1. A sphere with a mass of 10 kg and radius of 0.5 m moves in free fall at sea level (where the air density is 1.22 kg/m3). If the object has a drag coefficient of 0.8, what is the object’s terminal velocity? What is the terminal velocity at an altitude of 5,000 m, where the air density is 0.736 kg/m3? Show all calculations in your answer.

Answers

Answer 1

Answer:

The value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.

Given data:

The mass of sphere is, m = 10 kg.

The radius of sphere is, r = 0.5 m.

The density of air is, $\rho = 1.22 \;\rm kg/m^(3)$ .

The drag coefficient of object is, $C_(d)=0.8$ .

The altitude is, h = 5000 m.

The density of air at altitude is, $\rho' =0.736 \;\rm kg/m^(3)$ .

The mathematical expression for the terminal velocity of an object is,

$v_(t)=\sqrt(2mg)/(\rho * A * C_(d))$

here,

g is the gravitational acceleration.

A is the area of sphere.

Solving as,

$v_(t)=\sqrt{(2 * 10 * 9.8)/(1.22 * (4 \pi r^(2)) * C_(d))}\n\n\nv_(t)=\sqrt{(2 * 10 * 9.8)/(1.22 * (4 \pi * 0.5^(2)) * 0.8)}\n\n\n\v_(t)=7.99 \;\rm m/s$

Now, the terminal velocity at the altitude of 5000 m is given as,

$v_(t)'=\sqrt(2mg)/(\rho' * A * C_(d))$

Solving as,

$v_(t)'=\sqrt{(2 * 10 * 9.8)/(0.736 * (4 \pi * 0.5^(2)) * 0.8)}\n\n\nv_(t)'=10.30 \;\rm m/s$

Thus, we can conclude that the value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.

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Answer 2

Answer:

Answer:

The terminal velocity at sea level is 7.99 m/s

The terminal velocity at an altitude of 5000 m is 10.298 m/s

Explanation:

mass of sphere m = 10 kg

radius of sphere r = 0.5 m

air density at sea level p = 1.22 kg/m^3

drag coefficient Cd = 0.8

terminal velocity = ?

Area of the sphere A = $4\pi r^(2)$ = 4 x 3.142 x $0.5^(2)$ = 3.142 m^2

terminal velocity is gotten from the relationship

$Vt = \sqrt{(2mg)/(pACd) }$

where g = acceleration due to gravity = 9.81 m/s^2

imputing values into the equation

$Vt = \sqrt{(2*10*9.81)/(1.22*3.142*0.8) }$ = 7.99 m/s

If at an altitude of 5000 m where air density = 0.736 kg/m^3, then we replace value of air density in the relationship as 0.736 kg/m^3

$Vt = \sqrt{(2mg)/(pACd) }$

$Vt = \sqrt{(2*10*9.81)/(0.736*3.142*0.8) }$ = 10.298 m/s

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Answers

Answer:

The force is greater if the light is absorbed instead of being reflected

Explanation:

Light could either be reflected or absorbed. Reflection takes place where light is concentrated back to another surface whereas absorption takes place when light is incorporated into a surface thereby providing kinetic energy. The kinetic energy produced by absorption provides more force than reflection which just involves concentrating back to another surface.

A bug flying horizontally at 1.7 m/s collides and sticks to the end of a uniform stick hanging vertically from its other end. After the impact, the stick swings out to a maximum angle of 7.0Â° from the vertical before rotating back. If the mass of the stick is 16 times that of the bug, calculate the length of the stick (in m).

Answers

Answer:12.11 m

Explanation:

Given

Bug speed =1.7 m/s

Let mass of bug is m

mass of rod 16m

maximum angle turned by rod is 7^{\circ}[/tex]

From Energy conservation

kinetic energy of bug =Gain in potential energy of rod

$(1)/(2)mv^2=16mgL(1-cos\theta )$

$L=(1.7^2)/(2* 16(1-cos7))$

L=12.11 m

The 9-inch-long elephant nose fish in the Congo River generates a weak electric field around its body using an organ in its tail. When small prey, or even potential mates, swim within a few feet of the fish, they perturb the electric field. The change in the field is picked up by electric sensor cells in the skin of the elephant nose. These remarkable fish can detect changes in the electric field as small as 3.00 μN/C. How much charge, modeled as a point charge, in the fish would be needed to produce such a change in the electric field at a distance of 63.5 cm ?

Answers

Answer:

$1.34\cdot 10^(-16) C$

Explanation:

The strength of the electric field produced by a charge Q is given by

$E=k(Q)/(r^2)$

where

Q is the charge

r is the distance from the charge

k is the Coulomb's constant

In this problem, the electric field that can be detected by the fish is

$E=3.00 \mu N/C = 3.00\cdot 10^(-6)N/C$

and the fish can detect the electric field at a distance of

$r=63.5 cm = 0.635 m$

Substituting these numbers into the equation and solving for Q, we find the amount of charge needed:

$Q=(Er^2)/(k)=((3.00\cdot 10^(-6) N/C)(0.635 m)^2)/(9\cdot 10^9 Nm^2 C^(-2))=1.34\cdot 10^(-16) C$

A long, thin superconducting wire carrying a 17 A current passes through the center of a thin, 3.0-cm-diameter ring. A uniform electric field of increasing strength also passes through the ring, parallel to the wire. The magnetic field through the ring is zero.a. At what rate is the electric field strength increasing?
b. is the electric field in the direction of the current or opposite to the current?

Answers

Answer:

$(dE)/(dt) =- 2.72 *10^(15) \ N/C \cdot s$

The direction of the electric field is opposite that of the current

Explanation:

From the question we are told that

The current is $I = 17\ A$

The diameter of the ring is $d = 3.0 \ cm = 0.03 \ m$

Generally the radius is mathematically represented as

$r = (d)/(2)$

$r = (0.03)/(2)$

$r = 0.015 \ m$

The cross-sectional area is mathematically represented as

$A = \pi r^2$

=> $A = 3.142 * (0.015^2)$

=> $A = 7.07 *10^(-4 ) \ m^ 2$

Generally according to ampere -Maxwell equation we have that

$\oint \= B \cdot \= ds = \mu_o I + \epsilon_o \mu _o( d \phi )/(dt )$

Now given that $\= B = 0$ it implies that

$\oint \= B \cdot \= ds = 0$

$\mu_o I + \epsilon_o \mu _o( d \phi )/(dt ) = 0$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^(-12 ) \ m^(-3) \cdot kg^(-1)\cdot s^4 \cdot A^2$

$\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^(-7) N/A^2$

$\phi$ is magnetic flux which is mathematically represented as

$\phi = E * A$

Where E is the electric field strength

$\mu_o I + \epsilon_o \mu _o ( d [EA] )/(dt ) = 0$

=> $(dE)/(dt) =- (I)/(\epsilon_o * A )$

=> $(dE)/(dt) =- (17)/(8.85*10^(-12) * 7.07*10^(-4) )$

=> $(dE)/(dt) =- 2.72 *10^(15) \ N/C \cdot s$

The negative sign shows that the direction of the electric field is opposite that of the current

A 60.0-kg boy is surfing and catches a wave which gives him an initial speed of 1.60 m/s. He then drops through a height of 1.57 m, and ends with a speed of 8.50 m/s. How much nonconservative work was done on the boy

Answers

Answer:

Work = 1167.54 J

Explanation:

The amount of non-conservative work here can be given by the difference in kinetic energy and the potential energy. From Law of conservation of energy, we can write that:

Gain in K.E = Loss in P.E + Work

(0.5)(m)(Vf² - Vi²) - mgh = Work

where,

m = mass of boy = 60 kg

Vf = Final Speed = 8.5 m/s

Vi = Initial Speed = 1.6 m/s

g = 9.8 m/s²

h = height drop = 1.57 m

Therefore,

(0.5)(60 kg)[(8.5 m/s)² - (1.6 m/s)²] - (60 kg)(9.8 m/s²)(1.57 m) = Work

Work = 2090.7 J - 923.16 J

Work = 1167.54 J

Susan’s 10 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30° above the floor. The tension is a constant 30 N and the coefficient of friction is 0.20. Use work and energy to find Paul’s speed after being pulled 3.0 m.

Answers

Final answer:

In this problem, we have calculated the work done by Susan pulling her baby brother on a mat and the work done against friction. The net work done, which is the work done by Susan's pulling minus the work done against friction, is transformed into kinetic energy, giving us the baby's speed after being pulled 3m, which is approximately 1.95 m/s.

Explanation:

To answer this question, we first need to calculate the work done by Susan when she pulls the mat over the distance of 3.0 meters. The angle at which the rope is pulled does make a difference in this calculation. The force that is actually contributing to the work is the horizontal component of the tension, which can be determined by the equation Fh = F cos θ which equals 30N * cos30 = 25.98N.

The work done, W, is equal to this force multiplied by the displacement, so W = Fd = 25.98N * 3m = 77.94 Joules.

Next, we need to calculate the work done against friction. The force of friction is calculated as Ff = µN. Here N is the normal force, which is equal to the weight of the baby, so N = mg = 10kg * 9.8m/s² = 98N. The force of friction then is Ff = µN = 0.20 * 98N = 19.6N. The work done against friction is Wf = Ff * d = 19.6N * 3m = 58.8 Joules.

The net work done on the baby is the work done by Susan minus the work done against friction, so Wnet = W - Wf = 77.94J - 58.8J = 19.14 Joules. This net work is equal to the change in kinetic energy of the baby, ∆K, since Kinitial = 0 (Paul starts at rest), the work done is all transformed into final kinetic energy. So ∆K = 19.14J.

The kinetic energy of an object is given by the equation K = 1/2 mv², so we have 19.14J = 1/2 * 10kg * v². Solving for v gives us roughly v = 1.95 m/s. Therefore, the speed of the baby after being pulled 3 meters is approximately 1.95 m/s.

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Final answer:

To determine Paul's speed, we must calculate the net work done on him using the work-energy theorem. This includes the work done by Susan and the work done against friction. Paul’s speed after being pulled 3.0 m is approximately 1.96 m/s.

Explanation:

Solving this problem involves understanding the work-energy theorem and forces. First, let's calculate the work done. The work done by the force Susan applies (W1) is the product of the tension (T), the distance (d), and the cosine of the angle (θ). W1 = T * d * cos(θ) = 30N * 3.0m * cos(30) = 77.94J.

Next, the work done against friction (W2) is the product of the frictional force and the distance, which is µmgd. Here, µ is the coefficient of friction (0.20), m (10kg) is the mass of the baby, g (9.8m/s2) is the acceleration due to gravity, and d is the distance (3.0 m). W2 = µmgd = 0.20 * 10kg * 9.8m/s2 * 3.0m = 58.8J.

According to the work-energy theorem, the net work done on an object is equal to the change in its kinetic energy. Therefore, the final kinetic energy (and thus the final speed) of Paul will be the initial kinetic energy plus the net work done on him. His initial speed is assumed to be zero, hence the initial kinetic energy is zero. The net work done on him is W = W1 - W2= 77.94J - 58.8J = 19.14J. Setting this equal to the final kinetic energy, (1/2)mv2, allows us to solve for the final speed, v = sqrt((2 * W)/m) = sqrt((2 * 19.14J)/10kg) = 1.96 m/s approximately.

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