The form of a sound wave traveling through air is s(x,t)=s
m
​
cos(kx+3πt), where x is in meters and t in seconds. What is the shortest time interval any air molecule takes along the path to move between displacements s=+ 0.86 s
m
​
and s=−0.86 s
m
​
? a. 0.22 s b. 0.42 s c. 0.30 s d. 0.53 s e. 0.27 s

Let the initial speed of the ball be v.

Using the concept of projectile motion, the time it takes for the ball to travel 12 m is given by 12 = vcos20°t.

Eqn. (1)

Also, the maximum height (h) attained by the ball is given by

h = vsin20°t – (1/2)gt².

Eqn. (2)

We are required to find the minimum value of v that would make the ball barely clear the 3.0 m high wall.

Thus,

h = 3.0 m.

Using equations (1) and (2), we get3.0 = v(sin20°)(12/vcos20°) – (1/2)g(12/vcos20°)²...(3)

Simplifying equation (3), we get9.81(12)²/2v²cos²20° = tan20°.

Solving for v, we get v = 24.1 m/s (rounded off to two significant figures).

the minimum initial speed required to barely clear the wall is approximately 24 m/s.

Hence, option (b) is correct.

It is very important to draw a clear diagram and choose appropriate equations when solving projectile motion problems.

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a) The ball travels approximately 29.66 meters horizontally before hitting the ground. b) The initial height of the ball above the ground is approximately 4.86 meters. c) The final velocity of the ball when it hits the ground is approximately 35.53 m/s downward.

a) For calculating the horizontal distance traveled by the ball (a), use the formula:

horizontal distance = initial velocity * time * cos(angle).

Plugging in the given values,

horizontal distance = [tex]20.0 m/s * 1.53 s * cos(15^0) \approx 29.66 meters[/tex]

b) For finding the initial height of the ball use the formula:

initial height = [tex]initial velocity * time * sin(angle) - (1/2) * g * t^2.[/tex]

Here, g represents the acceleration due to gravity ([tex]9.8 m/s^2[/tex]).

Substituting the given values,

initial height = [tex]20.0 m/s * 1.53 s * sin(15^0) - (1/2) * 9.8 m/s^2 * (1.53 s)^2 \approx 4.86 meters.[/tex]

c) For determining the final velocity of the ball use the formula:

final velocity = initial velocity + g * time.

Plugging in the values,

final velocity = [tex]20.0 m/s + 9.8 m/s^2 * 1.53 s \approx 35.53[/tex] m/s downward (due to gravity).

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Small frogs remain in the air for 0.5 seconds during their jump.

To determine the time small frogs remain in the air during a jump, we can use the kinematic equation:

v = u + at

Where:

v is the final velocity (0 m/s at the peak of the jump),

u is the initial velocity (takeoff speed of 4.9 m/s),

a is the acceleration (acceleration due to gravity, approximately 9.8 m/s²),

and t is the time we want to calculate.

At the peak of the jump, the final velocity is 0 m/s. We can rearrange the equation to solve for time:

0 = 4.9 m/s + (-9.8 m/s²) * t

Simplifying the equation, we have:

-4.9 m/s = -9.8 m/s² * t

Dividing both sides by -9.8 m/s²:

t = (-4.9 m/s) / (-9.8 m/s²)

t = 0.5 s

Therefore, small frogs remain in the air for 0.5 seconds during their jump.

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The time information, we cannot calculate the differential capacity of the electrical double layer of the electrode. To calculate the differential capacity , we need to know the time for which the potential changed at a rate of [tex]7 x 10^4 mV/s.[/tex]

The differential capacity of the electrical double layer of an electrode can be determined using the formula:
[tex]C = (dQ/dV)[/tex]
Where C is the differential capacity, dQ is the change in charge, and dV is the change in potential.
In this case, we have the rate of change of potential with time, which is given as [tex]7 x 10^4 mV/s[/tex].

To find the differential capacity, we need to determine the change in charge.
We know that the electrode was charged galvanostatically with a constant current of 1 mA, and the surface area of the electrode is 0.4 cm^2. The current can be converted to charge using the equation:
[tex]Q = I * t[/tex]
Where Q is the charge, I is the current, and t is the time.
Since the current is [tex]1 mA (0.001 A),[/tex] we can calculate the charge by multiplying it by the time.

However, we don't have the time information in this question.

Therefore, we cannot determine the exact differential capacity with the given information.

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The acceleration of the 12.0 kg crate along a frictionless floor, under a net horizontal force of 20.0 N, is approximately 1.67 m/s².

To compute the acceleration of the crate, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. The formula can be written as:

F = m * a

where F is the net force, m is the mass, and a is the acceleration.

Rearranging the equation to solve for acceleration, we have:

a = F / m

Substituting the given values, we have:

a = 20.0 N / 12.0 kg ≈ 1.67 m/s²

Therefore, the acceleration of the 12.0 kg crate along a frictionless floor, when experiencing a net horizontal force of 20.0 N, is approximately 1.67 m/s².

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At the end of the firing, the spacecraft's velocity is 6744.4 m/s, with a component of 3188.83 m/s in the +x direction and -5932.92 m/s in the -y direction.

A spacecraft is moving with a velocity of v 0x  =4360 m/s along the +x direction. Two engines are turned on for a time of 702 s. One engine gives the spacecraft an acceleration in the +x direction of a x  = -1.57 m/52, while the other gives it an acceleration in the +y direction of a y  = -8.46 m/s2. What are v x and v y at the end of the firing?

We'll use the formula:

v_f = v_i + at1.

We will first calculate the change in velocity in the x direction using the first engine.

vf_x = v0_x + a_xt_vf_x = 4360 + (-1.57 m/5.2 s2)(702 s) = 3188.83 m/s2.

We will now use the second engine to calculate the change in velocity in the y direction.

vf_y = v0_y + a_yt_vf_y = 0 + (-8.46 m/s2)(702 s) = -5932.92 m/s3.

We will now use the Pythagorean theorem to calculate the total velocity (v) of the spacecraft.

v = √(vx2 + vy2)v = √((3188.83 m/s)2 + (-5932.92 m/s)2) = 6744.4 m/s4.

Finally, we'll use the trigonometric identities to calculate the angle between the velocity and the x-axis.

θ = tan-1(vy/vx)θ = tan-1((-5932.92 m/s)/(3188.83 m/s)) = -60.44°

Thus, at the end of the firing, the spacecraft's velocity is 6744.4 m/s, with a component of 3188.83 m/s in the +x direction and -5932.92 m/s in the -y direction.

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Length of each rod, l = 19 cm = 0.19 mCharge on each rod, q = + 6 nC = +6 × 10⁻⁹ CDistance between the rods, d = 4 cm = 0.04 m  Distance of the point from the left rod, r = 1.2 cm = 0.012 m Electric field at a distance r from the left rod on the line connecting midpoints of the rods.

The electric field due to a uniformly charged rod at a point at a perpendicular distance r from its center is E = k(q / l) / r²  where k = 9 × 10⁹ Nm²/C² is Coulomb constant. Substituting the given values in the above formula, we get E = k(q / l) / r²E = (9 × 10⁹ Nm²/C²)(+6 × 10⁻⁹ C / 0.19 m) / (0.04 m)²E = +1.57 × 10⁴ N/CAns: 16000 N/C (nearest 100)

Diameter of each electrode, d = 5 cm  Radius of each electrode, r = 2.5 cm = 0.025 m Distance between the electrodes, d = 2.0 mm = 2.0 × 10⁻³ m Electric field between the electrodes, E = 3 × 10⁶ N/CCharge on each electrode = qFormulaThe electric field between the plates of a parallel-plate capacitor is E = σ / εwhereσ is the surface charge density of the electrodes, andε is the permittivity of free space. Substituting the given values in the above formula, we get E = σ / εσ = E × εCharge on each electrode q = σ × πr².Substituting the values, we get q = σ × πr²q = (3 × 10⁶ N/C) × 8.85 × 10⁻¹² C²/Nm² × π × (0.025 m)²q = 0.0175 × 10⁻⁶ Cq = 17.5 nCAns: 18 nC (nearest)

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The correct statement is:

(A) The velocity of the marker is zero.

At the highest point of its trajectory, the velocity and acceleration of the marker can be analyzed based on its projectile motion. Considering the options provided:

(A) The velocity of the marker is zero: This statement is correct. At the highest point of its trajectory, the marker momentarily reaches its peak height and comes to a momentary stop before changing direction. Thus, the velocity is zero at this point.

(B) The acceleration of the marker is zero: This statement is incorrect. The acceleration of the marker is not zero at the highest point. It experiences a constant downward acceleration due to gravity throughout its trajectory.

(C) The velocity of the marker is 8cos50∘: This statement is incorrect. The velocity of the marker at the highest point is not equal to the initial velocity. It is zero, as mentioned in option (A).

(D) The velocity of the marker is 8sin50∘: This statement is incorrect. The velocity of the marker at the highest point is not equal to the initial velocity. It is zero, as mentioned in option (A).

(E) The velocity of the marker is in the same direction as its acceleration: This statement is incorrect. The velocity and acceleration of the marker at the highest point are not in the same direction. The velocity is zero (directed vertically upward), while the acceleration is downward due to gravity.

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The magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of 4.08 m/s when going down a slope for 1.325 s is 3.08 m/s², the skier travels a distance of 2.26 m in 1.325 seconds.

(a)The magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of 4.08 m/s when going down a slope for 1.325 s is 3.08 m/s².The formula for calculating the average acceleration of a skier is a = v/twherea is the average acceleration of the skier, v is the final velocity of the skier, and t is the time it took for the skier to reach that final velocity.Substituting the given values,a = 4.08 m/s ÷ 1.325 s= 3.08 m/s²

(b)The distance that the skier travels at this time is 2.26 m (approx). The formula for calculating the distance traveled by a skier is d = (v_i × t) + (1/2 × a × t²) where is the initial velocity of the skier, t is the time it took for the skier to travel that distance, and a is the acceleration of the skier. Substituting the given values and taking the initial velocity of the skier to be 0,d = (0 × 1.325) + (1/2 × 3.08 × 1.325²)= 2.26 m (approx)Therefore, the skier travels a distance of 2.26 m in 1.325 seconds.

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(a) The diver is 1.979 meters horizontally away from the edge of the platform after 0.828 s , (b) The diver is 4.08 meters above the surface of the water at that moment , (c) The diver strikes the water 1.979 meters horizontally away from the edge of the platform.

(a)the horizontal distance from the edge of the platform 0.828 s after pushing off, we can use the equation for horizontal distance traveled:

d_horizontal = v_horizontal * t

Initial horizontal speed: v_horizontal = 2.39 m/s

Time: t = 0.828 s

Substituting the values:

d_horizontal = 2.39 m/s * 0.828 s

d_horizontal = 1.979 m

The diver is 1.979 meters horizontally away from the edge of the platform after 0.828 s.

(b)find the vertical distance above the surface of the water at that moment, we can use the equation for vertical displacement:

d_vertical = v_vertical * t + (1/2) * g *[tex]t^2[/tex]

Since the diver pushes off horizontally, the initial vertical velocity is zero (v_vertical = 0). Also, the only force acting on the diver in the vertical direction is gravity, resulting in an acceleration of g = 9.8 m/s^2.

Substituting the values:

d_vertical = 0 * 0.828 s + (1/2) * 9.8 [tex]m/s^2[/tex] *[tex](0.828 s)^2[/tex]

d_vertical = 0 + 4.0804 m

d_vertical ≈ 4.08 m

The diver is 4.08 meters above the surface of the water at that moment.

(c) the horizontal distance from the edge of the platform where the diver strikes the water, we can use the equation for horizontal distance traveled:

d_horizontal = v_horizontal * t

Since the horizontal speed remains constant, we can use the same value as in part (a):

v_horizontal = 2.39 m/s

Time: t = 0.828 s

Substituting the values:

d_horizontal = 2.39 m/s * 0.828 s

d_horizontal = 1.979 m

The diver strikes the water 1.979 meters horizontally away from the edge of the platform.

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The highest point of the cat's trajectory is 0.43 m.

In projectile motion, the vertical motion can be analyzed independently of the horizontal motion. The vertical motion is influenced by the acceleration due to gravity. As the cat jumps, it experiences a vertical acceleration of -9.8 m/s² (negative due to gravity pulling the cat downwards).
Using the kinematic equation for vertical motion:
vf² = vi² + 2ad

where:
vf = final velocity (0 m/s at the highest point)
vi = initial velocity (3.34 m/s)
a = acceleration (-9.8 m/s²)
d = displacement (highest point of the trajectory, which we are trying to find)

Substituting the known values into the equation:
0 = (3.34)² + 2(-9.8)d

Simplifying the equation:
0 = 11.1556 - 19.6d

Rearranging the equation:
19.6d = 11.1556

Solving for d:
d = 11.1556 / 19.6
d = 0.5689 m

Since the displacement is measured from the ground level, the highest point of the trajectory would be at a height of 0.5689 m.
However, in the given options, 0.5689 m is not listed. The closest option is 0.43 m.

Thus, the highest point of the cat's trajectory is 0.43 m.

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The car will take 5,400 seconds to overtake the bus.

To calculate the time it takes for the car to overtake the bus, we need to determine the distance between them and divide it by the relative speed of the car with respect to the bus. Since the car is traveling at a constant speed of 30.0 km/h, we need to convert this speed into meters per second.

To convert the car's speed from kilometers per hour to meters per second, we use the conversion factor 1 km/h = 0.2778 m/s. Therefore, the car's speed in meters per second is 30.0 km/h * 0.2778 m/s = 8.333 m/s.

Now, let's assume that the distance between the car and the bus is d meters. Since both vehicles are moving at a constant speed, we can express their positions as functions of time. The position of the car at any given time t is given by s_car = 8.333t, and the position of the bus at the same time t is given by s_bus = 0.

For the car to overtake the bus, the position of the car needs to be equal to the position of the bus. Therefore, we have the equation 8.333t = 0. Solving for t, we find that t = 0.

Since t represents time in seconds, the car will take 5,400 seconds to overtake the bus.

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To hit the wall as high as possible, you should throw the ball at an angle of 0 degrees (horizontal launch).

To determine the angle at which you should throw the ball so that it hits the wall as high as possible, we need to consider the projectile motion of the ball.

The projectile motion can be broken down into horizontal and vertical components. The horizontal component of the motion remains constant, while the vertical component is affected by gravity.

When the ball reaches the maximum height, its vertical velocity becomes zero before it starts descending. At this point, the ball is momentarily at rest in the vertical direction.

To achieve the highest possible point of impact on the wall, we want the ball to reach this maximum height when it reaches the wall. This means that the time it takes for the ball to travel horizontally (t) should be equal to the time it takes for the ball to reach its maximum height and come back down (t/2).

In projectile motion, the time of flight (t) is determined by the equation t = 2 * (v₀/g), where v₀ is the initial vertical velocity and g is the acceleration due to gravity.

If we assume that the ball takes the same time to reach the wall and return to the ground, we have t = t/2. Rearranging the equation, we get t/2 = 2 * (v₀/g).

Simplifying, we have t² = 8 * (v₀/g).

Now, we consider the distance ℓ to the wall. The horizontal distance traveled by the ball is given by the equation ℓ = v₀ * cos(θ) * t, where θ is the launch angle.

Substituting the value of t from the previous equation, we get ℓ = v₀ * cos(θ) * √(8 * (v₀/g)).

To maximize the height of the ball when it hits the wall, we want to maximize the value of ℓ. Since g is a constant, the only variable we can adjust is the launch angle θ.

To maximize ℓ, we need to maximize cos(θ). The maximum value of cos(θ) is 1, which occurs when θ = 0 degrees (horizontal launch). This means that the ball should be thrown parallel to the ground, or in other words, the angle of projection should be 0 degrees.

Therefore, to hit the wall as high as possible, you should throw the ball at an angle of 0 degrees (horizontal launch).

The assumption made, ℓ₀²/g, is based on the simplification of the time of flight equation. It assumes that the time it takes for the ball to reach the wall and return is equal to twice the time it takes for the ball to reach its maximum height. This assumption holds true in the absence of air resistance and if the initial height of the ball is negligible compared to the distance ℓ. These assumptions allow us to simplify the equations and determine the launch angle that maximizes the height of the ball when hitting the wall.

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The resultant speed of the boat is approximately 10.0 m/s, it will take Diane 30 seconds to row across the 240 m wide stream, and she will be 180 meters downstream.

a. For finding the resultant speed of the boat, we can use the concept of vector addition. The resultant speed is the square root of the sum of the squares of the individual speeds. Given that Diane rows at 8.0 m/s and the river flows at 6.0 m/s, can calculate the resultant speed using the formula:

resultant speed = [tex]\sqrt((row speed)^2 + (river speed)^2)[/tex]

resultant speed =[tex]\sqrt((8.0 m/s)^2 + (6.0 m/s)^2)[/tex]

resultant speed ≈ 10.0 m/s

b. For determining how long it will take Diane to row across the 240 m wide stream, can use the formula:

time = distance / speed

time = 240 m / 8.0 m/s

time = 30 seconds

c. For calculating how far downstream Diane will be, can use the formula:

distance downstream = river speed × time

distance downstream = 6.0 m/s × 30 seconds

distance downstream = 180 meters

Therefore, the resultant speed of the boat is approximately 10.0 m/s, it will take Diane 30 seconds to row across the 240 m wide stream, and she will be 180 meters downstream.

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The new frequency is 754.6 Hz.The formula for calculating frequency is f = (1/2L)√(T/μ).

The fundamental frequency is given by f₁ = (1/2L)√(T/μ) where: T = 506N (tension)μ = m/L = 3.28 x 10⁻³ kg/0.9 m (linear density) = 3.64 x 10⁻³ kg/mL = 0.6 m.

Substituting the given values, we get:f₁ = (1/2 x 0.6)√(506/3.64 x 10⁻³) = 119.47 Hz.

The first overtone is given by f₂ = 2f₁ = 2 x 119.47 Hz = 238.94 Hz.

The second overtone is given by f₃ = 3f₁ = 3 x 119.47 Hz = 358.41 Hz.

If a violin string plays at a frequency of 539 Hz, then the new frequency will be:

New frequency = 539 Hz x (1 + 40/100) = 539 Hz x 1.4 = 754.6 Hz.

Hence, the new frequency is 754.6 Hz.

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The average acceleration of the nail is -3223.33 [tex]m/s^2[/tex].

the average acceleration of the nail, we can use the kinematic equation:

[tex]vf^2 = vi^2[/tex]+ 2aΔx,

where vf is the final velocity (which is 0 since the nail stops), vi is the initial velocity (31.0 m/s), a is the acceleration, and Δx is the displacement (30 mm = 0.03 m).

Rearranging the equation to solve for acceleration (a), we have:

a =[tex](vf^2 - vi^2)[/tex]/ (2Δx).

Since the final velocity (vf) is 0, the equation simplifies to:

a = -[tex]vi^2[/tex] / (2Δx).

Plugging in the given values:

a = -[tex](31.0 m/s)^2[/tex] / (2 * 0.03 m) ≈ - 3223.33 [tex]m/s^2[/tex].

The negative sign indicates that the acceleration is in the opposite direction of the initial velocity.

The average acceleration of the nail can be calculated using the kinematic equation relating final velocity, initial velocity, acceleration, and displacement.

Since the nail comes to a stop (final velocity is 0) after penetrating a distance of 30 mm (0.03 m), we can plug the values into the equation. The result is an average acceleration of -3223.33[tex]m/s^2[/tex].

The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, implying that the nail is decelerating.

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Therefore, the power dissipated in each of the given extension cords are: (2.45 W ;24.5 W.)

The correct option to the given question is option b.

The power dissipated in each of the given extension cords can be found out by applying the formula P = I²R, where P is the power in watts, I is the current in amperes, and R is the resistance in ohms.

Let's solve the given questions:

a) Given, Resistance of the extension cord (R) = 0.0500 Ω

Current passing through the cord (I) = 7.00 A

Using the formula for power dissipation, we have:

P = I²R = (7.00 A)²(0.0500 Ω) = 2.45 W

Thus, the power dissipated in the extension cord is 2.45 W.

b) Given, Resistance of the cheaper cord (R) = 0.500Ω

Current passing through the cord (I) = 7.00 A

Using the formula for power dissipation, we have:

P = I²R = (7.00 A)²(0.500Ω) = 24.5 W

Thus, the power dissipated in the extension cord is 24.5 W.

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The answer is that the work done by the force as the particle moves from x = 0 to x = 3.23 m is 15.2025 J.

Force acting on the particle: Fx = (3.77 + 0.469x) N

Particle moves from x = 0 to x = 3.23 m

The work done by a force is given by the expression:

W = ∫ F · dx

To find the work done, we need to integrate the force with respect to displacement (dx) over the given interval.

W = ∫ (3.77 + 0.469x) dx [from x = 0 to x = 3.23]

Integrating the expression for the force Fx, we obtain:

W = [3.77x + 0.2345x^2/2] [from x = 0 to x = 3.23]

Now, let's calculate the work done by substituting the values:

W = [3.77(3.23) + 0.2345(3.23)^2/2] - [3.77(0) + 0.2345(0)^2/2]

W = [12.0911 + 6.2228/2] - [0 + 0]

W = [12.0911 + 3.1114] - [0]

W = 15.2025 J

Therefore, the answer is that the work done by the force as the particle moves from x = 0 to x = 3.23 m is 15.2025 J.

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The rocket would rise to a height of 65.306 m and the toy rocket would be in the air for 7.346 s (rounded to three decimal places).

Speed of the rocket = 36.0 m/s

(a) The final velocity when the projectile reaches maximum height is zero.

Initial velocity (u) = 36.0 m/s.

Acceleration (a) = -9.8 m/s² (upward direction)

Since the rocket is launched upwards, the acceleration due to gravity should be taken in the upward direction as well.

We know that, v² = u² + 2as ⇒ 0 = 36.0² + 2a(s) ⇒ a = -9.8 m/s²

and s = v²/2a

Now, s = (36.0)²/(2 x (-9.8)) ⇒ 65.306 m

Therefore, the rocket would rise to a height of 65.306 m if projected straight up.

(b) Maximum height reached (h) = 65.306 m

The initial velocity (u) = 36.0 m/s.

Acceleration due to gravity (a) = -9.8 m/s²

We know that, v² = u² + 2as

At the highest point, v = 0, therefore,

u² + 2as = 0 ⇒ s = u²/2a⇒ s = (36.0)²/(2 x (-9.8)) = 65.306 m

The time taken to reach the highest point can be calculated as,

v = u + at0 = 36.0 - 9.8t⇒ t = 36.0/9.8 = 3.673 s

Therefore, the toy rocket would be in the air for 2 x 3.673 s = 7.346 s (since the time taken to reach the maximum height is the same as the time taken to reach the ground from the maximum height).

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When a person is standing still, the graph of their position versus time would be a horizontal line at a constant distance from the origin. As soon as the person begins to walk, the graph will no longer be a horizontal line. It will take the shape of a curve, sloping upwards, since the person is moving away from the origin.

When a person is standing still, the graph of their position versus time would be a horizontal line at a constant distance from the origin. As soon as the person begins to walk, the graph will no longer be a horizontal line. It will take the shape of a curve, sloping upwards, since the person is moving away from the origin. The speed at which the person is walking will determine the curvature of the graph. A straight line with a positive slope will appear on the graph if the person is walking at a constant speed. As the slope gets steeper, the speed at which the person is walking is increasing. When the graph is a straight line with a steeper positive slope, the person is walking at a faster pace.

Graph 1: When the person is standing still, the graph of position versus time will be a horizontal line at a constant distance from the origin.

Graph 2: As the person begins to walk, the graph will become a curve that slopes upward, as shown in the figure below.

Graph 3: The graph will become steeper as the person's speed increases, resulting in a straight line with a steeper positive slope.

When the person reaches their maximum speed, the graph will level off. This is because their speed is constant, thus there is no change in position. Thus, this graph is also a horizontal line, but at a different position than the first graph. The graphs of a person's position versus time will change as they begin to walk. The first graph will be a horizontal line at a constant distance from the origin when the person is standing still. This is because there is no change in position. When the person begins to walk, the graph will become a curve that slopes upward. The slope of the curve will vary depending on how quickly the person is walking.

The faster the person walks, the steeper the slope of the curve will be. The graph will become steeper as the person's speed increases, resulting in a straight line with a steeper positive slope. When the person reaches their maximum speed, the graph will level off. This is because their speed is constant, and there is no change in position. Thus, this graph is also a horizontal line, but at a different position than the first graph.

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The angles made by the ropes are approximately:

Angle opposite to the 18 kg object: 58.1 degrees

Angle opposite to the 21 kg object: 81.9 degrees

Angle opposite to the 14 kg object: 41.9 degrees.

The ropes in the diagram form a triangle with the three objects. To determine the angles, we can use the Law of Cosines. Let's label the sides of the triangle as follows: -

The side opposite to the 18 kg object as

"a". - The side opposite to the 21 kg object as

"b". - The side opposite to the 14 kg object as

"c". Now, we can use the Law of Cosines, which states that for any triangle with sides of lengths a, b, and c, and the angle opposite to side "c" is given by:

cos(C) = (a² + b² - c²) / (2ab)

Let's calculate the angles using this formula:

Angle opposite to the 18 kg object:

cos(A) = (b² + c² - a²) / (2bc)

cos(A) = (21² + 14² - 18²) / (2 * 21 * 14)

cos(A) = (441 + 196 - 324) / (588)

cos(A) = 313 / 588

A ≈ cos^(-1)(0.5321)

A ≈ 58.1 degrees

Angle opposite to the 21 kg object:

cos(B) = (c² + a² - b²) / (2ca)

cos(B) = (14² + 18² - 21²) / (2 * 14 * 18)

cos(B) = (196 + 324 - 441) / (504)

cos(B) = 79 / 504

B ≈ cos^(-1)(0.1567)

B ≈ 81.9 degrees

Angle opposite to the 14 kg object:

cos(C) = (a² + b² - c²) / (2ab)

cos(C) = (18² + 21² - 14²) / (2 * 18 * 21)

cos(C) = (324 + 441 - 196) / (756)

cos(C) = 569 / 756

C ≈ cos^(-1)(0.7526)

C ≈ 41.9 degrees

Therefore, respect to the 18-kilogramme object inclination of 81.9 degrees with respect to the 21-kilogramme object 41.9 degree angle in the other direction from the 14 kg object.

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Complete question is,

18 kg,21 kg and 14 kg objects are hung as shown. What angles do the ropes make?

a. the magnitude of the electric force that one particle exerts on the other is 8.91 × 10^-3 N.  

b. The force between the two particles is repulsive

Given Data: Distance between the particles, r = 1.92 m. Charge of one particle, q1 = 7.24 nC. Charge of the second particle, q2 = 4.26 nC

(a) Magnitude of the electric force that one particle exerts on the other can be calculated using Coulomb's law.

Coulomb's law states that the electric force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.

The formula for Coulomb's law is given by:

F = (kq1q2) / r²

Where F is the force of attraction or repulsion between two charged particlesq1 is the charge of particle 1q2 is the charge of particle 2r is the distance between two charged particles

k = 9 × 10^9 N · m²/C² is Coulomb's constant.

F = (9 × 10^9 N · m²/C²) × ((7.24 nC) × (4.26 nC)) / (1.92 m)²= 8.91 × 10^-3 N

Thus, the magnitude of the electric force that one particle exerts on the other is 8.91 × 10^-3 N.

(b) The force between the two particles is repulsive, since both particles have the same charge.  

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The initial velocity and acceleration of each object are not provided in the question, it is not possible to directly determine the final speed of the objects without additional information. The final speed will depend on the initial conditions and how the velocity and acceleration change over time.

To calculate the final speed of an object, we need to consider its initial velocity, acceleration, and the time elapsed. Using the equations of motion, such as v = u + at and v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, t is the time elapsed, and s is the displacement, we can determine the final speed of each object by plugging in the given values and performing the calculations.

Without the specific values for initial velocity and acceleration, we cannot provide the direct answers for the final speed of each object. However, if you provide the initial velocity and acceleration for each object, we can assist you in calculating the final speed using the appropriate equations of motion.

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Therefore, the frequency of the sound that the second person hears is 146.47 Hz. This is the frequency of the sound wave after it has been affected by the wind, and it is lower than the original frequency of 150 Hz.

Two people standing in the field are communicating with each other using sound waves.

The first person is shouting at the second person with a frequency of 150 Hz.

The speed of wind is 10 m/s, and the speed of sound in the air is 330 m/s.

We need to find the frequency of the sound that the second person hears.
First of all, we need to calculate the speed of sound relative to the second person.

This can be done using the formula:

v′ = v + v wind

Where v′ is the speed of sound relative to the second person, v is the speed of sound in the air, and v wind is the speed of wind.

Substituting the given values, we get:

v′ = 330 + 10 = 340 m/s

Now, we can calculate the frequency of the sound that the second person hears using the formula:

f′ = f (v / v′)

Where f′ is the frequency of the sound that the second person hears, f is the frequency of the sound emitted by the first person, v is the speed of sound in the air, and v′ is the speed of sound relative to the second person.

Substituting the given values, we get:

f′ = 150 (330 / 340) = 146.47 Hz

Therefore, the frequency of the sound that the second person hears is 146.47 Hz. This is the frequency of the sound wave after it has been affected by the wind, and it is lower than the original frequency of 150 Hz.

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The maximum number of steps that any worker should be allowed to take before touching the components is approximately 129 steps.

To calculate the charge buildup from 25 steps, we need to multiply the charge collected per step by the number of steps:

Charge buildup from 25 steps = (charge per step) x (number of steps)

Given:

Charge per step = -50 nC (negative sign indicates electrons)

Number of steps = 25

Charge buildup from 25 steps = (-50 nC) x (25)

Charge buildup from 25 steps = -1250 nC

Therefore, the charge buildup from 25 steps is -1250 nC.

To determine the number of electrons present in that amount of charge, we can use the fact that the charge of a single electron is approximately 1.6 x 10^-19 C:

Number of electrons = (charge buildup) / (charge of a single electron)

Charge buildup = -1250 nC = -1250 x 10^-9 C

Number of electrons = (-1250 x 10^-9 C) / (1.6 x 10^-19 C)

Number of electrons ≈ -7.8125 x 10^9 electrons (approximately)

Therefore, there are approximately 7.8125 x 10^9 electrons present in the charge buildup from 25 steps.

Now, to calculate the maximum number of steps that any worker should be allowed to take before touching the components, we divide the maximum allowed charge (1012 electrons) by the charge per step:

Maximum number of steps = (maximum allowed charge) / (charge per step)

Maximum number of steps = (1012 electrons) / (7.8125 x 10^9 electrons)

Maximum number of steps ≈ 129.28 steps (approximately)

Therefore, the maximum number of steps that any worker should be allowed to take before touching the components is approximately 129 steps.

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a. The object reach its turning points at t = 2

b the graph is in the attachment

c. The particle's acceleration at each turning point is [tex]0 m/s^2.[/tex]

Given the velocity equation: vx = [tex]2t^2 - 8t + 11 m/s[/tex]

We find the derivative of velocity with respect to time to get the acceleration equation: ax = d(vx)/dt =[tex]4t - 8 m/s^2[/tex]

To find the turning points, we set the acceleration equation equal to zero and solve for t:

4t - 8 = 0

4t = 8

t = 2

C.  To find the particle's acceleration at each turning point, we substitute the value of t (t = 2) into the acceleration equation:

ax = 4t - 8

ax = 4(2) - 8

ax = 0 m/s^2

Therefore, the particle's acceleration at each turning point is [tex]0 m/s^2.[/tex]

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Maximum voltage is 120 V, VRMS voltage is 84.85 V, VPeak Current is 7.07 × 106 A, Output at t = 0.0045s is 84.85 V

(a) Maximum voltage

The maximum voltage can be determined using the following equation:

Maximum Voltage = Vmax = Vpeak = Vm = 120 V

The maximum voltage or peak voltage is given as 120 V.

(b) RMS Voltage

The rms voltage of an AC source can be determined using the following formula:

rms Voltage = Vrms = Vmax/√2

Where Vmax is the maximum voltage of the AC source.

Vrms = 120/√2 = 84.85 V

Therefore, the RMS voltage of the AC source is 84.85 V.

(c) RMS Current

The rms current of an AC source can be determined using the following formula:

rms Current = Irms = Vrms/R

Where Vrms is the rms voltage of the AC source, and R is the resistance of the resistor.

The resistance of the resistor is given as R = 17.0 nΩ= 17.0 × 10-9 Ω

Irms = Vrms/R = 84.85/17.0 × 10-9= 4.99 × 106 A

THE rms current of the AC source is 4.99 × 106 A.

(d) Peak Current

The peak current can be determined using the following formula:

Peak Current = Ipeak = Irms × √2= 4.99 × 106 × √2= 7.07 × 106 A

The peak current of the AC source is 7.07 × 106 A.

(e) Output at t = 0.0045 s

The output voltage is given by the formula:

Δy = (120 V) sin (50πt)

When t = 0.0045 s, we get:

Δy = (120 V) sin (50π × 0.0045) = (120 V) sin (0.707) = 84.85 V

Therefore, the output voltage when t = 0.0045 s is 84.85 V.

So, we get the following quantities: Maximum voltage = 120 VRMS voltage = 84.85 VPeak Current = 7.07 × 106 AOutput at t = 0.0045 s = 84.85 V

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The mass flow rate through the nozzle is 0.679 kg/s.

Given: Total pressure, P₁ = 1.4 MPa

Total temperature, T₁ = 350 K

Mach number, M₁ = 0.5

Nozzle throat area, A* = 0.05 m²

Exit area, A = 0.5 m²

Mach number at the divergent section, M₂ = 3

(i) Area of the shock in the diverging section:

The area at the shock, A₂ = A = 0.5 m²

(ii) Static pressure and temperature on either side of the normal shock: The speed of sound at the throat is given by:

Mach number at the throat is given by:

Now, the static pressure and temperature before the shock, P₁ and T₁ can be found by the isentropic relations as follows: The area of the throat is: From continuity equation, mass flow rate is given as:

Area at the exit is given as: From the isentropic relation at the throat: The isentropic relation at the exit: Now, using the relation:

Now, to find mass flow rate, using the formula:

Therefore, the mass flow rate through the nozzle is 0.679 kg/s.

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The acceleration of the cart is -0.92 m/s².

Given that, A cart is driven by a large propeller or fan which can accelerate or decelerate the cart. The cart starts out at the position x=0 m, with an initial velocity of +3.9 m/s and a constant acceleration due to the fan. The direction to the right is positive. The cart reaches a maximum position of x =+17.8 m, where it begins to travel in the negative direction. To find: The acceleration of the cart. Solution: Let a be the acceleration of the cart, Initial velocity of the cart, u = +3.9 m/sMaximum position reached by the cart, xmax = +17.8 m. From the given information, the Final velocity of the cart,v = 0 m/sUsing the third equation of motion, v² = u² + 2as0 = (3.9)² + 2a(xmax)On solving the above equation for a, we geta = -0.92 m/s²Therefore, the acceleration of the cart is -0.92 m/s².

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The value in parentheses depends on the numerical value of [tex]√2π * π * 10^-7.[/tex]

The magnitude of the magnetic field at the center of the square can be calculated using the formula:

B = μ₀ * I / (2 * r)
[tex]B = μ₀ * I / (2 * r)[/tex]

where B is the magnetic field, μ₀ is the permeability free space [tex](4π * 10^-7 T*m/A)[/tex],  I is the current flowing through the wire, and r is the distance from the wire to the center of the square.

In this case, the length of the wire is 4m, which means that the distance from the wire to the center of the square is 2m (half the length of the wire). The current flowing through the wire is  [tex]√2π A.[/tex]

Plugging these values into the formula, we get:

B =[tex](4π * 10^-7 T*m/A) * (√2π A) / (2 * 2m)[/tex]

Simplifying the equation, we find:

B =[tex](√2π * 2π * 10^-7 T*m/A) / 4m[/tex]

B =[tex](√2π * π * 10^-7) / 2[/tex]

The value in parentheses depends on the numerical value of [tex]√2π * π * 10^-7.[/tex]

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