Block 1, of mass m1 = 2.50 kg , moves along a frictionless air track with speed v1 = 27.0 m/s. It collides with block 2, of mass m2 = 33.0 kg , which was initially at rest. The blocks stick together after the collision.A. Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically.B. Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically.C. what is the change deltaK= Kfinal- K initial in the two block systems kinetic energy due to the collision ? Express your answer numerically in joules.

Answer:

Explanation:

According to snell's law which states that the ratio of the sin of incidence (i) to the angle of refraction(n) is a constant for a given pair of media.

sini/sinr = n

n is the constant = refractive index

Since the diver shines light up to the surface of a flat glass-bottomed boat, the refractive index n = nw/ng

nw is the refractive index of water and ng is that of glass

sini/sinr = nw/ng

given i = 30°, nw = 1.33, ng = 1.5, r = angle the light leave the glass

On substitution;

sin 30/sinr = 1.33/1.5

1.5sin30 = 1.33sinr

sinr = 1.5sin30/1.33

sinr = 0.75/1.33

sinr = 0.5639

r = arcsin0.5639

r ≈35°

angle the light leave the glass is 35°

Answer:

  a) -5.26 m/s

  b) 27.91 m

Explanation:

a) The acceleration due to gravity makes the velocity increase in magnitude in a linear way. The average velocity over the interval will be equal to the actual velocity halfway through the interval. The velocity at the beginning of the interval will be higher (less negative) by the amount velocity changes in the first half of the interval.

  average velocity = (0 -(26.5 m))/(1.85 s) ≈ -14.324 m/s

The change in velocity in the first half of the interval is ...

  Δv = (Δt/2)×(-9.8 m/s²) = (1.85 s)(-4.9 m/s²) = -9.065 m/s

So, the initial velocity (at the beginning of the last 1.85 s interval) is ...

  v1 = (average velocity) -Δv = (-14.324 m/s) -(-9.065 m/s)

  v1 = -5.259 m/s

__

b) The velocity when the object hits the ground is ...

  v2 = average velocity +Δv = -14.324 m/s -9.065 m/s = -23.389 m/s

This is related to the distance traveled by ...

  v² = 2dg . . . . . where g is the acceleration and d is the distance traveled

  d = v²/(2g) = 23.389²/(2·9.8) = 27.911 . . . . meters

The object travels a total distance of about 27.911 meters.

_____

The attached graph shows height vs. time.

Answer:

a. about 20 years younger

b. Malcolm sits around for 49.94 years

c. 2.268x[tex]10^{17}[/tex] m

Explanation:

light travels 3x[tex]10^{8}[/tex] m in one seconds

in 20 years that will be 3x[tex]10^{8}[/tex] x 20 x 60 x 60 x 24 x 365 = 1.89x[tex]10^{17}[/tex] m

for the to and fro journey, total distance covered will be 2 x 1.89x[tex]10^{17}[/tex]  = 3.78x[tex]10^{17}[/tex] m

Flo's speed = 80% of speed of light = 0.8 x 3x[tex]10^{8}[/tex]  = 2.4x[tex]10^{8}[/tex]  m/s

time that will pass for Malcolm will be  distance/speed = 3.78x[tex]10^{17}[/tex] /2.4x[tex]10^{8}[/tex]  

= 1575000000 s = 49.94 years

the relativistic time t' will be

t' = t x [tex]\sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]

t' = 49.94 x [tex]\sqrt{1 - 0.8^{2} }[/tex]

t' = 49.94 x 0.6 = 29.96 years       this is the time that has passed for Flo

this means that Flo will be about 20 years younger than Malcolm when she returns

relativistic distance is

d' = d x [tex]\sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]

d' = 3.78x[tex]10^{17}[/tex] x [tex]\sqrt{1 - 0.8^{2} }[/tex]

d' = 3.78x[tex]10^{17}[/tex] x 0.6

d' = 2.268x[tex]10^{17}[/tex] m     this is how far it is to Flo

Answer:

Stretch in the spring = 0.1643 (Approx)

Explanation:

Given:

Mass of the sled (m) = 9 kg

Acceleration of the sled (a) = 2.10 m/s ²

Spring constant (k) = 115 N/m

Computation:

Tension force in the spring  (T) = ma

Tension force in the spring  (T) = 9 × 2.10

Tension force in the spring  (T) = 18.9 N

Tension force in the spring = Spring constant (k) × Stretch in the spring

18.9 N = 115 N  × Stretch in the spring

Stretch in the spring = 18.9 / 115

Stretch in the spring = 0.1643 (Approx)

Explanation:

Complete the first and second sentences, choosing the correct answer from the given ones.

1. T = 100 K

[tex]^{\circ}C=K-273[/tex]

Put T = 100 K

[tex]T=100-273=-173^{\circ} C[/tex]

A temperature of 100 K corresponds on a Celsius scale to (-173 °C)

2. T = 50 °C

[tex]K=^{\circ}C+273\\\\K=50+273\\\\T=323\ K[/tex]

So, At 50 °C, it corresponds to a Kelvin scale of 323 K.

Answer:

I = 4.189 mA    V = 0.338 V

Explanation:

In order to do this, we need to apply the following expression:

I = Is[exp^(qV/kT) - 1]   (1)

However, as the junction of the diode is illuminated, the above expression changes to:

I = Iopt + Is[exp^(qV/kT) - 1]   (2)

Now, as the shunt resistance becomes infinite while the current becomes zero, we can say that the leakage current is small, and so:

I ≅ Iopt

Therefore:

I ≅ I₀Aλq / hc  (3)

Where:

I₀A = Area of diode (radiation)

λ: wavelength

q: electron charge (1.6x10⁻¹⁹ C)

h: Planck constant (6.62x10⁻³⁴ m² kg/s)

c: speed of light (3x10⁸ m/s)

Replacing all these values, we can get the current:

I = (8x10⁻³) * (0.65x10⁻⁶) * (1.6x10⁻¹⁹) / (6.62x10⁻³⁴) * (3x10⁸)

I = 4.189x10⁻³ A or 4.189 mA

Now that we have the current, we just need to replace this value into the expression (2) and solve for the voltage:

I = Is[exp^(qV/kT) - 1]

k: boltzman constant (1.38x10⁻²³ J/K)

4.189x10⁻³ = 9x10⁻⁹ [exp(1.6x10⁻¹⁹ V / 1.38x10⁻²³ * 300) - 1]

4.189x10⁻³ / 9x10⁻⁹ = [exp(38.65V) - 1]

465,444.44 + 1  = exp(38.65V)

ln(465,445.44) = 38.65V

13.0508 = 38.65V

V = 0.338 V

Answer:

Angle = 18.41°

Explanation:

Torque = F•r•sin θ

where;

F = force

r = distance from the rotation point

θ = the angle between the force and the radius vector.

We are given;

Torque = 15 N.m

F = 95 N

r = 0.5 m

Thus, plugging in the relevant values ;

15 = 95 × 0.5 × sin θ

sin θ = 15/(95 × 0.5)

sin θ = 0.3158

θ = sin^(-1)0.3158

θ = 18.41°

Explanation:

The answer is B because Nuclear family mean a family with two kids and Mrs. Parker have two kids

Answer:

About 3.06 seconds

Explanation:

[tex]v_f=v_o+at[/tex]

Since at the peak of its trajectory, the ball will have no velocity, you can write the following equation:

[tex]0=30+(-9.81)t\\\\-30=-9.81t\\\\t\approx 3.06s[/tex]

Hope this helps!

Complete Question

A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.65 kg and its length is 0.55 m. You may want to review (Pages 314 - 318) .

Part A What is the rotational energy of the blade at its operating angular speed of 3510 rpm

Part B

If all of the rotational kinetic energy of the blade could be converted to gravitational potential energy, to what height would the blade rise?

Answer:

Part A  

    [tex]R = 1081 \ J[/tex]

Part B  

     [tex]h = 169.7 \ m[/tex]

Explanation:

From the question we are told that

  The mass of the blade is  [tex]m_b = 0.65 \ kg[/tex]

   The length is  [tex]l = 0.55 \ m[/tex]

   The angular speed is  [tex]w = 3510 rpm = 3510 * \frac{2 \pi }{60} = 367.6 \ rad/sec[/tex]

Generally the moment of inertia of the of this mower is mathematically evaluated as

         [tex]I = \frac{m_b * l^2 }{12}[/tex]

substituting values

         [tex]I = \frac{0.65 * 0.55^2 }{12}[/tex]

         [tex]I = 0.016 \ kg m^2[/tex]

Generally the rotational kinetic energy of the bland is  

        [tex]R = \frac{1}{2} * I * w^2[/tex]

substituting values

       [tex]R = \frac{1}{2} * 0.016 * 367.6^2[/tex]

     [tex]R = 1081 \ J[/tex]

At point where the gravitational potential energy is equal to the rotational kinetic energy  we have that

       [tex]P = R = m_b * h * g[/tex]

Where P is the  gravitational potential energy

substituting values

          [tex]1081 = 0.65 * 9.8 * h[/tex]

=>       [tex]h = 169.7 \ m[/tex]

Answer:

The entropy change of the Universe that occurs is 19.346 J/K

Explanation:

Given;

temperature of the sun, [tex]T_s[/tex] = 5,300 K

temperature of the Earth, [tex]T_E[/tex] = 293 K

radiation energy transferred by the sun to the earth, E = 6000 J

The sun loses Q of heat and therefore decreases its entropy by the amount

[tex]\delta S_{sun} = \frac{-Q}{T_s}[/tex]

The earth gains Q of heat and therefore increases its entropy by the amount

[tex]\delta S_{Earth} = \frac{-Q}{T_E}[/tex]

The total entropy change is:

[tex]\delta S_{Earth} + \delta S_{sun} = \frac{Q}{T_E} -\frac{Q}{T_S} \\\\ = Q(\frac{1}{T_E} -\frac{1}{T_S} )\\\\= 6000(\frac{1}{293} -\frac{1}{5300} )\\\\=6000(0.0032243)\\\\= 19.346 \ J/K[/tex]

Therefore, the entropy change of the Universe that occurs is 19.346 J/K

Answer:

Gravity

Explanation:

The solar system is held together by rotational forces and gravity. This can be seen from billions of years ago when the solar system was a cloud of dust and gas. This cloud of dust and gas is known as the solar nebula. All of these dust and gas were brought together by the rotational movement as well as the action of gravity which brought all the particles together to form a larger one. This alone brought about the sun's formation in the center of the nebula as well the formation of other planetary bodies, etc.

Cheers.

Answer:

C) less than 129 lb.

Explanation:

Let the elevator be slowing up with magnitude of a . That means it is accelerating downwards  with magnitude a .

If R be the reaction force

For the elevator is going downwards with acceleration a

mg - R = ma

R = mg - ma

R measures its apparent weight . Spring scale will measure his apparent weight.

So its apparent weight is less than 129 lb .

Answer:

The final speed of the ball is 9 m/s.

Explanation:

We have,

A steel ball is dropped from a diving platform. It is required to find the velocity of the ball 0.9 seconds after its released. It will move under the action of gravity. Using equation of motion to find it as :

[tex]v=u+at[/tex]

u = 0 (at rest), a = g

[tex]v=gt\\\\v=10\times 0.9\\\\v=9\ m/s[/tex]

So, the final speed of the ball is 9 m/s.

Answer:

A light wave will not stop if it enters a new medium perpendicular to the surface.

Explanation:

A light wave will not have any deviation if it enters a new medium perpendicular to the surface.

Refraction is defined as an optical phenomenon by which the direction of a light wave gets changed when it travels from one medium to another. This is because of the change in speed.

Here,

The light wave is entering a new medium such that it enters perpendicular to the surface. Angle of incidence is the angle between the incident ray and the line perpendicular to the surface at the point of incidence. Since, here the light ray is incident normal to the surface that means the angle of incidence is 0.

According to Snell's law,

sin i = μ sin r

where i is the angle of incidence, r is the angle of refraction and μ is the constant called refractive index.

As i = 0, sin i = 0

So, μ sin r = 0

Since μ is a constant, we can say that sin r = 0 or the angle of refraction,

r = 0

This means that there is no refraction and hence the light wave won't get deviated when it enters the medium normally.

Hence,

A light wave will not have any deviation if it enters a new medium perpendicular to the surface.

To learn more about refraction, click:

https://brainly.com/question/14397443

#SPJ3

Answer:

18 N

Explanation:

Force can be found using the following formula.

f= m*a

where m is the mass and a is the acceleration.

We know the desk has a mass of 36 kilograms. We also know that its acceleration is 0.5 m/s^2.

m= 36 kg

a= 0.5 m/s^2

Substitute these values into the formula.

f= 36 kg * 0.5 m/s^2

Multiply 36 and 0.5

f=18 kg m/s^2

1 kg m/s^2 is equivalent to 1 Newton, or N.

f= 18 Newtons

The force being applied is 18 kg m/s^2, Newtons, or N

Answer

The rate at which the magnetic field is changing is  [tex][\frac{dB}{dt} ] = 0.000467 T/s[/tex]

Explanation

From the question we are told that

   The electric field strength is [tex]E = 3.5mV/m = 3.5 *10^{-3} \ V/m[/tex]

    The radius is  [tex]r = 1.5 \ m[/tex]

The rate of change of the  magnetic  field  is mathematically represented as

        [tex]\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}[/tex]

Where [tex]dl[/tex] is change of a unit length

     [tex]\frac{d \phi}{dt} = A * \frac{dB}{dt}[/tex]

Where A is the area which is mathematically represented as

     [tex]A = \pi r^2[/tex]

    So

    [tex]E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )[/tex]  

  [tex]E L = ( \pi r^2) (\frac{dB}{dt} )[/tex]  

where L is the circumference of the circle which is mathematically represented as

     [tex]L = 2 \pi r[/tex]

So

     [tex]E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ][/tex]

      [tex]E = \frac{r}{2} [\frac{dB}{dt} ][/tex]

       [tex][\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }[/tex]

substituting values

      [tex][\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }[/tex]

      [tex][\frac{dB}{dt} ] = 0.000467 T/s[/tex]    

Answer:

d

Explanation:

good question. now the bus is moving in constant velocity . a student in front tosses a ball to the student in back. but we dont know the speed at which the student tosses a ball. we have to assume the speed

assume the speed of ball is slightly less than the speed of bus. in this case the stationary observer sees the ball in slower speed than the one inside the bus.

so a is correct

now assume the speed of ball is 1/2 the speed of bus. here stationary observer sees the ball the same speed as the one in bus observe

b is correct

assume the speed of ball is very small than the speed of bus . in this case the stationary observer see in grater speed than the student in bus

e also correct

so correct answer is d. it depends on the speed of ball tossed by the student in front.

Answer:

The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].

Explanation:

As CD is acceleration uniformly, the following equation of motion can be used to determine the angular acceleration:

[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot \Delta n[/tex]

Where:

[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.

[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.

[tex]\ddot n[/tex] - Angular acceleration, measured in revolution per square minute.

[tex]\Delta n[/tex] - Change in angular position, measured in revolutions.

The angular acceleration is cleared and calculated:

[tex]\ddot n = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \Delta n}[/tex]

Given that [tex]\dot n_{o} = 0\,\frac{rev}{min}[/tex], [tex]\dot n = 430\,\frac{rev}{min}[/tex] and [tex]\Delta n = 4\, rev[/tex], the angular acceleration is:

[tex]\ddot n = \frac{\left(430\,\frac{rev}{min} \right)^{2}-\left(0\,\frac{rev}{min} \right)^{2}}{2\cdot (4\,rev)}[/tex]

[tex]\ddot n = 23112.5\,\frac{rev}{min^{2}}[/tex]

The angular accelaration measured in radians per square second is:

[tex]\alpha = \left(23112.5\,\frac{rev}{min^{2}} \right)\cdot \left(2\pi\,\frac{rad}{rev}\right)\cdot \left(\frac{1}{3600}\,\frac{min^{2}}{s^{2}} \right)[/tex]

[tex]\alpha \approx 40.339\,\frac{rad}{s^{2}}[/tex]

Net torque experimented by the CD during its accleration is equal to the product of its moment of inertia with respect to its axis of rotation and angular acceleration:

[tex]\tau = I \cdot \alpha[/tex]

Where:

[tex]I[/tex] - Moment of inertia, measured in [tex]kg \cdot m^{2}[/tex].

[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.

In addition, a CD has a form of a uniform disk, whose moment of inertia is:

[tex]I = \frac{1}{2}\cdot m \cdot r^{2}[/tex]

Where:

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

[tex]r[/tex] - Radius of the CD, measured in meters.

If [tex]m = 0.017\,kg[/tex] and [tex]r = 0.07\,m[/tex], then:

[tex]I = \frac{1}{2}\cdot (0.017\,kg)\cdot (0.07\,m)^{2}[/tex]

[tex]I = 4.165\times 10^{-5}\,kg\cdot m^{2}[/tex]

Now, the net torque exerted on CD is:

[tex]\tau = (4.165\times 10^{-5}\,kg\cdot m^{2})\cdot \left(40.339\,\frac{rad}{s^{2}} \right)[/tex]

[tex]\tau = 1.680\times 10^{-3}\,N\cdot m[/tex]

The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].

Answer:

The voltage is  [tex]V = 418.60 \ Volts[/tex]  

Explanation:

From the question we are told that

    The area of the both plate is  [tex]A = 7.00 *10^{-3} \ m^2[/tex]

    The distance between the plate is [tex]d = 4.80*10^{-4}\ m[/tex]

     The magnitude of the charge is  [tex]q = 5.40 *10^{-8} \ C[/tex]

The capacitance of the capacitor that consist of the two plates is mathematically represented as

        [tex]C = \frac{\epsilon _o A}{d}[/tex]

Where [tex]\epsilon_o[/tex] is the permitivity of free space with a value  [tex]e = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

So

       [tex]C = \frac{8.85*10^{-12} * (7* 10^{-3})}{ 4.8*10^{-4}}[/tex]

        [tex]C = 1.29 *10^{-10} \ F[/tex]

The potential difference between the plate is mathematically represented as

      [tex]V = \frac{ Q}{C }[/tex]

     [tex]V = \frac{ 5.4*10^{-8}}{1.29 *10^{-10}}[/tex]

     [tex]V = 418.60 \ Volts[/tex]

Answer:

b and e

Explanation:

r x F is the formula for torque.

The "turning effect" or torque happens when concentric forces rotate an object along said center.

a) False because T = Fr = Ia (a = angular acceleration)

b) True

c) False. L = Iw (w = angular velocity), which does not equal Ia

d) False. It is torque, not the product of torque and something else

e) True.

Answer:

(a) q = 2.357 x 10⁻⁵ C

(b) Φ = 2.66 x 10⁶ N.m²/C

Explanation:

Given;

diameter of the sphere, d = 1.1 m

radius of the sphere, r = 1.1 / 2 = 0.55 m

surface charge density, σ = 6.2 µC/m²

(a)  Net charge on the sphere

q = 4πr²σ

where;

4πr² is surface area of the sphere

q is the net charge on the sphere

σ is the surface charge density

q = 4π(0.55)²(6.2 x 10⁻⁶)

q = 2.357 x 10⁻⁵ C

(b) the total electric flux leaving the surface of the sphere

Φ = q / ε

where;

Φ is the total electric flux leaving the surface of the sphere

ε is the permittivity of free space

Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)

Φ = 2.66 x 10⁶ N.m²/C

Answer:(a)

Explanation:

Student must have known that insulators can only be charged when they are rubbed against each other. In this process, one becomes electrically negative while other becomes electrically positive such that both have the same magnitude. The one which gains electrons becomes electrically negative due to the transfer of electrons while others lose the electron becomes positive due to the transfer of an electron to another body.

Answer:

Explanation:

The man comes to halt due to reaction force acting on him in opposite direction . If R be the reaction force

impulse by net  force = change in momentum

Net force = R - mg , mg is weight of the man .

( R-mg ) x 2. 07 x 10⁻³ = 73 x 6.46 - 0

R - mg = 227.81 x 10³

Average net force = 227.81 x 10³ N .

Answer:

A. F = 0.06 N

B. t = 6.37 s

Explanation:

A)

First we need to find the constant acceleration of the cart. For this purpose, we use 2nd equation of motion:

s = (Vi)(t) + (0.5)at²

where,

s = distance traveled = 1.62 m

Vi = 0 m/s   (Since, it starts from rest)

t = Time Taken = 4.34 s

a = acceleration = ?

Therefore,

1.62 m = (0 m/s)(4.34 s) + (0.5)(a)(4.34 s)²

1.62 m/9.4178 s² = a

a = 0.172 m/s²

Now, from Newton's Second law, we know that:

F = ma

where,

F = Net Force of the combination = ?

m = Mass pf combination = 354 g = 0.354 kg

Therefore,

F = (0.354 kg)(0.172 m/s²)

F = 0.06 N

B)

Now, for the same force, but changed mass = 762 g = 0.762 kg, we have the acceleration to be:

F = ma

a = F/m

a = 0.06 N/0.762 kg

a = 0.08 m/s²

Now, using  2nd equation of motion:

s = (Vi)(t) + (0.5)at²

1.62 m = (0 m/s)(t) + (0.5)(0.08 m/s²)t²

t² = 1.62 m/(0.04 m/s²)

t = √40.54 s²

t = 6.37 s

Answer:

a = 1.72 m/s²

Explanation:

The given kinematic equation is the 2nd equation of motion. The equation is as follows:

xf = xi + (Vi)(t) + (1/2)(a)t²

where,

xf = the final position =  5000 m

xi = the initial position = 1000 m

Vi = the initial velocity = 15 m/s

t = the time taken = 60 s

a = acceleration = ?

Therefore,

5000 m = 1000 m + (15 m/s)(60 s) + (1/2)(a)(60 s)²

5000 m = 1000 m + 900 m + a(1800 s²)

5000 m = 1900 m + a(1800 s²)

5000 m - 1900 m = a(1800 s²)

a(1800 s²) = 3100 m

a = 3100 m/1800 s²

a = 1.72 m/s²

Answer:

B. F/2

Explanation:

The radiation force per unit area (radiation pressure Prad) exerted by an electromagnetic wave on a perfectly absorbing body has been found by experiment to be equal to the energy density of the wave

i.e Prad = u

For a reflecting body, this force exerted per unit area has been found to be twice the energy density of the wave.

i.e Prad = 2u.

Therefore, if the force exerted on a perfectly reflective body is F, then the force exerted on a perfectly absorbing body will be F/2

Answer:

* air resistance.

*the direction of the rotation of the Earth

rotation of the thrown body

Explanation:

The projectile launch is described by the expressions

x-axis         x = v₀ₓ t

y-axis         y = [tex]v_{oy}[/tex] t - ½ gt²

When the things that affect this movement are analyzed, in order of importance we have:

* air resistance. This significantly changes the body's horizontal position, so it introduces a horizontal acceleration that is not contained in the equations.

* air resistance. At the height that the body reaches, since air resistance has the same direction as the gravity of gravity and therefore the relationship is more challenging.

* to a lesser extent the direction of launch, in the direction of the rotation of the Earth against. Since this creates an operational on the x and y axis that changes the initial assumption

* The possible rotation of the thrown body, since this rotation creates a lift that is not taken in the equations, this value is more noticeable the lighter the body, this effect has to keep the body longer in the air achieving more reach and height

Answer:

E = -4556.18 N/m

Explanation:

Given data

u = 3.6×10^6 m/sec

angle = 34°

distance x = 1.5 cm = 1.5×10^-2 m  (This data has been assumed not given in

Question)

from the projectile motion the horizontal distance traveled by electron is

x = u×cosA×t

⇒t = x/(u×cos A)

We also know that force in an electric field is given as

F = qE

q= charge , E= strength of electric field

By newton 2nd law of motion

ma = qE

⇒a = qE/m

Also, y = u×sinA×t - 0.5×a×t^2

⇒y = u×sinA×t - 0.5×(qE/m)×t^2

if y = 0 then

⇒t = 2mu×sinA/(qE) = x/(u×cosA)

Also, E = 2mu^2×sinA×cosA/(x×q)

Now plugging the values we get

E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})

E = -4556.18 N/m

The value of Ey such that the particle will cross the x axis at x=1.5 cm is -4556.18 N/m.

The field developed when a charge is moved. In this field, a charge experiences an electrostatic force of attraction or repulsion depending on the nature of charge.

Given is a particle leaves the origin with a speed of 3.6 x 10⁶ m/s at 34 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis.

The distance x = 1.5 cm = 1.5×10⁻² m (assumed, not given in question)

The horizontal distance traveled by particle is

x = ucosθt

t = x/ucosθ

The force in an electric field is F = qE...................(1)

where, q is charge , E is the strength of electric field

From, newton 2nd law of motion, Force F = ma.................(2)

Equating both the equations, we get

ma = qE

a = qE/m..................(3)

The vertical distance, y =usinθt - 1/2at²

From equation 3, we have

y = usinθt  -  1/2 (qE/m) t²

if y = 0, t = 2musinθ/(qE) = x / (ucosθ)

The electric field is represented as

Also, E = 2mu²×sinθ×cosθ/(xq)

Plug the values, we get

E = 2×(9.1×10⁻³¹)×(3.6 x 10⁶)²×sin34°×cos34°/( 1.5×10⁻² ×(-1.6)×10⁻¹⁹)

E = -4556.18 N/m

Thus, the electric field of the particle is  -4556.18 N/m.

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Answer:

General Expression: E = kql/(l² + r²)^(3/2)

(a) 6.3 MN/C

(b) 22.8 MN/C

(c) 6.1 MN/C

(d) 0.63 MN/C

Explanation:

The general expression for electric field along axis of a uniformly charged ring is:

E = kqL/(L² + r²)^(3/2)

where,

E = Electric Field Strength = ?

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q = Total Charge = 71 μC = 71 x 10⁻⁶ C

L = Distance from center on axis

r = radius of ring = 10 cm = 0.1 m

(a)

L = 1 cm = 0.01 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.01 m)/[(0.01 m)² + (0.1 m)²]^(3/2)

E = (6390 N.m³/C)/(0.00101 m³)

E =  6.3 x 10⁶ N/C = 6.3 MN/C

(b)

L = 5 cm = 0.05 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.05 m)/[(0.05 m)² + (0.1 m)²]^(3/2)

E = (31950 N.m³/C)/(0.00139 m³)

E =  22.8 x 10⁶ N/C = 27.4 MN/C

(c)

L = 30 cm = 0.3 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.3 m)/[(0.3 m)² + (0.1 m)²]^(3/2)

E = (191700 N.m³/C)/(0.03162 m³)

E =  6.1 x 10⁶ N/C = 6.1 MN/C

(d)

L = 100 cm = 1 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(1 m)/[(1 m)² + (0.1 m)²]^(3/2)

E = (639000 N.m³/C)/(1.015 m³)

E =  0.63 x 10⁶ N/C = 0.63 MN/C