A machinist with normal vision has a near point at 25 cm. This machinist wears +4.25-diopter eyeglasses in order to do very close work. With these eyeglasses, what is the near point of the machinist? Assume that he wears the glasses extremely close to his eyes. 12 cm 10 cm 7 cm 15 cm 17 cm

Adding globes in parallel makes no difference to their brightness because all the bulbs glowed with the same brightness indicating that the current flowing through the bulbs had the same value.

The current in each light bulb was the same.

An electric current is a flow of electrons as a result of a potential difference between two points in conductor.

Electrons are negatively-charged particles that are one of the fundamental particles of an atom. The flow of electrons through a conductor is able to do work.

Two or more globes in a simple parallel circuit do not experience any drop in the voltage, thus allowing the maximum flow of electric current. Also, connecting devices in a parallel circuit ensures that if one of the component lops is disconnected, the flow of current through the other loops remains intact.

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The acceleration of Thomas is 0.233 m/s^2.

Acceleration is the rate of change of velocity. Thomas the Train chugs along at a velocity of 2 m/s.

Thomas needs to go faster so more coal is shoveled into his engine and he accelerates for 10 seconds until he is going 4.33 m/s.

We are to find the acceleration of Thomas.

The formula for acceleration is given as :

acceleration = (final velocity - initial velocity) / time

In the given problem, the initial velocity of Thomas, u = 2 m/s.

The final velocity of Thomas, v = 4.33 m/s The time for which Thomas accelerates, t = 10 s.

Therefore, the acceleration of Thomas will be given as:

a = (v - u) / ta = (4.33 - 2) / 10s => 2.33 / 10s => 0.233 m/s^2

Thus, the acceleration of Thomas is 0.233 m/s^2.

To summarize, the acceleration of Thomas is 0.233 m/s^2.

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The equilibrium position of the electron at the center of the nucleus is  R  = 3.16 x10⁻¹⁰m.

Thomson discovered the electron and proposed a structural model of the atom. His research also led to the invention of the mass spectrometer. Thomson announced that he discovered that atoms are made up of smaller components.

Calculation:-

K = 2.304 × 20^-²⁸ / 3

f = 1/2π√k/m

4.63 × 10¹⁴ = 1/2× 3.141 √2.304 × 20^-²⁸ / R³ (9.1 × 10-31)

R = 3.16 x10⁻¹⁰m.

This discovery revolutionized the way scientists think about atoms and had a major impact on the field of physics. Thompson called them corpuscles but what he found are now commonly known as electrons. The plum pudding model is defined by an electron surrounded by a volume of positive charge like a negatively charged plum embedded in a positively charged pudding.

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Answer:

5 cm - 53 degrees

Explanation:

\( \sqrt{ {4}^{2} + {3}^{2} } = 5cm\)

\( \tan ^{ - 1} (4 \div 3) \) = 53°

Answer:

"\(2.829\times 10^{31} \ photon/s\)" is the appropriate solution.

Explanation:

The given values are:

Power,

P = 3.47 W

Frequency,

f = 185 Hz

Planck's constant,

h = 6.63 x 10⁻³⁴ J.s

As we know,

⇒  \(n=\frac{P}{E}\)

Or,

⇒  \(n=\frac{P}{hf}\)

On substituting the given values, we get

⇒     \(=\frac{3.47}{6.63\times 10^{-34}\times 185}\)

⇒     \(=\frac{3.47}{1.22655\times 10^{-31}}\)

⇒     \(=2.829\times 10^{31} \ photon/s\)

The potential energy of the roller coaster is 176,400 J (joules).

The potential energy of an object is given by the formula PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical position of the object.

In this case, the roller coaster is at a peak of 20m and has a mass of 900kg. The acceleration due to gravity, g, is approximately 9.8 \(m/s^2\).

Using the formula, we can calculate the potential energy:

PE = mgh

= (900 kg)(9.8 \(m/s^2\))(20 m)

= 176,400 J

Therefore, the potential energy of the roller coaster is 176,400 J (joules).

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The charge of the spheres can be determined using Coulomb's law of electrostatic forces. Here, the separation is 1 m and the electric force between them is 1 N. Then their charge is 1.05 × 10⁻⁵ C.

Coulomb's law states the relation between the charges of two charged particles  and the distance between them to the force of attraction or repulsion between them as written below:

F = k q1 q2 / r²

where q1 and q2 be the charges and r is the distance.

k = 9 × 10⁹ N m²/C²

Given r = 1m

F = 1 N

here q1 = q2 = q.

then 1 N =  9 × 10⁹ N m²/C² q²/1 m²

q² = 1.11 × 10⁻¹⁰ C²

then q = 1.05 × 10⁻⁵ C.

Therefore, the charge of the tow tiny spheres is 1.05 × 10⁻⁵ C.

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Answer:

The organisms are widely dispersed in the habitat and rarely encounter each other...

Explanation:

As asexual reproduction needs only 1 parent.. they can do it on their own.. without encountering their mates

Answer:

It isn't moving

Explanation:

Between Kinetic energy and speed: The relationship is quadratic or parabolic. According to the kinetic energy formula, KE = 1/2 mv^2, the kinetic energy is proportional to the square of the speed.

Between potential energy and mass: The relationship is linear. The potential energy is directly proportional to the mass. In simple cases, the potential energy is given by PE = mgh, where mass (m) and height (h) are directly proportional.

Between pressure and volume: The relationship is inverse. According to Boyle's law, the pressure and volume of a gas are inversely proportional when temperature is constant. Mathematically, P1V1 = P2V2.

Between pressure and temperature: The relationship is linear. According to Charles's law, the pressure and temperature of a gas are directly proportional when volume is constant. Mathematically, P1/T1 = P2/T2.

Between displacement and velocity: The relationship is linear. Velocity is the rate of change of displacement with respect to time, so the two are directly proportional.

Answer:

Explanation: because Keller concluded that the circular obits of the Copernican model weren't correct paths of planets were elleptical

To determine the magnetic force on a straight wire carrying a current in a uniform magnetic field, we can use the formula for the magnetic force:

F = I * L * B * sin(θ)

where:

F is the magnetic force,

I is the current in the wire,

L is the length of the wire,

B is the magnitude of the magnetic field, and

θ is the angle between the wire and the magnetic field.

In this case, the values are:

I = 30 A (current in the wire)

L = 2.0 m (length of the wire)

B = 55 mT = 0.055 T (magnitude of the magnetic field)

θ = 20° (angle between the wire and the magnetic field)

Substituting the values into the formula:

F = 30 A * 2.0 m * 0.055 T * sin(20°)

Calculating sin(20°):

F = 30 A * 2.0 m * 0.055 T * 0.3420

F ≈ 1.5714 N

Therefore, the magnetic force on the 2.0-meter length of wire carrying a current of 30 A in a region with a uniform magnetic field of magnitude 55 mT and at an angle of 20° away from the wire is approximately 1.5714 N.

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The magnetic field of the wire will be directed towards west. Using right thumb rule one can get the direction of field lines.

Answer:

The velocity is  \(v_2= 0.45 \ m/s\)

Explanation:

From the question we are told that

      The initial speed of the hot water is  \(v_1 = 0.85 \ m/s\)

     The pressure from the heater  \(P_1 = 450 \ KPa = 450 *10^{3} \ Pa\)

      The height of the hot water before flowing is  \(h_1 = 0 \ m\)

      The height of bathtub above the heater is \(h_2 = 3.70 \ m\)

       The pressure in the pipe is \(P_2 = 414 KPa = 414 *10^{3} \ Pa\)

       The density of water is \(\rho = 1000 \ kg/m^3\)

Apply Bernoulli equation

      \(P_1 + \rho gh_1 +\frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2}\rho v_2 ^2\)

Substituting values

     \((450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) = (1000 * *9.8*3.70) + (0.5*1000*v_2^2 )\)

=>   \(v_2^2 = \frac{ (450 *10^{3}) + (1000 * 9.8 *0 ) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}\)

=>   \(v_2= \sqrt{ \frac{ (450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}}\)

=>    \(v_2= 0.45 \ m/s\)

The electric potential energy stored in the capacitor is 4.70 x 10^-8 Joules.

The electric potential energy stored in a capacitor is given by the formula:

U = (1/2) * C * V^2

where U is the potential energy in Joules, C is the capacitance in Farads, and V is the voltage across the capacitor in Volts.

In this case, we are given that the capacitor stores 9.40 x 10^-10 C of charge at 50.0 V. However, we are not given the capacitance value. Therefore, we cannot calculate the potential energy directly using the above formula.

To find the capacitance value, we can use the formula:

C = Q / V

where Q is the charge stored in the capacitor and V is the voltage across the capacitor.

Substituting the given values, we get:

C = 9.40 x 10^-10 / 50.0

= 1.88 x 10^-11 F

Now we can use the formula for electric potential energy to find the energy stored in the capacitor:

U = (1/2) * 1.88 x 10^-11 * (50.0)^2

= 4.70 x 10^-8 J

Therefore, the electric potential energy stored in the capacitor is 4.70 x 10^-8 Joules.

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The wavelengths are given as

firsts harmonic = 40 m

second harmonic = 20 m  

third harmonic = 13.33 m

Wavelength refers to the distance between two consecutive points of a wave that are in phase, or the distance it takes for a wave to complete one full cycle. It is commonly denoted by the Greek letter lambda (λ).

Using the wavelength of the fourth harmonics as reference, we have that the full scale is 20 m

firsts harmonic = 40 m (double of the full scale)

second harmonic = 20 m (covers the full scale)

third harmonic = 20 m * 2/3 = 13.33 m (covers 2/3 of the full scale)

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a. Total energy is 0.098 J

b. Potential and Kinetic Energies is 0.032 sin^2(8.08t) J

c. Velocity at x is -0.808 sin(8.08t) m/s

d. Potential and Kinetic Energies at x is 0.016 sin^2(8.08t) J

We can use the following formulas for the energy, velocity, and potential and kinetic energies of a simple harmonic oscillator:

where ω = √(k/m) is the angular frequency.

Given that k = 19.6 N/m, A = 0.100 m, x = -(0.100 m) cos 8.08t, and v = (0.808 m/s) sin 8.08t, we can find the values of E, U, and K as follows:

(a) Total Energy:

E = 1/2 k A^2 = 1/2 * 19.6 * 0.1^2 = 0.098 J

(b) Potential and Kinetic Energies:

U = 1/2 k x^2 = 1/2 * 19.6 * (-0.1 cos(8.08t))^2 = 0.098 cos^2(8.08t) J

K = 1/2 m v^2 = 1/2 * (0.1) * (0.808 sin(8.08t))^2 = 0.032 sin^2(8.08t) J

(c) Velocity at x = 0.050 m:

When x = 0.050 m, cos(8.08t) = -0.5, so we have:

v = -ωA sin(ωt) = -ω(0.1) sin(8.08t) = -0.808 sin(8.08t) m/s

(d) Potential and Kinetic Energies at x = A/2:

When x = A/2 = 0.050 m, cos(8.08t) = -0.5, so we have:

U = 1/2 k x^2 = 1/2 * 19.6 * (0.050)^2 = 0.0245 J

K = 1/2 m v^2 = 1/2 * (0.1) * (0.808 sin(8.08t))^2 = 0.016 sin^2(8.08t) J

Note that the sum of potential and kinetic energies at any point in time is equal to the total energy, which is constant.

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Answer:

Explanation:

It's graph A because the pressure in the tire is increasing as the amount of air going into it increases. B says the pressure drops exponentially as air goes in, and C says that the pressure stays the same as air goes in. Pressure in a tire increases proportionally to the amount of air in it.

Answer: what characterizes moral development in mid to late childhood is

People at this level of moral development base their decisions on what their parents and/or law enforcement says is right. Stage 3 is about social conformity. ... Stage 4 is all about law and order for all. For example, someone may think, 'If I steal, I will break the law and breaking the law is wrong.Feb 17, 2016

Explanation:

The sound from a tuning fork is not a transverse wave but a longitudinal wave

A common example of longitudinal waves is sound waves, which are pressure waves or vibrations that travel through air or other materials, like water. Tuning forks produce a single note, or a specific sound, when struck by making the arms move in and out very rapidly (hundreds or thousands of times a second).

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Answer:

Stars are the light bulbs of the universe.

Answer:

Stars are the recycling centers of the universe

Explanation:

I just did on edge C

The bicyclist is applying more force on the pedals at Time 2. Option C

If we look at the image that have been shown, we can be able to see that the force that is acting have been shows by the arrows that have been used to label the movement of the cyclist in the image that is shown here.

We can see that the forward arrow at the time 2 is seen to be larger than the forward arrow that is shown for time 1. The implication of this is that the cyclist is cycling harder and applying more force at time 2 than at time 1.

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Answer:

1.21 m/s

Explanation:

From the law of conservation of energy,

U₁ + K₁ + E₁ = U₂ + K₂ + E₂

where U₁ = initial potential energy of system =initial potential energy of still ball = mgh where m = mass of still ball = 0.514 kg, g = acceleration due to gravity = 9.8 m/s² and h = height = length of cord = 68.7 cm = 0.687 m.

K₁ = initial kinetic energy of system = 0

E₁ = initial internal energy of system = unknown and

U₂ = final potential energy of system = 0

K₁ = final kinetic energy of system = final kinetic energy of ball + steel block = 1/2(m + M)v² where m = mass of still ball, M = mass of steel block = 2.63 kg and v = speed of still ball + steel block

E₁ = final internal energy of system = unknown

So,

U₁ + K₁ + E₁ = U₂ + K₂ + E₂

mgh + 0 + E₁ = 0 + 1/2(m + M)v² + E₂

mgh = 1/2(m + M)v² + (E₂ - E₁)

Given that (E₂ - E₁) = change in internal energy = ΔE = 1/2ΔK where ΔK = change in kinetic energy. So, ΔE = 1/2ΔK = 1/2(K₂ - K₁) = K₂/2 = 1/2(m + M)v²/2 = (m + M)v²/4

Thus, mgh = 1/2(m + M)v² + (E₂ - E₁)

mgh = 1/2(m + M)v² + (m + M)v²/4

mgh = 3(m + M)v²/4

So, making v subject of the formula, we have

v² = 4mgh/3(m + M)

taking square root of both sides, we have

v = √[4mgh/3(m + M)]

Substituting the values of the variables into the equation, we have

v = √[4 × 0.514 kg × 9.8 m/s² × 0.687 m/{3(0.514 kg + 2.63 kg)}]

v = √[13.8422/{3(3.144 kg)}]

v = √[13.8422 kgm/s²/{9.432 kg)}]

v = √(1.4676 m²/s²)

v = 1.21 m/s

The final speed of the given ball law of conservation of energy. The final speed of the given ball is 1.21 m/s.

The law of conservation of energy,  

U₁ + K₁ + E₁ = U₂ + K₂ + E₂  

where

U₁ = initial potential energy  

K₁ = initial kinetic energy  

E₁ = initial internal energy  

U₂ = final potential energy  

K₁ = final kinetic energy  

E₁ = final internal energy  

So,    

mgh + 0 + E₁ = 0 + 1/2(m + M)v² + E₂  

mgh = 1/2(m + M)v² + (E₂ - E₁)  

Given that (E₂ - E₁) = change in internal energy,

ΔE = 1/2ΔK

Where

ΔK = change in kinetic energy.

So,

ΔE = 1/2ΔK = 1/2(K₂ - K₁)

ΔE = K₂/2

ΔE = 1/2(m + M)v²/2

ΔE = (m + M)v²/4  

Thus,

mgh = 1/2(m + M)v² + (E₂ - E₁)  

mgh = 1/2(m + M)v² + (m + M)v²/4  

mgh = 3(m + M)v²/4    

v² = 4mgh/3(m + M)  

Take square root of both sides,  

v = √[4mgh/3(m + M)  

put the values in the formula,  

v = √[4 × 0.514 kg × 9.8 m/s² × 0.687 m/3(0.514 kg + 2.63 kg)    

v = 1.21 m/s

Therefore, the final speed of the given ball is 1.21 m/s.

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a) The tension in the rope is 123.9 N.

b) The moment of inertia of the wheel is 0.09 kg m².

c) The angular speed of the wheel 3.50 s after it begins rotating, starting from rest, is 58.5 rad/s.

(a) To determine the tension in the rope, we need to analyze the forces acting on the object. There are two forces: the force of gravity pulling the object down the incline and the tension force pulling the object up the incline.

The force of gravity can be broken down into two components: one parallel to the incline and one perpendicular to the incline.

The parallel component causes the object to accelerate down the incline, while the perpendicular component is balanced by the normal force of the incline.

The tension force is responsible for the object's acceleration down the incline, so we can set up the following equation:

T - mg sin(theta) = ma

where T is the tension force, m is the mass of the object, g is the acceleration due to gravity, theta is the angle of the incline, and a is the acceleration of the object down the incline.

Putting in the given values, we get:

T - (12.5 kg)(9.81 m/s²)(sin(37°)) = (12.5 kg)(2.00 m/s²)

Solving for T, we get:

T = 123.9 N

Therefore, the tension in the rope is 123.9 N.

(b) To determine the moment of inertia of the wheel, we can use the following equation:

I = (1/2)MR²

where I is the moment of inertia, M is the mass of the wheel, and R is the radius of the wheel.

Putting in the given values, we get:

I = (1/2)(12.5 kg)(0.12 m)²

 = 0.09 kg m²

Therefore, the moment of inertia of the wheel is 0.09 kg m².

(c) To determine the angular speed of the wheel after 3.50 s, we can use the following equation:

ω = ω₀ + αt

where ω is the final angular speed, ω₀ is the initial angular speed (which is zero in this case), α is the angular acceleration, and t is the time.

We can find the angular acceleration using the following equation:

α = a/R

where a is the acceleration of the object down the incline (which we already know) and R is the radius of the wheel.

Putting in the given values, we get:

α = 2.00 m/s² / 0.12 m

  = 16.7 rad/s²

Putting in the values for α and t, we get:

ω = 0 + (16.7 rad/s²)(3.50 s)

   = 58.5 rad/s

Therefore, the angular speed of the wheel 3.50 s after it begins rotating, starting from rest, is 58.5 rad/s.

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Answer:

\(\theta \approx 59.036^{\circ}\), \(T_{2} \approx 23.324\,N\)

Explanation:

First we build the Free Body Diagram (please see first image for further details) associated with the mass, we notice that system consist of a three forces that form a right triangle (please see second image for further details): (i) The weight of the mass, (ii) two tensions.

The requested tension and angle can be found by the following trigonometrical and geometrical expressions:

\(\theta = \tan^{-1} \frac{W}{T_{2}}\) (1)

\(T_{1} = \sqrt{W^{2}+T_{2}^{2}}\) (2)

Where:

\(W\) - Weight of the mass, measured in newtons.

\(T_{1}\), \(T_{2}\) - Tensions from the mass, measured in newtons.

If we know that \(W = 20\,N\) and \(T_{2} = 12\,N\), then the requested values are, respectively:

\(\theta = \tan^{-1} \frac{20\,N}{12\,N}\)

\(\theta \approx 59.036^{\circ}\)

\(T_{2} = \sqrt{(20\,N)^{2}+(12\,N)^{2}}\)

\(T_{2} \approx 23.324\,N\)

Answer:

The horizon has constant surface gravity for a stationary black hole.

Explanation:

Answer:

all parts of the event horizon of a black hole at equilibrium have the same surface gravity g

Explanation:

Block B, which has a mass of 4t kg, is moving at a constant velocity across a flat surface, the magnitude of the force is  F=100 N. This is further explained below.

Generally, Force is simply defined as a function of mass and time.

In conclusion, A requirement of balance is that the left side exerts the same force as the right, therefore if 100 N is applied to the right, then 100 N must be exerted on the left.

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Answer:

On that line segment between the two charges, at approximately \(0.7\; \rm m\) away from the smaller charge (the one with a magnitude of \(5 \times 10^{-19}\; \rm C\),) and approximately \(1.3\; \rm m\) from the larger charge (the one with a magnitude of \(20 \times 10^{-19}\; \rm C\).)

Explanation:

Each of the two point charges generate an electric field. These two fields overlap at all points in the space around the two point charges. At each point in that region, the actual electric field will be the sum of the field vectors of these two electric fields.

Let \(k\) denote the Coulomb constant, and let \(q\) denote the size of a point charge. At a distance of \(r\) away from the charge, the electric field due to this point charge will be:

\(\displaystyle E = \frac{k\, q}{r^2}\).

At the point (or points) where the electric field is zero, the size of the net electrostatic force on any test charge should also be zero.

Consider a positive test charge placed on the line joining the two point charges in this question. Both of the two point charges here are positive. They will both repel the positive test charge regardless of the position of this test charge.

When the test charge is on the same side of both point charges, both point charges will push the test charge in the same direction. As a result, the two electric forces (due to the two point charges) will not balance each other, and the net electric force on the test charge will be non-zero.  

On the other hand, when the test charge is between the two point charges, the electric forces due to the two point charges will counteract each other. This force should be zero at some point in that region.

Keep in mind that the electric field at a point is zero only if the electric force on any test charge at that position is zero. Therefore, among the three sections, the line segment between the two point charges is the only place where the electric field could be zero.

Let \(q_1 = 5\times 10^{-19}\; \rm C\) and \(q_2 = 20 \times 10^{-19}\; \rm C\). Assume that the electric field is zero at \(r\) meters to the right of the \(5\times 10^{-19}\; \rm C\) point charge. That would be \((2 - r)\) meters to the left of the \(20 \times 10^{-19}\; \rm C\) point charge. (Since this point should be between the two point charges, \(0 < r < 2\).)

The electric field due to \(q_1 = 5\times 10^{-19}\; \rm C\) would have a magnitude of:

\(\displaystyle | E_1 | = \frac{k\cdot q_1}{r^2}\).

The electric field due to \(q_2 = 20 \times 10^{-19}\; \rm C\) would have a magnitude of:

\(\displaystyle | E_2 | = \frac{k\cdot q_2}{(2 - r)^2}\).

Note that at all point in this section, the two electric fields \(E_1\) and \(E_2\) will be acting in opposite directions. At the point where the two electric fields balance each other precisely, \(| E_1 | = | E_2 |\). That's where the actual electric field is zero.

\(| E_1 | = | E_2 |\) means that \(\displaystyle \frac{k\cdot q_1}{r^2} = \frac{k\cdot q_2}{(2 - r)^2}\).

Simplify this expression and solve for \(r\):

\(\displaystyle q_1\, (2 - r)^2 - q_2 \, r^2 = 0\).

\(\displaystyle 5\times (2 - r)^2 - 20\, r^2 = 0\).

Either \(r = -2\) or \(\displaystyle r = \frac{2}{3}\approx 0.67\) will satisfy this equation. However, since this point (the point where the actual electric field is zero) should be between the two point charges, \(0 < r < 2\). Therefore, \((-2)\) isn't a valid value for \(r\) in this context.

As a result, the electric field is zero at the point approximately \(0.67\; \rm m\) away the \(5\times 10^{-19}\; \rm C\) charge, and approximately \(2 - 0.67 \approx 1.3\; \rm m\) away from the \(20 \times 10^{-19}\; \rm C\) charge.

Answer:

a)y = 485 m ,  v = 220 m / s , b)  y = 2954.39 m , c)   t_total = 51 s ,

d) v = 240.59 m / s

Explanation:

a) We can use vertical launch ratios for this exercise

the speed of the rocket the run out the fuel is

        v = v₀ + a t

the rocket departs with initial velocity v₀ = 0

        v = a t

        v = 55 4

        v = 220 m / s

the height at this point is

        y = y₀ + v₀t + ½ a t²

        y = y₀ + 1/2 a t²

        y = 45 + ½ 55 4²

        y = 485 m

b) the maximum height of the rocket is when its speed is zero

for this part we will use as the initial speed the speed at the end of the fuel (v₀´ = 220 m / s) and the height of y₀´ = 485 m

        v² = v₀´² + 2 g (y-y₀´ )

         0 = v₀´² + 2 g (y-y₀´ )

         y = y₀´ + v₀´² / 2g

         y = 485 + 220 2/2 9.8

         y = 2954.39 m

c) the time that the rocket is in the air is the acceleration time t₁ = 4 s, plus the rise time (t₂) plus the time to reach the ground (t₃)

let's calculate the rise time

           v = v₀´- g t

           v = 0

            t₂ = v₀´ / g

            t₂ = 220 / 9.8

            t₂ = 22.45 s

Now let's calculate the time it takes to get from this point (y₀´´ = 2954.39 m) to the floor

           y = y₀´´ + v₀´´ t - ½ g t²

           0 = y₀´´ - ½ g t²

          t = √ (2 y₀´´ / g)

          t = √ (2 2954.39 / 9.8)

          t = 24.55 s

the total flight time is

       t_total = t₁ + t₂ + t₃

       t_total = 4 + 22.45 + 24.55

        t_total = 51 s

d) the veloicda right now

       v = vo + g t

       v = 9.8 24.55

       v = 240.59 m / s

The velocity of the man given the data from the question is 0.45 m/s

Velocity is simply defined as the rate of change of displacement with time. Mathematically, it can be expressed as:

Velocity = Change of displacement / time

From the question given above, the following data were obtained:

The velocity of the man can be obtained as follow:

Velocity = Change of displacement / time

Velocity = 2 / 4.4

Velocity = 0.45 m/s

Thus, the velocity of the man is 0.45 m/s

Learn more about velocity:

https://brainly.com/question/3411682

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