda = 2 * L[/tex]

=>        [tex]\lambda = 2 * 0.590[/tex]

=>         [tex]\lambda = 1.18 \ m[/tex]

Generally the wave speed is  

          [tex]v = \lambda * f[/tex]

=>       [tex]v = 1.18 * 65.4[/tex]

=>       [tex]v = 77.172 \ m/s[/tex]

Generally the tension on the wire is mathematically represented as

        [tex]T = v^2 * \frac{ m }{L }[/tex]

=>      [tex]T = 77.172 ^2 * \frac{ 0.015 }{0.590}[/tex]

=>      [tex]T = 151.41 \ N[/tex]

a burning piece of wood converts ____ energy into ____ energy​

Answers

Answer:

a burning piece of converts Chemical energy into Heat(Thermal) and Light energy

A box at rest on a ramp at an incline of 22°. The normal force on the box is 538 N

Answers

Answer:

F = 580.25 N

Explanation:

It is given that,

A box at rest on a ramp at an incline of 22°

The normal force on the box is 538 N

We need to find the gravitational force on the box.

The force of gravity acting on an object is equal to its weight. So,

F = mg

But here the box is incline at an angle of 22 degrees

It means that mg will resolve in rectangular components such that, normal force is :

[tex]N=mg\cos\theta\\\\mg=F=\dfrac{N}{\cos\theta}\\\\F=\dfrac{538}{\cos22}\\\\F=580.25\ N[/tex]

So, the gravitational force on the box is 580.25 N.

An electron is projected with horizontal speed 105m / s in a downwardly directed 304N / C electric field. Find the vertical position (in m) after 2.0 μs.

Answers

Answer:

-107 m

Explanation:

Sum of forces in the y direction:

∑F = ma

-qE = ma

a = -qE/m

a = -(1.60×10⁻¹⁹ C) (304 N/C) / (9.11×10⁻³¹ kg)

a = -53.4×10¹² m/s²

Given in the y direction:

v₀ = 0 m/s

a = -53.4×10¹² m/s²

t = 2×10⁻⁶ s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (2×10⁻⁶ s) + ½ (-53.4×10¹² m/s²) (2×10⁻⁶ s)²

Δy = -107 m

During the class prize-giving ceremony, Anand clapped his hands hard while Kumar clapped his hands softly. Everybody could hear Anand's clapping while only a few could hear Kumar's clapping. This was because the sound produced by Anand was of____________.
A - higher pitch
B - lower frequency
C - higher volume
D- lower pitch

Answers

Answer:

C - higher volume

Explanation:

The pitch or frequency of sound that an object can produce depends upon its size and configuration . The shape of hand of all are same so the frequency of sound produced by hands of all will be almost same . Hence frequency of sound produced by the hands of Anand and Kumar would have been almost the same .

But the intensity of sound produced by them would have been different . Intensity represents energy a sound carries . Hard hitting clap will produce sound of higher intensity . Intensity of sound is also called high volume sound . So Kumar's clap will carry greater energy and hence greater volume of sound .

The acceleration of an object is always in the direction of the net force acting on it.

Answers

Answer:

yes

Explanation:

An object which moves in a circle is accelerating. Accelerations are caused by an unbalanced or net force. The net force is always in the same direction as the acceleration. For objects moving in circles at constant speed, the net force is directed towards the center of the circle about which the object moves.

the kinetic energy of a body executing simple harmonic motion of amplitude A is equal to the potential energy when its displacement is​

Answers

Answer:

My answer to the question is "when its displacement is zero.

A bungee cord stretches 25 meters and has a spring constant of 140 N/m. How much energy is stored in the bungee

Answers

Answer:

87.5 kJ

Explanation:

The potential energy stored in a stretch spring is given by the expression below

PE = Work = force * distance

So:

[tex]PE = (kx) * x[/tex]

This then simplifies to:

[tex]PE = kx^2[/tex]

where k is the spring constant= 140 N/m

Given that x= 25 m

Substituting our data into the expression for P.E stored we have

[tex]PE = 140*25^2\\\\PE= 140*625\\\\PE= 87500 J[/tex]

Hence the energy stored in the spring is 87.5kJ

A series RLC circuit connected across an ac voltage source has minimum current flowing through the circuit when operating at the resonant frequency. is this statement true or false and why?

Answers

Answer:

False

Explanation:

This is because Since the current flowing through a series resonance circuit is the product of voltage divided by impedance, at resonance the impedance, Z is at its minimum value, ( =R ). So, the circuit current at this frequency will be at its maximum value of V/R

An object is moving with constant non-zero velocity. Which of the following statements about it must be true?
(A) A constant force is being applied to it in the direction of motion.
(B) A constant force is being applied to it in the direction opposite of motion.
(C) A constant force is being applied to it perpendicular to the direction of motion.
(D) The net force on the object is zero.
(E) Its acceleration is in the same direction as its velocity.

Answers

Answer:

The net force on the object is zero.

Explanation:

An object is moving with constant non-zero velocity. If velocity is constant, it means that the change in velocity is equal to 0. As a result, acceleration of the object is equal to 0. Net force is the product of mass and acceleration. Hence, the correct option is (d) "The net force on the object is zero".

For a constant non-zero velocity the net force on the object is zero.

Velocity is defined as the change in displacement per change in time of motion.

[tex]v = \frac{\Delta x}{\Delta t}= \frac{x_2 - x_1}{t_2 - t_1}[/tex]

where;

Δx is the change in displacement

Δt is the change in time motion

When the change in the displacement is same for equal time interval, the resulting velocity will be constant but non-zero in magnitude.

Also, when the net force acting on an object is zero, the object will move with a constant velocity.

[tex]\Sigma F_x = 0[/tex]

Thus, for a constant non-zero velocity the net force on the object is zero.

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: The maximum theoretical efficiency of a Carnot engine operating between reservoirs at the steam point and at room temperature is about A :

Answers

Answer:

The value is   [tex]\eta = 0.2145[/tex]  or  21.45%

Explanation:

From the question we are told that

    The first reservoir is at steam point  [tex]T_s = 100^o C = 100 + 273 = 373 \ K[/tex]  

    The  second reservoir is at room temperature [tex]T_r = 20^o C = 293 \ K[/tex]

Generally the  maximum theoretical efficiency of a Carnot engine  is mathematically evaluated as

     [tex]\eta = 1- \frac{T_r}{T_s}[/tex]

=>    [tex]1 - \frac{ 293}{373}[/tex]

=>    [tex]\eta = 0.2145[/tex]

A superball has a coefficient of restitution of .90. The ball is dropped from an intitial height of 1.60m.
a) if the ball is allowed to bounce 3 times, how high will it rebound after the third bounce?
b) If the ball has a mass of 48g, calculate the amount of the ball's orignal energy that was lost in the three impacts with the floor.

Answers

Answer:

a

  [tex]H_3 = 0.85 \ m[/tex]

b

 [tex]PE = 0.3528 \ J[/tex]

Explanation:

From the question we are told that  

   The  coefficient of resolution is  [tex]C_r = 0.90[/tex]

    The  initial height is  [tex]H_i = 1.60 \ m[/tex]

   

Considering question a

  The  number of times the ball bounced is [tex]k = 3[/tex]

    Generally the height attained after the 3rd bounce is mathematically represented as

       [tex]H_3 = H_i * (C_r)^{2 * k }[/tex]

 =>    [tex]H_3 = 1.60 * 0.90^{2 * 3 }[/tex]

=>    [tex]H_3 = 0.85 \ m[/tex]

Considering question b

The  mass is  m =  48 g  =  0.048  kg

Generally the amount of potential energy that was lost is mathematically represented as

        [tex]PE = mg [ H_i - H_3 ][/tex]

=>    [tex]PE = 0.048 * 9.8 [1.60 - 0.85 ][/tex]

=>     [tex]PE = 0.3528 \ J[/tex]

A helicopter blade spins at exactly 180 revolutions per minute. Its tip is 10.00 m from the center of rotation. What is its average velocity over one revolution

Answers

Answer:

5.8E-3m/s

Explanation:

Using

V= d/t

V= velocity

d= distance

t= time

But d= 2πr

But 180rev= 1min

So 1min/180= 60s/180

So

Vavr= 2π(10m)/180*60

=5.8E-3m/s

How many (whole number of) 87 kg people
can safely occupy an elevator that can hold a
maximum mass of exactly 1 metric ton? A
metric ton is 1.000 x 103 kg.
Answer in units of people.

Answers

Answer:

11

Explanation:

1. You are going to be rounding down.

2. change the metric ton to kg.

1.000 * 10^3 kg = 1000 kg

1000 / 87 = 11.49 = 11 people

In what way is the study of psychology similar to other “hard sciences”? A. use of the scientific method B. use of memory C. use of emotion D. use of behavior

Answers

Answer:

The correct option is a

Explanation:

Psychology can be defined as the "scientific study" of the mind and behavior of humans and other animals. It involves the use of scientific methods in determining or understanding the social, physiological and biological behaviors in different groups or individual organisms

The correct option is a

The following information should be considered:

Psychology refers to the "scientific study" of the mind & behavior of humans and other animals. It  includes the use of scientific methods in measurement or understanding the social, physiological ,and biological behaviors in different groups or individual organisms.

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A standard gold bar stored at Fort Knox, Kentucky, is 7.00 inches long, 3.63 inches wide, and 1.75 inches tall. Gold has a density of 19,300 kg/m3. What is the mass of such a gold bar?

Answers

Answer:

14.1 kg

Explanation:

Given:

Length=7.00inches

Width=3.63 inches

Height=1.75 inches

density = 19,300 kg/m3.

We can convert the given parameters to metre for unit consistency

But we know 1 inches= 0.0254 metre

✓Then Length l=7.00inches

=7×0.0254 metre=0.1778m

✓Width w =3.63 inches

==3.63 ×0.0254 metre=0.092m

✓Height h =1.75 inches

=1.75 ×0.0254 metre=0.0445 m

But Mass= density × volume

Volume= Length× width×height

Mass= density× Length× width×height

= 19300kg/m³×0.1778×0.0922×0.0445

=14.1 kg

Therefore, the mass of the gold bar is 14.1 kg

The mass of such a gold bar is of 13.89 kg.

Given data:

The length of gold bar is, [tex]L=7.00 \;\rm in =7.00 \times 0.0254=0.1778 \;\rm m[/tex].

The width of gold bar is, [tex]w= 3.63 \;\rm in =3.63 \times 0.0254 = 0.092 \;\rm m[/tex].

The height if gold bar is, [tex]h = 1.75 \;\rm in =0.044 \;\rm m[/tex].

The density of the gold bar is,  [tex]\rho =19,300 \;\rm kg/m^{3}[/tex].

The given problem is based on the concept of density. The density of any substance is equal to the ratio of mass and volume. Considering the gold bar to rectangular shape, the volume of gold bar is calculated as,

[tex]V= L \times w \times h\\\\V = 0.1778 \times 0.092 \times 0.044\\\\V=7.197 \times 10^{-4} \;\rm m^{3}[/tex]

Now, use the formula of density to calculate the mass of gold bar as,

[tex]\rho =\dfrac{m}{V}\\\\m = \rho \times V\\\\m = 19300 \times (7.197 \times 10^{-4})\\\\m= 13.89 \;\rm kg[/tex]

Thus, we can conclude that the mass of such a gold bar is of 13.89 kg.

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Which type of graph uses wedges to show the amount of data points in a
category?
A. Circle graph
O B. Line graph
O C. Box-and-whisker plot
D. Stemplot

Answers

A. Circle graph. A pie chart is a circular graph that shows the relative contribution that different categories contribute to an overall total. A wedge of the circle represents each category's contribution, such that the graph resembles a pie that has been cut into different sized slices.

The type of graph that uses wedges to show the amount of data points in a category is a Circle graph, also known as a Pie chart. In a Pie chart, the circle represents the whole data set, and each wedge represents a specific category or data point.

In a Circle graph, the entire circle represents the total amount or the whole data set being analyzed. The circle is divided into wedges or slices, with each wedge representing a specific category or data point. The size of each wedge is proportional to the amount or percentage of data points that belong to that category.

The wedges are typically labeled with the category they represent and are often accompanied by a numerical value or percentage to indicate the exact proportion of data points in that category. The labels and values help the reader interpret the graph and understand the distribution of data across different categories.

Therefore, the correct answer is  A. Circle graph.

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#SPJ2

What is charging by contact ​

Answers

Answer:

Charging by contact is when the charged object is brought near but never contacted to the object being charged

Explanation:

Unpolarized light with intensity 370W/m2 passes first through a polarizing filter with its axis vertical, then through a second polarizing filter. It emerges from the second filter with intensity 121W/m2 .
What is the angle from vertical of the axis of the second polarizing filter?
Express your answer to two significant figures and include the appropriate units

Answers

Answer:

Answer:

36°

Explanation:

Using

I= (1/2Io) cos²စ

Then Substituting

121W/m² = 1/2*370W/m²cos²စ

0.8086= cosစ

Cos^-1 0.8086= 36°

A boy weighing 70 kg is standing on the sand. Calculate the pressure on if you are resting on your feet, whose surfaces add up to 0.035 m2 and if you are resting on diving fins of 0.300 × 0.42 m each

Answers

Explanation:

Pressure = force / area

If you are standing on your feet:

P = (70 kg × 10 m/s²) / (0.035 m²)

P = 20,000 Pa

If you are standing on diving fins:

P = (70 kg × 10 m/s²) / (2 × 0.30 m × 0.42 m)

P ≈ 2,800 Pa

A remote-controlled toy car travels at a constant speed of 3 m/s. What distance does the car travel in 6 seconds?

Answers

Answer:

speed=d/t.

3=d/6.

d=3×6=18m

An electron as q = 1.602 * 10-19 C is placed .03m away from spherical object with a net charge of -7.2 C.
A. What is the force exerted on the electron?
B. How strong is the electric field at the electron’s location?
C. How much work would be done on the electron if it was moved so that it’s .001m away from the sphere?
D. Now replace the electron with a positron (q = +1.602 × 10-19C). Explain.

Answers

Answer:

See explanation

Explanation:

The electric force exerted on the electron is given by Coulomb's law;

F= KqQ/r^2

F= force on the electron

K= constant of Coulomb's law

q= charge on the electron

Q=charge on the spherical object

r= distance between the charges

F= 9×10^9 × -1.602 × 10^-19 × (-7.2)/(0.03)^2

F= 103.8 × 10^-10/9×10^-4

F= 11.53 ×10^-6

F= 1.153 ×10^-5 N

The force in this case is repulsive

E= F/q

E= 1.153 ×10^-5 N/ 1.602 * 10-19 C

E= 0.719 × 10^14

E= 7.19 × 10^13 NC-1

The electro field around the electron is very strong.

c)

V= Kq/r

V= 9×10^9 × 1.602 × 10^-19/0.001

V= 14418 × 10^-10

V=1.4418 ×10^-6 V

W= qV

W= 1.602×10^-19 C × 1.4418 ×10^-6 V

W= 2.31 ×10^-25 J

Replacing the electron with a positron

F= 9×10^9 × 1.602 × 10^-19 × (-7.2)/(0.03)^2

F= 103.8 × 10^-10/9×10^-4

F= 11.53 ×10^-6

F= -1.153 ×10^-5 N

The force in this case is attractive

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