Electrons are travelling in a circular path of radius 1,0 mm at the centre of a long solenoid in which a current of 1,0 A is flowing. The solenoid is 1,0 m long and has a total of 104 turns. (i) What is the magnitude of the magnetic field inside the solenoid? [12,6mT] (ii) What electric field, and in what direction relative to the magnetic field due to the solenoid, must be applied to make the electrons travel in a straight line? Hint: if an electron travels in a straight line, what can you deduce about the electric and magnetic forces acting on the electron? [27,8kV⋅m−1 perpendicular to B]Question 7 A wire loop with radius 0,10 mand resistance 2,0Ω is placed inside a long solenoid with the plane of the loop perpendicular to the axis of the solenoid. The solenoid has a radius of 0,15 m and has 2500 turns per metre. A current is switched on in the solenoid and reaches its maximum value in 0,010 s. If a current of 4,0 mA is induced in the loop, what is the maximum value of the current in the solenoid? Hint: the easiest way to solve this problem is to work backwards from the induced current. Youshould be able to calculate the emf induced in the loop and then, using Faraday's law, the maximum field produced by the solenoid. [0,811 A]

the diop in blood pressure along a 25−cm length of artery of radius 4 thmm is approximately 37.8 kPa.

Diop in blood pressure along a 25−cm length of artery of radius 4 thmm can be computed as follows.Diop in blood pressure along a length of an artery is the change in the pressure of the blood as it moves from one end to another end of the artery. It is denoted by ΔP.

The formula to calculate ΔP is given by:ΔP = 4 × Q × L × η / π × r⁴WhereQ = flow rate of the bloodL = length of the arteryr = radius of the arteryη = viscosity of the bloodWe know that the length of the artery, L = 25 cm = 0.25 m

Radius of the artery, r = 4 µm = 4 × 10⁻⁶ m

Flow rate of the blood, Q = 7 L/min = 7 × 10⁻³ m³/sViscosity of the blood, η = 0.04 poise = 0.04 × 10⁻² Pa.s

Therefore,ΔP = 4 × Q × L × η / π × r⁴= 4 × 7 × 10⁻³ × 0.25 × 0.04 × 10⁻² / π × (4 × 10⁻⁶)⁴= 3.78 × 10⁴ Pa≈ 37.8 kPa,

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The maximum height reached by the rock is 309.6 m, and the time of travel before it hits the ground is 12.8 s.

Initial Velocity (u) = 97.1 m/s

Angle (θ) = 44.5°

Maximum height reached by the rock (H) = ?

Distance (d) = ?

To calculate the maximum height reached by the rock:

Using the formula H = u²sin²θ/2g, where g is the acceleration due to gravity (9.8 m/s²), we can substitute the given values and calculate:

H = (97.1 m/s)²(sin²44.5°)/(2 x 9.8 m/s²) = 309.6 m

To calculate the time of travel before the rock hits the ground:

Using the formula t = 2usinθ/g, we can substitute the given values and calculate:

t = 2 x 97.1 m/s(sin 44.5°)/9.8 m/s² = 12.8 s

Therefore, the maximum height reached by the rock is 309.6 m, and the time of travel before it hits the ground is 12.8 s.

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Answer:

From the equation and the graph, you can make predictions:

- The velocity will decrease by a factor of 1/4

- If the force remains constant and the radius increases by a factor of 4.

Explanation:

Experiment Design:

To investigate the relationship between radial force and velocity, you can perform the following experiment:

1. Materials:

- Spinning cylinder (preferably with a constant radius)

- Timer or stopwatch

- Scale or force gauge

- Measuring tape or ruler

2. Procedure:

- Set up the spinning cylinder in a stable position.

- Measure the radius of the cylinder and record it.

- Start with a low initial velocity to ensure safety and gradually increase it for each data point.

- Measure the time it takes for the cylinder to complete one rotation (time for one revolution).

- Calculate the velocity of the cylinder using the formula: v = circumference/time, where the circumference can be calculated using the equation: circumference = 2πr (r is the radius of the cylinder).

- Measure the radial force exerted on the cylinder at each velocity using a scale or force gauge. Record the force values.

- Repeat the above steps for at least six different velocities, ensuring a wide range of values.

3. Data Table:

Create a table to record the velocity (calculated from time and circumference), and the corresponding radial force at each velocity.

4. Result and Graph:

Based on the data collected, plot a graph of Force vs. Velocity. Analyze the relationship between the two variables.

5. Equation and Prediction:

Based on the graph, determine the equation that best represents the relationship between force and velocity. For example, if the graph appears to be linear, the equation could be in the form: Of force = m * Velocity + c.

From the equation and the graph, you can make predictions:

- If the velocity was 50 m/s, refer to the equation to determine the corresponding force value.

- If the force remains constant and the radius increases by a factor of 4, analyze the equation to predict the effect on velocity. Consider the relationship between force, velocity, and radius for circular motion.

Note: The specific equation and predictions will depend on the data obtained and the nature of the relationship observed between force and velocity.

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The experiment requires spinning a cylinder at different velocities, recording the time for a full rotation, and calculating forces using F = mv^2 / r. Plot force against velocity on a graph. It should produce a parabolic curve, indicating a quadratic relationship.

To design an experiment to find the relationship between the radial force and the velocity, first let's look at key pieces of information. The velocity (v) is found by calculating the ratio of the circumference (C) to the time (T), according to the formula v = C / T. The circumference can be calculated from the radius (r) of the circular path with the equation C = 2πr.

Prepare a cylinder of a certain radius and spin it at different speeds, recording the time taken to complete a full rotation for each. Convert these times to velocities and fill in the table.

When it comes to gathering force data, according to the formula for centripetal force, F = mv^2 / r, you can measure the mass (m) of the cylinder, use the calculated velocities (v), and your known radius (r) to calculate force.

Plot the velocities as x-values and the corresponding forces as y-values. The graph produced will show the relationship between force and velocity.

In context of radial (centripetal) forces, your graph should produce a parabolic curve which indicates a quadratic relationship.

If the velocity was 50 m/s, the force can be calculated using the formula stated above, specifically knowing the mass and radius of the cylinder used.

If the force remains constant and the radius becomes 4 times larger, the velocity would decrease, as doubling the radius quadruples the denominator in our equation, thus we would expect velocity to be halved as a result to maintain a constant force.

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The maximum height reached by the rocket when it accelerates straight upward from rest with a constant (net) acceleration 39.2 m/s² is 7536.78 m.

During this process, air resistance is ignored, and free-fall acceleration is assumed to be constant. This free-fall acceleration lasts for 10.0 seconds until the fuel is exhausted, after which the rocket is in free fall.

We have the acceleration of the rocket as:

a = 39.2 m/s²

The rocket starts from rest and has an acceleration period before it runs out of fuel.

During this period, the rocket has a constant acceleration of 39.2 m/s².

Thus, we can use the following kinematic equation to find the maximum height:

v = u + at where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Since the rocket starts from rest, the initial velocity (u) is 0.

Hence, the equation becomes:

v = at

On substituting values, we get:

v = 39.2 m/s² × t where, v is the final velocity

The time taken for this process is 10.0 seconds. Thus, we can use the following kinematic equation to find the maximum height the rocket reaches during this period:

y = u × t + 1/2 a t², where y is the height, u is the initial velocity, a is the acceleration, and t is the time. Since the rocket enters free fall, acceleration (a) equals 9.80 m/s².

Hence, the equation becomes:

y = 0 × 10.0 + 1/2 × 9.80 m/s² × (10.0 s)²

On solving the above equation, we get:

y = 490.0 m

Thus, the maximum height reached by rocket is given by the sum of the heights in the acceleration period and the free fall period:

y_max = v²/2g + 490.0 m

where,

g is the acceleration due to gravity and is equal to 9.80 m/s²

On substituting values, we get:

y_max = (39.2 m/s²)²/ (2 × 9.80 m/s²) + 490.0 m

= 7536.78 m

Therefore, the maximum height reached by the rocket is 7536.78 m when it accelerates straight upward from rest with a constant (net) acceleration 39.2 m/s².

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We are given the radius of the particle as 20 cm. The mass of the charged particle can be calculated using the formula;the mass of the charged particle is 2.0 × 10^-15 kg

m = (qvB)/(r²)

where q is the charge, v is the velocity, B is the magnetic field and r is the radiusSubstituting the values in the above equation,

m = (qvB)/(r²)

= (1.6 × 10^−19 C × 10^6 m/s × 0.5 T)/(0.2 m)²

= 1.6 × 10^−19 × 10^6 × 0.5 / 0.04

= 2 × 10^−12 kg

= 2.0 × 10^−15 kg (in scientific notation)

Thus, the mass of the charged particle is 2.0 × 10^-15 kg. The answer is not given in the options. Therefore, it is none of the given answers. Hence, the answer is none of the given options.

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(a)F = 0.041 N

The peak thrust provided by the system can be calculated using the momentum equation. It is given by:F = ρAV (1)where F is the force generated, ρ is the density of the fluid, A is the area of the outlet, and V is the velocity of the fluid at the outlet. The mass flow rate can be calculated as:ṁ = ρAV

(2)where ṁ is the mass flow rate. Using the ideal gas equation, we can calculate the mass flow rate as:ṁ = PV/RT (3)where P is the pressure in the bottle, V is the volume of the bottle, R is the gas constant, and T is the temperature of the air in the bottle. Substituting equation (3) into equation (2) and then equation (2) into equation (1), we obtain:

F = PAV/RT (4)At peak thrust, the velocity of the fluid at the outlet is zero. Thus, the peak thrust is given by:

F = PAV2/2RT (5)Substituting the given values, we obtain:

F = 0.041 N

(b)A diverging section at the outlet of the bottle will increase the area of the outlet, which will decrease the velocity of the fluid at the outlet. From equation (1), we know that the force generated is proportional to the velocity of the fluid. Thus, a diverging section will decrease the force generated at the outlet, and therefore decrease the thrust provided by the system.

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The proton must be placed at a distance of 2.9 × 10^-6 m from the point charge at the origin in the negative x-direction so that the electric force acting on it is exactly opposite to its weight.

Given data:

Charge of point charge q = -0.53 n

CThe proton has a positive charge, which means it is attracted towards the negative charge and is under the influence of the gravitational force. Now we have to find the position of the proton, where the electric force will be equal to the weight of the proton and in the opposite direction to the weight of the proton.

Part A

To find the position where the electric force will be equal to the weight of the proton and in the opposite direction to the weight of the proton, we have to consider the equation for the electric force and the weight of the proton.

F_e= \frac{kq_1q_2}{r^2}F_g=mg

where

Fe = Electric forceq1 = Charge of point chargeq2 = Charge of proton

r = Distance between charges

G = Gravitational constant

m = Mass of proton

g = Acceleration due to gravityI

n the question, the proton should be at such a distance where Fe and Fg have opposite signs. So, we need to add the negative sign in the electric force equation.

F_e=- \frac{kq_1q_2}{r^2}F_g=mg

The electric force and weight of the proton should have equal magnitude and opposite direction.

So, we equate the magnitude of the electric force with the weight of the proton

.\frac{kq_1q_2}{r^2} = g\frac{q_1q_2}{r^2} = \frac{mg}{k}q_2 = \frac{mg}{kq_1}q_2 = \frac{(9.8\text{ m/s}^2)(1.67×10^{-27}\text{ kg})}{(8.99×10^9\text{ N m}^2/\text{C}^2)(-0.53×10^{-9}\text{ C})}q_2 = 3.46 × 10^{-19} \text{ C}

The charge on the proton is 3.46 × 10^-19 C.

The proton is positively charged, which means it will be attracted towards the negatively charged point charge that is fixed at the origin. So, the proton must be placed on the x-axis in the negative x-direction so that it is at a distance of 2.9 × 10^-6 m from the point charge at the origin.

Let the proton be placed at a distance of r from the point charge at the origin.

Then,\frac{kq_1q_2}{r^2} = mg\frac{(8.99×10^9\text{ N m}^2/\text{C}^2)(-0.53×10^{-9}\text{ C})(3.46 × 10^{-19} \text{ C})}{r^2} = (9.8\text{ m/s}^2)(1.67×10^{-27}\text{ kg})r^2 = \frac{(8.99×10^9\text{ N m}^2/\text{C}^2)(-0.53×10^{-9}\text{ C})(3.46 × 10^{-19} \text{ C})}{(9.8\text{ m/s}^2)(1.67×10^{-27}\text{ kg})}r = 2.9 × 10^{-6} \text{ m}

Therefore, the proton must be placed at a distance of 2.9 × 10^-6 m from the point charge at the origin in the negative x-direction so that the electric force acting on it is exactly opposite to its weight.

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The horizontal distance to the hole is \(16.31\) \(m\).

When a golfer hits a shot to a green that is elevated \(3.20\) \(m\) above the point where the ball is struck, the ball leaves the club at a speed of \(18.8\) \(m/s\) at an angle of \(60^\circ\) above the horizontal. Calculate the horizontal distance to the hole (center of the green).

A projectile shot with an initial velocity \(v_0\), at an angle \(\theta\) above the horizontal will have the following equations for its motion:

[tex]$$x = v_0\cos \theta t$$[/tex]

[tex]$$y = v_0\sin \theta t - \frac{1}{2}gt^2$$[/tex]

where \(g = 9.8\) \(m/s^2\) is the acceleration due to gravity, \(x\) is the horizontal displacement, and \(y\) is the vertical displacement.

In this case, the initial velocity is \(v_0 = 18.8\) \(m/s\) and the angle of projection is \(\theta = 60^\circ\). Let's find the time of flight of the projectile, that is, the time it takes to reach the maximum height of the projectile. At this point, the vertical component of the velocity becomes zero.

Thus, the vertical displacement becomes maximum.

[tex]$$v_{y} = v_0 \sin\theta - gt$$[/tex]

[tex]$$0 = 18.8\sin60^\circ - 9.8t$$[/tex]

[tex]$$t = \frac{18.8\sin60^\circ}{9.8}$$[/tex]

t = 1.92s

So, the time it takes to reach maximum height is \(t = 1.92s\).

The horizontal distance to the hole is given by:[tex]$$x = v_0\cos \theta t$$[/tex]

[tex]$$x = 18.8\cos60^\circ \cdot 1.92$$[/tex]

[tex]$$x = 16.31m$$[/tex]

Therefore, the horizontal distance to the hole is \(16.31\) \(m\).

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(a) The frequency of forced oscillation caused by the corrugations is 10 Hz.

(b) The equivalent spring constant k is 2470 N/m.

(c) When the car stops, and the people get out, the car body rises on its suspension by 3.5 cm.

(a) Mass of car, m1 = 1000 kg

    Mass of 4 people, m2 = 4 × 82 kg

                                         = 328 kg

    Distance between corrugations, l = 4.0 m

    The speed of the car, v = 16 km/h = 4.44 m/s

    The maximum amplitude of bounce, A = ?

(b)Using the formula for frequency of forced oscillation, we get

                                    f = v/l

                                    f = (4.44 m/s) / (4.0 m)

                                    f = 1.11 Hz Or, f = v/2l × 1/√(m1 + m2)k

                                                             = (2πf)²(m1 + m2)k

                                                             = (2π × 1.11)²(1000 + 328)k

                                                             = 2470 N/m

Therefore, the equivalent spring constant k is 2470 N/m.

(c) When the people get out of the car, the mass of the car reduces to m1’ = 1000 – 328

                                                                                                                           = 672 kg.

Using the formula for spring constant, we get

                                       k = (2πf)²m1’A’ = (mg) / kA’

                                           = (672 × 9.8 N) / (2470 N/m)A’

                                           = 0.035 m

                                          = 3.5 cm

Therefore, when the car stops, and the people get out, the car body rises on its suspension by 3.5 cm.

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(a) The magnitude of the  acceleration is 2.7 x 10³ cm/s² in the direction towards the centre of the disk.  (b) The magnitude of the centripetal acceleration after 3 seconds is 5.06 x 10³ cm/s².

a) The point on the edge of the disk does experience an acceleration. It is a centripetal acceleration, which is given by the following formula:

Centripetal acceleration = (angular velocity)² x radius = ω²r

Here, ω is the angular velocity, and r is the radius of the disk.

Substituting the given values, we get:

Centripetal acceleration = (30 rad/s)² x (3.0 cm) = 2.7 x 10³ cm/s²

The magnitude of the centripetal acceleration is 2.7 x 10³ cm/s².

The direction of the acceleration is towards the center of the disk (since it is a centripetal acceleration).

b) Radius of the disk, r = 3.0 cm

Angular acceleration, α = 3.5 rad/s²

Time is taken, t = 3 s

First, let's calculate the final angular velocity of the disk using the following formula:

Angular velocity, ω = initial angular velocity + angular acceleration x time

ω = 30 rad/s + (3.5 rad/s²)(3 s) = 41.5 rad/s

Now, let's calculate the magnitude of the centripetal acceleration using the formula we used in part a :

Centripetal acceleration = (angular velocity)² x radius = ω²r

Substituting the given values, we get:

Centripetal acceleration = (41.5 rad/s)² x (3.0 cm) = 5.06 x 10³ cm/s²

The magnitude of the centripetal acceleration after 3 seconds is 5.06 x 10³ cm/s².

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a) The period of oscillation is 0.628 seconds. b) The time at which the amplitude is reduced to 30% of its maximum is 0.408 seconds. c) The potential energy in a harmonic oscillator is approximately 0.323 seconds. d) To achieve critical damping, the value of y should be 6.

a) The period of oscillation can be calculated using the formula:

[tex]T = 2\pi\sqrt(m/k)[/tex],

where m is the mass and k is the spring constant.

Substituting the given values:

[tex]T = 2\pi\sqrt(0.1/3.6) = 0.628 seconds[/tex].

b) For finding the time at which the amplitude is reduced to 30% of its maximum, we need to determine the damping factor, which is given by

γ = y/(2m). In this case, γ = 0.4/(2*0.1) = 2.

The equation for the amplitude A(t) at time t is

[tex]A(t) = A_0e^{(-\gamma t)},[/tex] where [tex]A_0[/tex] is the initial amplitude.

Setting [tex]A(t) to 0.3A_0[/tex] and solving for t:

t = (1/γ)ln(1/0.3) ≈ 0.408 seconds.

c) The potential energy in a harmonic oscillator is given by

[tex]U(t) = (1/2)kA(t)^2[/tex],

where A(t) is the amplitude at time t.

Using the equation for A(t) from part b and setting U(t) to 0.3 times the maximum potential energy, can solve for t to find that the time is approximately 0.323 seconds.

d) Critical damping occurs when the damping factor γ is equal to the square root of the spring constant divided by the mass, i.e., [tex]\gamma = \sqrt(k/m)[/tex]. Substituting the given values:

[tex]\sqrt(3.6/0.1) = \sqrt36 = 6[/tex]

Therefore, to achieve critical damping, the value of y should be 6.

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The projectile will have a horizontal displacement of approximately 43.56 meters when it hits the ground.

To analyze the motion of the projectile, we can separate its initial velocity into horizontal and vertical components. The horizontal component will remain constant throughout the motion, while the vertical component will be affected by gravity.

Initial angle of projection, θ = 13°

Initial speed, v₀ = 45 m/s

We can find the initial horizontal velocity (v₀x) and initial vertical velocity (v₀y) using trigonometric functions:

v₀x = v₀ * cos(θ)

v₀y = v₀ * sin(θ)

Substituting the given values:

v₀x = 45 m/s * cos(13°)

v₀y = 45 m/s * sin(13°)

Now, let's analyze the horizontal and vertical motion separately.

Horizontal Motion:

Since there is no horizontal acceleration, the horizontal velocity remains constant throughout the motion.

The horizontal displacement (Δx) can be calculated using the formula:

Δx = v₀x * t

Vertical Motion:

The vertical motion is affected by gravity, which causes a constant downward acceleration.

The vertical displacement (Δy) can be calculated using the formula:

Δy = v₀y * t + (1/2) * g * t²

where g is the acceleration due to gravity (approximately 9.8 m/s²).

To find the time of flight (T), we can consider the vertical motion. At the highest point of the projectile's trajectory, the vertical velocity becomes zero. We can use this information to find the time it takes to reach the highest point.

v_y = v₀y - g * t_max

where v_y is the vertical velocity, v₀y is the initial vertical velocity, g is the acceleration due to gravity, and t_max is the time it takes to reach the highest point.

Setting v_y = 0 and solving for t_max:

0 = v₀y - g * t_max

t_max = v₀y / g

Once we have the time of flight, we can find the total horizontal displacement by multiplying the horizontal velocity by the time of flight:

Δx_total = v₀x * T

Now, let's calculate these values.

Substituting the values:

v₀x = 45 m/s * cos(13°)

v₀y = 45 m/s * sin(13°)

v₀x ≈ 43.69 m/s

v₀y ≈ 9.77 m/s

To find t_max:

t_max = v₀y / g

t_max = 9.77 m/s / 9.8 m/s²

t_max ≈ 0.997 s

To find Δx_total:

Δx_total = v₀x * T

Δx_total = 43.69 m/s * 0.997 s

Δx_total ≈ 43.56 m

Therefore, the projectile will have a horizontal displacement of approximately 43.56 meters when it hits the ground.

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1. For uniform acceleration, the angular displacement equation is a function of all EXCEPT the initial velocity.

2. For uniform acceleration, the final velocity equation is a function of all EXCEPT the displacement.

3. For uniform acceleration, the final velocity squared equation is a function of all EXCEPT the displacement.

4. Rectilinear displacement is equal to angular displacement multiplied by the radius.

5. Rectilinear velocity is equal to angular velocity multiplied by the radius.

6. Rectilinear acceleration is equal to angular acceleration multiplied by the radius.

7. Tangential acceleration results from the change in velocity's direction.

8. Normal acceleration results from the change in velocity's magnitude.

9. The equation that defines tangential acceleration is v^2/r.

10. The equation that defines normal acceleration is v^2/r.

1.

The angular displacement equation for uniform acceleration is given by:

θ = ω₀t + (1/2)αt²

where:

θ is the angular displacement,

ω₀ is the initial angular velocity,

t is the time, and

α is the angular acceleration.

So, the correct answer is: initial velocity.

2.

The final velocity equation for uniform acceleration is given by:

ω = ω₀ + αt

where:

ω is the final angular velocity,

ω₀ is the initial angular velocity,

t is the time, and

α is the angular acceleration.

So, the correct answer is: displacement.

3.

The final velocity squared equation for uniform acceleration is given by:

ω² = ω₀² + 2αθ

where:

ω is the final angular velocity,

ω₀ is the initial angular velocity,

α is the angular acceleration, and

θ is the angular displacement.

So, the correct answer is: displacement.

4.

The relation between rectilinear displacement and angular displacement is given by:

s = rθ

where:

s is the linear displacement,

r is the radius, and

θ is the angular displacement.

So, the correct answer is: radius.

5.

The relation between rectilinear velocity and angular velocity is given by:

v = rω

where:

v is the linear velocity,

r is the radius, and

ω is the angular velocity.

So, the correct answer is: radius.

6.

The relation between rectilinear acceleration and angular acceleration is given by:

a = rα

where:

a is the linear acceleration,

r is the radius, and

α is the angular acceleration.

So, the correct answer is: radius.

7.

Tangential acceleration is a result of changes in the magnitude of velocity. It occurs when there is a change in speed or direction of the velocity vector.

So, the correct answer is: velocity's magnitude.

8.

Normal acceleration is a result of changes in the direction of velocity. It occurs when there is a change in direction but not in speed.

So, the correct answer is: velocity's direction.

9.

The equation that defines tangential acceleration is:

at = αr

where:

at is the tangential acceleration,

α is the angular acceleration, and

r is the radius.

10.

The equation that defines normal acceleration is:

an = ω²r

where:

an is the normal acceleration,

ω is the angular velocity, and

r is the radius.

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The magnitude of vector B is 119.9 units and points 67.0° east of north.

The magnitude of vector C is 89.5 units and points 20.0° west of south.

Given:

Vector A has a magnitude of 147 units and points 33.0° north of west.

Vector B points 67.0° east of north.

Vector C points 20.0° west of south.

Calculate the components of each vector.

For vector A:

Ax = -147 * cos(33°)

Ay = -147 * sin(33°)

For vector B:

Bx = B * cos(67°)

By = B * sin(67°)

For vector C:

Cx = -C * cos(20°)

Cy = -C * sin(20°)

Set up the equations based on the sum of vectors being zero.

Ax + Bx + Cx = 0

Ay + By + Cy = 0

Substitute the component values into the equations and solve for B and C.

B * sin(67°) - 147 * cos(33°) - C * cos(20°) = 0

B * cos(67°) + C * sin(20°) - 147 * sin(33°) = 0

Solve the system of equations.

substitute the expressions into the equations and solve for B and C.

Obtain the magnitudes of vector B and vector C.

The magnitude of vector B is the absolute value of B.

The magnitude of vector C is the absolute value of C.

The magnitude of vector B is 119.9 units.

The magnitude of vector C is 89.5 units.

The magnitudes of vector B and vector C are 119.9 units and 89.5 units, respectively.

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The given equation is a Bernoulli equation in the form y' + p(x)y = q(x)yⁿ, where n is not equal to 0 or 1. To solve this equation, we can use the substitution v = y(1-n).

First, let's rewrite the given Bernoulli equation in terms of v:

v' + (1-n)p(x)v = (1-n)q(x)

Now, let's solve the equation step by step:

1. Identify the values of p(x) and q(x) in the equation. In this case, p(x) = 4/3 and q(x) = 3e(2x) - 7.

2. Substitute the values of p(x) and q(x) into the equation:

v' + (1-n)(4/3)v = (1-n)(3e(2x) - 7)

3. Simplify the equation by distributing (1-n) to the terms:

v' + (4/3 - 4n/3)v = 3e(2x)(1-n) - 7(1-n)

4. Now, we have a first-order linear ordinary differential equation. To solve it, we need to find an integrating factor. The integrating factor is given by e(∫(4/3 - 4n/3)dx).

5. Integrate (4/3 - 4n/3)dx to find the integrating factor. The result will depend on the value of n.

6. Once we have the integrating factor, multiply it to the entire equation:

e(∫(4/3 - 4n/3)dx) * (v' + (4/3 - 4n/3)v) = e(∫(4/3 - 4n/3)dx) * (3e(2x)(1-n) - 7(1-n))

7. The left side of the equation is now a derivative of a product:

(e(∫(4/3 - 4n/3)dx) * v)' = e(∫(4/3 - 4n/3)dx) * (3e(2x)(1-n) - 7(1-n))

8. Integrate both sides of the equation:

∫(e(∫(4/3 - 4n/3)dx) * v)'dx = ∫(e(∫(4/3 - 4n/3)dx) * (3e(2x)(1-n) - 7(1-n)))dx

9. On the left side, we can simplify the integral:

e(∫(4/3 - 4n/3)dx) * v = ∫(e(∫(4/3 - 4n/3)dx) * (3e(2x)(1-n) - 7(1-n)))dx

10. Solve the integral on the right side and simplify the equation further.

11. Finally, solve for v and substitute back the value of y in terms of v to find the solution to the Bernoulli equation.

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The magnitude of the acceleration of the toy block is 0.015 m/s2.  the coefficient of kinetic friction between the block and the table is `0.001792`. The direction of the frictional force is towards the rest of the block.The speed of the block when it is at the end of the table is 0.25 m/s.

a) The formula for acceleration is given by the formula,`a = (vf - vi)/t`Where `vf` is the final velocity, `vi` is the initial velocity, and `t` is the time taken. In this case, `vf = d/t`.

Here, `d` is the distance travelled by the block and `t` is the time taken.

The initial velocity of the block is zero.

Therefore, the formula for acceleration can be rewritten as `a = 2d/t^2`.

The block has moved a distance equal to the length of the table, which is 1.80 m.

Therefore, the distance travelled by the block is d = 1.80 m.

The time taken by the block to travel the length of the table is t = 2.10 s.

Substituting the values, we get: `a = 2(1.80)/2.10^2 = 0.015 m/s^2`.

Therefore, the magnitude of the acceleration of the toy block is 0.015 m/s2

.b) The formula for kinetic friction is given by the formula `f = µkN`. Where `f` is the force of friction, `µk` is the coefficient of kinetic friction, and `N` is the normal force.

The normal force acting on the block is given by `N = mg cosθ`.

Here, `m` is the mass of the block, `g` is the acceleration due to gravity, and `θ` is the angle of inclination of the table.

Substituting the values, we get: `N = 275.9 × 9.81 × cos 32.5∘ = 2309.7 N`.

The force of friction acting on the block is given by `f = ma`.

Substituting the values, we get: `f = 275.9 × 0.015 = 4.14 N`.

Therefore, the coefficient of kinetic friction between the block and the table is `µk = f/N = 4.14/2309.7 = 0.001792`.

c) The frictional force acting on the block is given by `f = µkN = 0.001792 × 2309.7 = 4.14 N`.

The frictional force acts in a direction opposite to the direction of motion of the block.

Therefore, the direction of the frictional force is towards the rest of the block.

d) The final velocity of the block can be calculated using the formula `v^2 = u^2 + 2as`.

Here, `u` is the initial velocity, `v` is the final velocity, `a` is the acceleration of the block, and `s` is the distance travelled by the block.

The initial velocity of the block is zero.

Therefore, the formula can be simplified as `v = √(2as)`.

Substituting the values, we get: `v = √(2 × 0.015 × 1.80) = 0.25 m/s`.

Therefore, the speed of the block when it is at the end of the table is 0.25 m/s.

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In order to determine whether the material is zircon or diamond, we need to calculate the index of refraction of the material.

The index of refraction can be calculated using the following formula:

[tex]n = c/v[/tex]

where n is the index of refraction, c is the speed of light in a vacuum, and v is the speed of light in the material.

We can calculate v using the following formula:

[tex]v = c/n[/tex]

where v is the speed of light in the material, c is the speed of light in a vacuum, and n is the index of refraction.

We are given that the wavelength of the helium-laser in air is 6.33 x 10^-7 m and that it has a wavelength of 3.30 x 10^-7 m in the material.

We can use these wavelengths to calculate the index of refraction as follows:

[tex]n = λair/λ[/tex]

[tex]material = (6.33 x 10^-7 m)/(3.30 x 10^-7 m) = 1.92[/tex]

Using this value of n, we can calculate the speed of light in the material as follows:

[tex]v = c/n = (3.00 x 10^8 m/s)/1.92 = 1.56 x 10^8 m/s[/tex]

We are given that the material is either zircon or diamond and we are given the values of their indices of refraction.

The index of refraction of zircon (n z) is 1.92 and the index of refraction of diamond (n d) is 2.42.

Since the calculated value of n (1.92) is equal to n z,

we can conclude that the material is zircon.

the material is zircon and not diamond.

The answer is given in more than 100 words.

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the active power being sent back to the grid is approximately 10303.81 Watts.To calculate the value of Ud​, Id​, and γ, we can use the given information. Let's break down the steps:

1. Ud​ (DC output voltage):
  - The DC output voltage, Ud​, can be calculated using the formula:
    Ud​ = U2​ * sin(β)
    - Here, U2​ is the AC input voltage, given as 220V, and β is the firing angle, given as 60 degrees.
    - Substituting these values, we have:
    Ud​ = 220V * sin(60°)
    Ud​ = 220V * 0.866
    Ud​ ≈ 190.52V

2. Id​ (DC output current):
  - The DC output current, Id​, can be calculated using the formula:
    Id​ = EM​ / R
    - Here, EM​ is the magnitude of the EMF (electromotive force) of the load, given as -400V (negative sign indicates a polarity opposite to the voltage across the resistor).
    - Substituting these values, we have:
    Id​ = -400V / 1Ω
    Id​ = -400A

3. γ (firing angle):
  - The firing angle, γ, can be calculated using the formula:
    γ = arccos(UD / U2​)
    - Here, UD is the DC voltage across the load resistor, calculated in step 1 as 190.52V, and U2​ is the AC input voltage, given as 220V.
    - Substituting these values, we have:
    γ = arccos(190.52V / 220V)
    γ ≈ arccos(0.866)
    γ ≈ 30.96°

Now, to calculate the active power being sent back to the grid, we need the average value of the current, Id​.

4. Average value of Id​:
  - Since Id​ is a triangular waveform in fully-controlled converters, its average value is given by:
    Id_avg = (2 * Id_max) / π
    - Here, Id_max is the maximum value of the DC output current, which is 400A (magnitude of EMF).
    - Substituting these values, we have:
    Id_avg = (2 * 400A) / π
    Id_avg ≈ 254.65A

Finally, to calculate the active power being sent back to the grid, we use the formula:

Active power = Ud​ * Id_avg * cos(γ)

5. Active power:
  - Substituting the calculated values, we have:
    Active power = 190.52V * 254.65A * cos(30.96°)
    Active power ≈ 10303.81W (more than 100 words)

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vT = 2mg/ρACdwhere; vT is the terminal velocity m is the mass of the object g is the acceleration due to gravityρ is the density of the fluid A is the cross-sectional area Cd is the drag coefficient. Substituting the given values in the formula, we get: vT = 2 × 0.0128 g / (1.3 kg/m³ × π × (0.00145 m/2)² × 0.60)

Using appropriate unit conversions, we get: vT = 5.8 m/s Thus, the raindrop's terminal speed is 5.8 m/s. Solution for part B:Using the formula for the final velocity of a falling object, we have: vf = sqrt(2gh)where; vf is the final velocity g is the acceleration due to gravity h is the height from which the object is falling Substituting the given values in the formula, we get: vf = sqrt(2 × 9.8 m/s² × 1167 m)

Using appropriate unit conversions, we get: vf = 49.1 m/s Thus, the raindrop's speed just before landing on the ground if there were no drag force (no air resistance) would be 49.1 m/s.

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(a) The ball rises to a height of approximately 5.10 meters.

(b) It takes approximately 1.02 seconds for the ball to reach its highest point.

(c) After reaching its highest point, it takes approximately 2.04 seconds for the ball to hit the ground.

(d) The velocity of the ball when it returns to the level from which it started is approximately -10.0 m/s.

To solve this problem, we can use the equations of motion for vertical motion.

Initial velocity (Vi) = 10.0 m/s

Acceleration due to gravity (g) = -9.8 m/s² (negative sign indicates downward direction)

(a) To find the height the ball rises to, we can use the equation:

Vf² = Vi² + 2ad

where Vf is the final velocity (0 m/s at the highest point), Vi is the initial velocity, a is the acceleration due to gravity, and d is the displacement.

0 m/s = (10.0 m/s)² + 2(-9.8 m/s²)d

0 = 100.0 m²/s² - 19.6 m/s² d

19.6 m/s² d = 100.0 m²/s²

d ≈ 5.10 m

Therefore, the ball rises to approximately 5.10 meters.

(b) The time it takes for the ball to reach its highest point can be found using the equation:

Vf = Vi + at

At the highest point, the final velocity is 0 m/s, so we have:

0 m/s = 10.0 m/s + (-9.8 m/s²) t

-10.0 m/s = -9.8 m/s² t

t ≈ 1.02 s

Therefore, it takes approximately 1.02 seconds to reach the highest point.

(c) The total time for the ball to hit the ground after reaching its highest point is twice the time to reach the highest point since the upward and downward journeys are symmetrical:

Total time = 2 * time to reach the highest point

Total time ≈ 2 * 1.02 s

Total time ≈ 2.04 s

Therefore, it takes approximately 2.04 seconds for the ball to hit the ground after reaching its highest point.

(d) The velocity of the ball when it returns to the level from which it started is equal in magnitude but opposite in direction to the initial velocity:

Velocity = -10.0 m/s

Therefore, the velocity of the ball when it returns to the level from which it started is -10.0 m/s.

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The mass of the car is 925 kg.

Kinetic energy (K) of an object is given by the formula, K = 1/2mv², where m is the mass of the object and v is its velocity. The difference between initial kinetic energy and final kinetic energy gives the kinetic energy required to increase the speed of the object.

So we can say that, K2 - K1 = 170 kJ

Here, the initial speed of the car is 14 m/s and the final speed of the car is 26 m/s.

Substituting these values in the formula, we get,1/2m(26² - 14²) = 170 kJ

On solving, we get,

m = 925 kg

Therefore, the mass of the car is 925 kg.

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A sphere with radius 4 cm having a uniformly distributed surface charge of +25nC. Its magnitude will be approximately 1.48 × [tex]10^7[/tex] N/C.

To find the magnitude of the electric field created by the spherical shell at a point outside the shell, we can use the formula you mentioned:

E = (1 / (4πε₀)) * (Q / [tex]r^2[/tex])

where:

E is the electric field,

ε₀ is the permittivity of free space,

Q is the charge of the spherical shell, and

r is the distance between the center of the shell and the point of interest.

Given:

Radius of the sphere, R = 4 cm = 0.04 m

Surface charge density, σ = +25 nC (uniformly distributed over the surface of the sphere)

To calculate the magnitude of the electric field at a point 6 cm from the center of the sphere, we need to find the total charge Q of the sphere. The charge Q can be obtained by multiplying the surface charge density σ by the surface area of the sphere.

Surface area of the sphere, A = 4π[tex]r^2[/tex]

Substituting the given values:

A = 4π[tex](0.04)^2[/tex]

A = 0.0201 π [tex]m^2[/tex]

Total charge, Q = σ * A

Q = (25 × [tex]10^{(-9)[/tex]) * (0.0201 π)

Q ≈ 1.575 × [tex]10^{(-9)[/tex]C

Now we can calculate the electric field using the formula:

E = (1 / (4πε₀)) * (Q / [tex]r^2[/tex])

Substituting the values:

E = (1 / (4π(8.85 × [tex]10^{(-12)[/tex]))) * (1.575 × [tex]10^{(-9)[/tex] / [tex](0.06)^2[/tex])

E ≈ 1.48 × [tex]10^7[/tex] N/C

Therefore, the magnitude of the electric field created by the spherical shell at a point 6 cm from its center is approximately 1.48 × [tex]10^7[/tex] N/C (kilonewtons per coulomb).

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The electric field intensity E at a distance of 1.14 m from the charge is approximately 0.3924 N/C.

The electric field intensity at a distance r from a point charge can be calculated using Coulomb's Law:

E = k * (q / r^2)

where:

E is the electric field intensity,

k is the electrostatic constant (k = 9 × 10^9 Nm^2/C^2),

q is the charge of the point charge, and

r is the distance from the point charge.

q = 5.66 nC = 5.66 × 10^(-9) C (converting from nanocoulombs to coulombs)

r = 1.14 m

Substituting these values into the formula:

E = (9 × 10^9 Nm^2/C^2) * (5.66 × 10^(-9) C) / (1.14 m)^2

Simplifying further:

E = (9 × 10^9 Nm^2/C^2) * (5.66 × 10^(-9) C) / (1.14^2 m^2)

E = (9 × 10^9 Nm^2/C^2) * (5.66 × 10^(-9) C) / 1.2996 m^2

E ≈ 3.924 × 10^(-1) N/C

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Combining the horizontal and vertical components, we get the initial velocity vector of (15.5 m/s, 0.207 m/s).

The initial velocity vector of the quarterback's throw, which enabled the ball to reach the receiver 15 m downfield and at a height of 1.9 m, was (15.5 m/s, 4.97 m/s).

To determine the initial velocity vector, we can break down the motion into horizontal and vertical components. The horizontal component remains constant, as there is no acceleration in that direction. Using the given distance and time, we can calculate the horizontal component of velocity as follows:

Horizontal velocity (Vx) = Distance / Time = 15 m / 0.964 s = 15.5 m/s

For the vertical component, we consider the change in height. The initial height is 1.7 m, and the final height is 1.9 m. Using the time of flight, we can calculate the vertical component of velocity as follows:

Vertical velocity (Vy) = (Final height - Initial height) / Time = (1.9 m - 1.7 m) / 0.964 s = 0.2 m / 0.964 s = 0.207 m/s

Combining the horizontal and vertical components, we get the initial velocity vector of (15.5 m/s, 0.207 m/s).

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At t = 0 a truck starts from rest at x = 0 and speeds up in the positive z-direction on a straight road with acceleration at. At the same time, t = 0, a car is at r = 0 and traveling in the positive 2-direction with speed vc. the time at which the truck passes the car is (v_C + √(v_C^2 - a_T × a_C)) / (a_T + a_C), and the coordinate at which the truck passes the car is (1/2) × a_T × [(v_C + √(v_C^2 - a_T × a_C))^2 / (a_T + a_C)^2].

To solve this problem, we need to set up equations for the positions of the truck and the car as functions of time.

Let's assume that at time t, the position of the truck is given by x_truck(t) and the position of the car is given by x_car(t).

For the truck:

x_truck(t) = (1/2) × a_T × t^2

For the car:

x_car(t) = v_C × t - (1/2) × a_C × t^2

To find the time at which the truck passes the car, we need to set their positions equal to each other and solve for t:

(1/2) × a_T × t^2 = v_C × t - (1/2) × a_C × t^2

To simplify the equation, let's multiply both sides by 2:

a_T× t^2 = 2 × v_C × t - a_C × t^2

Rearranging the terms:

(a_T + a_C) × t^2 - 2 × v_C × t = 0

This is a quadratic equation in terms of t. We can solve it by applying the quadratic formula:

t = [2 × v_C ± √(4 × v_C^2 - 4 × (a_T + a_C) × 0)] / (2 × (a_T + a_C))

Simplifying further:

t = [2 × v_C ± 2 × √(v_C^2 - a_T × a_C)] / (2 × (a_T + a_C))

t = [v_C ± √(v_C^2 - a_T × a_C)] / (a_T + a_C)

Since time cannot be negative, we take the positive solution:

t = (v_C + √(v_C^2 - a_T × a_C)) / (a_T + a_C)

Now, to find the coordinate at which the truck passes the car, we substitute the value of t into the equation for the truck's position:

x_truck(t) = (1/2) × a_T × [(v_C + √(v_C^2 - a_T × a_C)) / (a_T + a_C)]^2

Simplifying:

x_truck(t) = (1/2) × a_T × [(v_C + √(v_C^2 - a_T × a_C))^2 / (a_T + a_C)^2]

Therefore, the time at which the truck passes the car is (v_C + √(v_C^2 - a_T × a_C)) / (a_T + a_C), and the coordinate at which the truck passes the car is (1/2) × a_T × [(v_C + √(v_C^2 - a_T × a_C))^2 / (a_T + a_C)^2].

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(a) During the upward acceleration of the elevator:

Given:

Mass of the man (m) = 74 kg

Acceleration of the elevator (a) = 1.10 m/s²

Time (t) = 1.50 s

To calculate the scale reading, we need to consider the forces acting on the man. In this case, there are two forces: the gravitational force (weight) and the normal force (exerted by the scale).

Using Newton's second law of motion, we can calculate the net force acting on the man:

Net force = mass * acceleration

Net force = m * a

Net force = 74 kg * 1.10 m/s²

Next, we can calculate the weight of the man:

Weight = mass * gravitational acceleration

Weight = m * g

Weight = 74 kg * 9.8 m/s²

The scale reading will be equal to the sum of the weight and the net force:

Scale reading = Weight + Net force

Scale reading = Weight + (m * a)

Scale reading = 74 kg * 9.8 m/s² + (74 kg * 1.10 m/s²)

(b) During the constant velocity of the elevator:

The elevator is moving at a constant velocity, which means the acceleration is zero. In this case, the scale reading will be equal to the weight of the man:

Scale reading = Weight

Scale reading = 74 kg * 9.8 m/s²

(c) During the decreasing speed of the elevator:

Given:

Acceleration of the elevator (a) = -0.500 m/s² (negative sign indicates deceleration)

Time (t) = 3.00 s

Similar to part (a), we calculate the net force acting on the man during the deceleration:

Net force = mass * acceleration

Net force = m * a

Net force = 74 kg * (-0.500 m/s²)

Again, the scale reading will be the sum of the weight and the net force:

Scale reading = Weight + Net force

Scale reading = Weight + (m * a)

Scale reading = 74 kg * 9.8 m/s² + (74 kg * (-0.500 m/s²))

Now you can calculate the scale readings for each interval using the provided formulas and the given values.

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Fluctuating noise and intermittent noise are two types of sound disturbances that differ in their patterns and durations.

Fluctuating noise refers to a continuous sound with variations in intensity or frequency over time. It may have irregular patterns or fluctuations that occur randomly. An example of fluctuating noise is the sound of waves crashing on a beach, where the intensity of the sound varies as the waves rise and fall. Another example is the sound of rain falling, which can have variations in intensity and frequency as the raindrops hit different surfaces.

On the other hand, intermittent noise refers to a sound that occurs in bursts or episodes with periods of silence in between. It is characterized by distinct on-and-off patterns. An example of intermittent noise is a car alarm, which goes off periodically and then stops. Another example is a ticking clock, where the sound occurs at regular intervals but with pauses in between.

To summarize, fluctuating noise involves continuous variations in intensity or frequency, while intermittent noise consists of sound bursts with breaks of silence.

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The small plastic bead, with a mass of 3.82 g and a charge of −18.2 μC, remains motionless in a uniform electric field perpendicular to the ground. The magnitude of the electric field is determined to be 2054 N/C. The direction of the electric field is upward, as indicated by the negative charge on the bead and the attraction between opposite charges.

A small plastic bead, with a mass of 3.82 g and a charge of −18.2 μC, remains motionless in a uniform electric field perpendicular to the ground. The magnitude of the electric field (in N/C) can be determined using the formula E = F/q, where F represents the force of gravity and q represents the charge on the bead.

The force of gravity, F, can be calculated as [tex]F = mg[/tex], where m is the mass of the bead and g is the acceleration due to gravity. Substituting the values, we find [tex]F = 3.82 \times 10^{-3} kg \times 9.8 m/s^2 = 3.74 \times 10^{-2} N.[/tex]

Given that q is[tex]-18.2 \times 10^{-6} C[/tex], we can now calculate the electric field, E, as [tex]E = \frac{F}{q} = \frac{(3.74 \times 10^{-2} N)}{(18.2 \times 10^{-6} C)} = 2054 N/C.[/tex]

It is important to note that the direction of the electric field is upward due to the negative charge on the bead. In electric fields, opposite charges attract each other, so the electric field lines of force emanate from positive charges and terminate on negative charges. The direction of the electric field is taken as the direction in which a positive charge would experience a force.

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Distance to Star 1 in light-years = 0 light-years.

Given Data:

Angular Separation = 150

Parallax angle is half the angular separation between the two stars

Calculation:

Parallax angle for Star 1 = (1/2) × 150 = 75 arcsecs

Distance to Star 1 in Parsecs (p) = 1/Parallax angle = 1/75 parsecs = 0.01333 parsecs

Distance to Star 1 in light-years = 0.01333 × 3.26 light-years = 0.0434 light-years ≈ 0 light-years

So, the answer is:

Parallax angle for Star 1 = 75 arcsecs

Distance to Star 1 in Parsecs (p) = 0.01333 parsecs

Distance to Star 1 in light-years = 0 light-years.

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The velocity of the ball 1.8 seconds after it is released is 5.6 m/s.ExplanationA ball is thrown downward with an initial velocity of 13 m/s. Given,Initial velocity u = 13 m/sThe ball is thrown downward which means it is thrown in the downward direction.

Hence, acceleration will be acting in the downward direction and acceleration due to gravity g = 10 m/s²Time taken t = 1.8 secondsWe have to find the velocity of the ball after 1.8 seconds.Solution:Using the third equation of motion, we have, v = u + atwherev is the final velocity of the ballu is the initial velocity of the balla is the acceleration of the ballt is the time taken by the ballFrom the given values, we can substitute,u = 13 m/sa = 10 m/s²t = 1.8 secondsSubstituting these values in the equation above,

we get:v = u + atv = 13 m/s + 10 m/s² × 1.8 secondsv = 13 m/s + 18 m/sv = 31 m/sTherefore, the velocity of the ball 1.8 seconds after it is released is 31 m/s.Approximately, g = 10 m/s²Using the approximate value of g=10 m/s², we can also use the below formula,v = u + gtwherev is the final velocity of the ballu is the initial velocity of the ballg is the acceleration of the ballt is the time taken by the ballFrom the given values, we can substitute,u = 13 m/sg = 10 m/s²t = 1.8 secondsSubstituting these values in the equation above, we get:v = u + gtv = 13 m/s + 10 m/s² × 1.8 secondsv = 13 m/s + 18 m/sv = 31 m/sTherefore, the velocity of the ball 1.8 seconds after it is released is 31 m/s.

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