The lowest-pitch tone to resonate in a pipe of length l that is closed at one end and open at the other end is 200 hz. Which frequencies will not resonate in the same pipe?

Answer:

Moment of inertia and angular velocity.

Explanation:

The translational kinetic energy of an object is possessed when the object is showing rotational motion. It can be given by the formula as :

[tex]KE=\dfrac{1}{2}I\omega^2[/tex]

Here,

I is the moment of inertia of the object

[tex]\omega[/tex] is the angular velocity of the object

So, the translational kinetic energy of an object is given by moment of inertia and angular velocity of the object. Hence, this is the required solution.

Answer:

The answer to the question is

3340800 m far

Explanation:

To solve the question, we note that acceleration = 29 m/s²

Time of acceleration = 8 minutes

Then if the shuttle starts from rest, we have

S = u·t+0.5·a·t² where u = 0 m/s = initial velocity

S = distance traveled, m

a = acceleration of the motion, m/s²

t = time of travel

S = 0.5·a·t² = 0.5×29×(8×60)² = 3340800 m far

Answer: D all of the above

Explanation:

All of the statements in the options are the character's exhibited by both longitudinal amd transverse waves.

Examples of transverse waves are water waves, light waves, radio waves, and also waves produced in strings and ropes.

Examples of longitudinal waves includes : vibrating turn fork,, drum head etc

Statement A is true for longitudinal waves only, statement C is true for transverse waves only, and statement B is true for both types of waves. These are based on the nature of particle movement in relation to the direction of wave energy flow.

In the given statements, Statement A, 'In longitudinal waves, the particles of the medium move parallel to the direction of the flow of energy', is true solely for longitudinal mechanical waves. This is because, in longitudinal waves, the disturbances (or oscillations) in the medium occur in the same direction as the wave propagation, with notable examples being sound waves in air and water.

Statement C, 'In transverse waves, the particles of the medium move perpendicular to the direction of the flow of energy', on the other hand, pertains exclusively to transverse mechanical waves. Transverse waves are characterized by disturbances in the medium that are perpendicular to the direction of the wave propagation, common examples of which are waves seen on stringed instruments and electromagnetic waves, such as visible light.

Statement B, 'Many wave motions in nature are a combination of longitudinal and transverse motion', applies to both longitudinal and transverse mechanical waves. Certain seismic waves generated during earthquakes, for instance, possess both longitudinal and transverse components, demonstrating that waves can indeed display composite behaviors. Thus, in essence, this statement is true for both types of mechanical waves.

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The radius R of the turn is 1.984 km.

Explanation:

As the falcon is experiencing a centripetal motion, the acceleration exhibited by the falcon will be centripetal acceleration. The formula for centripetal acceleration is

                 [tex]a=\frac{v^{2} }{R}[/tex]

Here a is the acceleration for centripetal motion, v is the velocity and R is the radius of the circular path.

As the centripetal acceleration is given as 0.6 g, the velocity is given as 108 m/s, then the radius of the path can be determined as

       [tex]0.6 \times 9.8=\frac{(108)^{2}}{R}[/tex]

      [tex]R=\frac{(108)^{2}}{0.6 \times 9.8}=\frac{11664}{5.88}=1983.67\ \mathrm{m}[/tex]

So, the radius of the turn is 1.984 km.

Answer:

THE ANSWER IS:  Rutherford, Marsden and Geiger discovered the dense atomic nucleus by bombarding a thin gold sheet with the alpha particles emitted by radium

Explanation:

Answer:

θ = 53.7°

Explanation:

Given:

- The mass of ball = M

- The mass of object = 3M

- The wire length L = 0.5 m

- The velocity of ball vi = 4.0 m/s

- The velocity of ball vf

- The velocity of object Vf

Find:

Find the maximum angle through which the block swings after it is hit.

Solution:

- When two objects collide with no external force acting on the system the linear momentum of the system is conserved. The initial (Pi) and final (Pf) linear momentum are equal:

                                  Pi = Pf

                                  M*vi = M*vf + 3M*Vf

                                  vi = vf + 3*Vf

                                  4 = vf + 3*Vf

- For elastic collision between two particles the relative velocities before and after collision have the same magnitude but opposite sign; so,

                                   vi - 0 = Vf - vf

                                   4 = Vf - vf

- Solve the above two equation simultaneously.

                                   8 = 4*Vf

                                   Vf = 2 m/s

                                    vf = -2 m/s

- When the ball hits the object it swing under the influence of gravity only. Hence, no external force acts on the object so we can apply the conservation of energy as the object attains a height h.

                                   ΔK.E = ΔP.E

                                   0.5*(3M)*Vf^2 = (3M)*(g)*(h)

                                   h = Vf^2 / 2*g

- Plug in the values:

                                   h = 2^2 / 2*9.81

                                   h = 0.2039 m

- We can see that the maximum angle can be given as θ according trigonometric relation as follows:

                                  θ = arccos [ ( L - h ) / L ]

                                  θ = arccos [ ( 0.5 - 0.2039 ) / 0.5 ]

                                  θ = 53.7°

The maximum angle through which the block swings after it is hit θ is = 53.7°

Given:

The mass of ball is = M

The mass of object is = 3M

The wire length L is = 0.5 m

The velocity of ball vi is = 4.0 m/s

Then The velocity of ball vf

After that The velocity of object Vf

Now we Find:

Find the maximum angle through which the block swings after it is hit that is :

When two objects collide with no external force acting on the system the linear momentum of the system is conserved. Then The initial (Pi) and also final (Pf) linear momentum are equal:

Pi is = Pf

M*vi is = M*vf + 3M*Vf

vi is = vf + 3*Vf

4 is = vf + 3*Vf

Now For elastic collision between two particles the relative velocities before and also after collision have the same magnitude but opposite signs; so,

vi - 0 is = Vf - vf

4 is = Vf - vf

Solve the above two equations simultaneously.

8 is = 4*Vf

Vf is = 2 m/s

vf is = -2 m/s

When the ball hits the object it swings under the influence of gravity only. Hence proof, no external force acts on the object so we can apply the conservation of energy as the object attains a height h.

ΔK.E is = ΔP.E

0.5*(3M)*Vf^2 is = (3M)*(g)*(h)

h is = Vf^2 / 2*g

Then we Plug in the values is:

h is = 2^2 / 2*9.81

h is = 0.2039 m

Now We can see that the maximum angle can be given as θ according trigonometric relation as follows:

θ is = arccos [ ( L - h ) / L ]

θ is = arccos [ ( 0.5 - 0.2039 ) / 0.5 ]

θ is = 53.7°

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Answer:

Explanation:

mass, m = 1 kg

specific heat, c = 1.5 J/g°C

rise in temperature, ΔT = 41 - 21 = 20

heat released, H = m x c x ΔT

H = 1 x 1.5 x 1000 x 20

H = 30,000 J

H = 30 kJ.

Answer:

Explanation:

on edg I got it right

Answer:

a. The pulses cancel each other resulting in zero displacement.

b. The pulses reinforce each other, having a displacement of 2× amplitude or 0.56 m.

Explanation:

When the pulse are sent down the attached string  it gets reflected and we have crossing pulses as the incident and reflected pulses cross each other.

a. If the string is rigidly attached to the post then the incident and reflected pulses will have the same amplitude but different direction. That is either the reflected will be going up and the incident down thereby resulting in a cancellation or zero displacement

Incident amplitude = 0.28 m

Reflected amplitude = -0.28 m

Sum = 0.28 m - 0.28 m = 0 m

b. If the end at which the relfection occurs is free to slide, then the incident and the reflected pulses will again have the same amplitude and in this case, the same direction. Therefore;

Incident amplitude = 0.28 m

Reflected amplitude = 0.28 m

Sum = 0.28 m + 0.28 m = 0.56 m

Answer:

The answer is zero displacement (0 m)

Explanation:

If the end of the string is not free, the reflected pulse has the same amplitude but opposite polarity to the incident pulse. Because of this, for A, the result equals zero (0 m).

Answer:

59.84 N

Explanation:

The net force is responsible for the upward acceleration of the bowling ball.

The net force acting on a body accelerates the body in the same direction as that in which the resultant is applied.

Net force = ma

ma = F - mg

In the first case,

F = 71.6 N, a = a m/s²

ma = 71.6 - mg (eqn 1)

In the second case,

F = 82.3 N, a = 1.91a m/s²

m(1.91a) = 82.3 - mg

1.91 ma = 82.3 - mg (eqn 2)

Substitute for ma in (eqn 2)

1.91 (71.6 - mg) = 82.3 - mg

136.756 - 1.91 mg = 82.3 - mg

136.756 - 82.3 = 1.91 mg - mg

0.91 mg = 54.456

mg = 54.456/0.91

mg = 59.84 N

Hence, the weight of the bowling ball = 59.84 N

Hope this Helps!!!

The weight of the bowling ball is 59.84 N

The net force should be held responsible where there is the upward acceleration of the bowling ball. When it should be acted on the body so it should be in a similar direction.

We know that

Net force = ma

So,

ma = F - mg

Now

In the first case,

F = 71.6 N, a = a m/s²

ma = 71.6 - mg (eqn 1)

And,

In the second case,

F = 82.3 N, a = 1.91a m/s²

m(1.91a) = 82.3 - mg

So,

1.91 ma = 82.3 - mg (eqn 2)

Now

1.91 (71.6 - mg) = 82.3 - mg

136.756 - 1.91 mg = 82.3 - mg

136.756 - 82.3 = 1.91 mg - mg

0.91 mg = 54.456

mg = 54.456/0.91

mg = 59.84 N

Hence, the weight of the bowling ball = 59.84 N

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Answer: Option (a) is the correct answer.

Explanation:

It is known that potential energy is the energy occupied by an object or substance due to its position is known as potential energy.

Therefore, more is the space occupied by an object more will be its position at a particular location. Hence, more will be its potential energy. On the other hand, smaller is the space occupied by an object, smaller will be the position holded by it.

Hence, smaller will be its potential energy.

Thus, we can conclude that for the given situation the statement, potential energy of the larger sphere is greater than that of the smaller sphere, is true.

Final answer:

The potential energy of the larger conducting sphere is less than that of the smaller sphere because electric potential is inversely proportional to radius and both spheres carry the same charge.

Explanation:

When two conducting spheres with the same excess charge but different radii are considered, the potential of each sphere is given by V = kQ/R, where V is the potential, k is Coulomb's constant, Q is the charge, and R is the radius of the sphere. Since the charge Q is the same for both spheres, but the radius R is larger for one of them, the electric potential energy will be higher for the smaller sphere since it has a smaller radius. This stems from the fact that electric potential is inversely proportional to the radius, meaning a smaller radius results in a higher potential for the same amount of charge.

The correct statement is: (d)he potential energy of the larger sphere is less than that of the smaller sphere.

Answer:

 = 238N

Explanation:

mass = 0.14g = 14 × 10⁻⁵kg

initial velocity = 218m/s

final velocity = 0

penetration depth (distance) = 14mm = 14 × 10⁻³m

v²(final) = v²(initial) + 2aΔx

0² = (218)² + 2(14 × 10⁻³)a

a = -(218)² /  2(14 × 10⁻³)

a = 16.97 × 10⁵m/s²

F = ma

 = (14 × 10⁻⁵)(16.97 × 10⁵)

 = 237.58N

 ≅ 238N

Explanation:

Below is an attachment containing the solution.

Answer:

Explanation:

Since momentum is a vector, you, indeed, in two dimension collisions, you can decompose it in two components, the x-direction and the y-direction, such as you do with the force, which is a vector too.

The law of conservation of momentum states that the total momentum before and after the collision are conserved.

Let's assume a collision in one dimension: x-direction.

If object A is moving to the right, its momentum is to the right. If objcet B is at rest its momentum is zero. Then, if when object A collides with object B, the first stops, the second must move to the right with a momentum in the x-direction equal to the momentum that object A initially had.

You can apply the same reasoning if object A is moving in two dimensions, and, a similar one, if object B is not at rest: at the end the momentum in each direction before the collision has to be equal to the momentum in each direction after the collision.

Answer:

1 m

Explanation:

P = Patm + ρgh

110,000 Pa = 100,000 Pa + (1020 kg/m³) (9.8 m/s²) h

h = 1 m

At depth of 1 m the sea would the total pressure be 110kpa.

The physical force applied to an object is referred to as pressure. Per unit area, a perpendicular force is delivered to the surface of the objects. F/A is the fundamental formula for pressure (Force per unit area). Pascals are a unit of pressure (Pa). Absolute, atmospheric, differential, and gauge pressures are different types of pressure.

Given that:

Atmospheric pressure at sea level has a value of  p = 100 kPa = 100000 Pa.

The density of sea water is 1020kg/m-3.

Let: at  depth of x m the sea would the total pressure be: P =110 kPa = 110000 Pa.

So, P = p + ρgh

⇒ 110,000 Pa = 100,000 Pa + (1020 kg/m³) (9.8 m/s²) h

⇒ h = 1 m

Hence, at depth of 1 m the sea would the total pressure be 110kpa.

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Answer:

Time spent on the greenway road  = 4.5 hours

Time spent on the 2 lane road = 1.5 hours

Explanation:

The distance of the trip  is 360 miles and the initial speed of the car is 62 miles/hr and after the road became 2 lane highway the car slowed to 54 miles/hr.

Let us divide the trip into two

Greenway

speed = distance/time

speed = 62 mph

time = a

distance = speed × time

distance = 62a

2 lane highway

speed = distance/time

speed = 54 mph

time = b

distance = speed × time

distance = 54b

Total distance

62a + 54b = 360......................(i)

Total time

a + b = 6..............................(ii)

a = 6-b

insert a in equation (i)

62(6-b) + 54b = 360

372  - 62b + 54b = 360

-8b = 360-372

-8b = - 12

b = 12/8

b = 1.5

from equation (ii)

a + 1.5 = 6

a = 6 - 1.5

a = 4.5

Answer:

Total force = 1257.5N

Explanation:

Acceleration = 3.50 m/s2

Total Mass = 245kg

Force = Mass x acceleration

Force = 245 x 3.50 = 857.5N

Opposing Force = 400N

Total Force = Force of Motorcycle + Opposing force

Total force = 857.5N + 400N

Total force = 1257.5N

Answer:

1. A Haploid

2. A. Somatic Cell production

3. B. Gametic cell production

4. C. Mitosis produces identical cells to parent cells.

5. B 4:0

6. C Both parent bees are heterozygous Ll

7. D DNA is made up of chromosomes

8. B Each hypothesis was tested and DNA supported the data better

9. D There may be an increase in pest populations that are resistant to control measures.

10. A They are proven to cause cancer.

11. B a form of a gene

Explanation:

Answer was too long, so is attached in a word document

Answer:

The magnitude of average force [tex]F_{avg} =[/tex] [tex]9.125 \times 10^{4} N[/tex]

Explanation:

Given :

Mass of person [tex]m = 73[/tex] Kg

Initial velocity of car  [tex]v_{o} = 20 \frac{m}{s}[/tex]

Average distance [tex]x = 0.16[/tex] m

According to the second law of newton.

 [tex]F_{avg} = m a[/tex]

From the kinematic equations,

 [tex]v^{2}- v_{o} ^{2} = 2ax[/tex]

Where [tex]v_{o} =[/tex] initial velocity, in our case final velocity is zero ([tex]v = 0[/tex])

 [tex]a =- \frac{400}{2 \times 0.16} = -1250 \frac{m}{s^{2} }[/tex]

So average force is given by,

[tex]F _{avg} = 73 \times -1250 = -91250[/tex] N

But magnitude of average force is,

[tex]F_{avg} = 9.125 \times 10^{4}[/tex] N

Answer:

Cover-braking technique

Explanation:

Cover braking involves the technique of taking off the foot from the accelerator and hovering it over the braking pedal. The foot must not be placed on the braking pedal to avoid the wear of brake unnecessarily.

It is done usually when an obstacle is expected in front of the moving car. This provides an edge in the reaction time of application of the brakes hence reducing the stopping time.

Regenerative braking and engine braking are techniques used for a smooth transition from acceleration to braking. These techniques convert kinetic and potential energy into electrical energy or use engine power to slow down the vehicle, respectively.

The technique that provides a smooth transition from acceleration to braking is commonly recognized in physics and engineering called regenerative braking. Regenerative braking is a central component of the operation of hybrid and electric vehicles. It functions by converting a vehicle's kinetic and gravitational potential energy into electrical energy that recharges the vehicle's battery when deceleration is initiated.

Another method is engine braking, usually applied in larger vehicles like trucks to avoid overheating of brakes specially when traveling downhill.

Both methods, whether it is engine braking or regenerative, provide a smooth progression from accelerated motion to a decrease in speed, or braking. This way, they manage the transition smoothly while conserving and efficiently using energy.

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Answer: moon mass = earth mass/96

Explanation:

The moon mass will be 1/96th of the earth mass. There is an attached detailed solution to this.

Answer:

2.55 m

Explanation:

The motion of the dancer is the motion of a projectile, which consists of 2 independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion (constant acceleration) along the vertical direction

The horizontal range of a projectile can be found by using the equations of motions along the two directions, and it is given by:

[tex]d=\frac{v^2 sin(2\theta)}{g}[/tex]

where

v is the intial velocity

[tex]\theta[/tex] is the angle of projection

[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity

For the dancer in this problem, we have:

v = 5 m/s

[tex]\theta=45^{\circ}[/tex]

Therefore, the horizontal range is:

[tex]d=\frac{(5)^2(sin 2\cdot 45^{\circ})}{9.8}=2.55 m[/tex]

Answer:

Both

A. Low tides are lowest at both full moon and new moon.

B. High tides are highest at both full moon and new moon.

Explanation:

Tides are formed as a consequence of the differentiation of gravity due to the moon across to the Earth sphere.

Since gravity variate with the distance:

[tex]F = G\frac{m1\cdot m2}{r^{2}}[/tex]  (1)

Where m1 and m2 are the masses of the two objects that are interacting and r is the distance Where m1 and m2 are the masses of the two objects that are interacting and r is the distance between them.

For example, see the image below, point A is closer to the moon than point b and at the same time the center of mass of the Earth will feel more attracted to the moon than point B. Therefore, that creates a tidal bulge in point A and point B.

On the other hand, a full moon it gets when Sun, the Earth and the moon are in a line and the moon is reflecting the sunlight.

When the Moon is between the Earth and the Sun it will be illuminated in its back, so it is not possible to see it from the Earth (that is called new moon).

In those two cases mentioned above, the Sun tidal force contributes to the tidal force of the moon over the earth making high tides higher and low tides lower.  

Answer:

k = 1755 N/m

Explanation:

Given:

- The length of the cord L = 23.4 m

- Weight of the student W = 818 N

- The elevation of balloon H = 31.3 m

Find:

Calculate the required force constant of the cord if the student is to stop safely 2.74 m above the river.

Solution:

- We know the potential energy of the student changes by  

                      ΔP.E = m*g*( H - 2.74 )

                      mg*(31.3 - 2.74) = 818*28.56 = 23362.08 J

- When he stops at 2.74 m above ground his KE = 0 so ALL his lost potential energy must be stored in the extended or stretched bungee cord.

- He falls 23.4 m before the bungee cord starts to stretch. That means it doesn't start stretching until he is 31.3 - 23.4 = 7.9 m above the ground.  

- It has to stop  stretching at 2.74 m above the ground so the

                                total stretch = 7.9 - 2.74 = 5.16 m

- Therefore his PE from 31.3 m to 2.74 m is stored in a 5.16 m stretch of the bungee cord.                      

                                ½kx² =  ΔP.E

                                k = 2*ΔP.E / x^2

                                k = 2*23362.08 / 5.16^2

                                k = 1755 N/m

Answer:

it would be longer

Answer:

Explanation:

Torque on smaller wheel

= F x r

50 x .30

= 15 Nm

Torque on larger wheel

= F x .5

For equilibrium

F x .5 = 15

F = 15 / .5

= 30 N

Answer:

Distance covered by Tevin before his tire went flat = 24 miles

Explanation:

Let x be the distance covered by Tevin before his tire went flat.

Given:

Tevin drives his bike in the town = 6 mph  

Tevin back to his house = 3 mph

Total taken time by Tevin = 12 hours

We need to find the distance covered by Tevin in 12 hours.

Solution:

Using speed formula    

[tex]Speed =\frac{Distance}{Time}[/tex]

We write the above formula for Time.

[tex]Time=\frac{Distance}{Speed}[/tex]-----------(1)

Time taken by Tevin when biking in town

Substitute speed = 6 mph and distance = x in equation 1.

[tex]Time=\frac{x}{6}[/tex]  ----------(2)

Time taken by Tevin when he is back to his house.

Substitute speed = 3 mph and distance = x in equation 1.

[tex]Time=\frac{x}{3}[/tex]   -------------(3)

Total time taken by Tevin.

Total taken time by Tevin = Time taken when biking in town + Time taken when Tevin back to his house.

Substitute time value from equation 2 and 3 in above equation and total time = 12 hours.

[tex]12=\frac{x}{6}+\frac{x}{3}[/tex]

Now, we solve the above equation for x.

[tex]12=\frac{x+2x}{6}[/tex]

[tex]12\times 6=x+2x[/tex]

[tex]72=3x[/tex]

[tex]x=\frac{72}{3}[/tex]

[tex]x=24\ mi[/tex]

Therefore, distance covered by Tevin in 12 hours is equal to 24 miles.

Answer:

b. slow-moving streams.

Explanation:

In Fluid Mechanics, the Reynolds numbers indicates the existence of turbulence in fluid streams. Low Reynolds numbers are related with laminar flow. The Reynolds formula is:

[tex]Re = \frac{\rho_{water} \cdot L_{c}}{\mu_{water}} \cdot v[/tex]

The Reynolds number is directly proportional to fluid speed. Hence, slow-moving streams are a sound example of laminar flow. The correct answer is B.

Answer:

Habitat manipulation

Explanation:

Habitat manipulation, otherwise known as ecological engineering, is a technique of promoting natural enemies within an ecosystem by making thriving conditions more suitable for them.

In this case, thriving conditions for the species (which happens to be predators and hence, natural enemies) were promoted via artificial introduction of food.

Answer:

Coefficient of friction between the book and floor is 0.582.

Explanation:

Using the velocity formula;

v^2 = 2as

a = v^2/(2s)

a = 1.6^2/(2*0.9)

a = 2.56/1.8

a = 1.42 m/s^2

the force necessary to give the book the acceleration is  

F = ma = 3.5*1.42 (m is mass of the book i.e. 3.5 kg)

F = 4.98 N

The difference in the force is the friction force, which is

Ff = 25 - 4.98 = 20 N

Ff = mgμ

where μ is coefficient of friction and g is acceleration due to gravity that is 9.8 m/s^2

μ = Ff/mg

μ = 20/(3.5*9.81)

μ = 0.582

Coefficient of friction between the book and floor is 0.582.

This question involves the concepts of the equation of motion and Newton's second law of motion.

The coefficient of kinetic friction between the book and the floor is "".

First, we will find the acceleration of the block by using the third equation of motion:

[tex]2as = v_f^2-v_i^2[/tex]

where,

a = acceleration = ?

s = distance covered = 0.9 m

vf = final speed = 1.6 m/s

vi = initial speed = 0 m/s

Therefore

[tex]a=\frac{(1.6\ m/s)^2-(0\ m/s)^2}{2(0.9\ m)}\\\\a=1.42\ m/s^2[/tex]

Hence, from Newton's second law of motion:

[tex]Net\ Force = Frictional\ Force + F\\Net\ Force = \mu mg+ma[/tex]

where,

Net Force = 25 N

μ = coefficient of kinetic friction = ?

m = mass of the book = 3.5 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]25\ N = \mu(3.5\ kg)(9.81\ m/s^2)+(3.5\ kg)(1.42\ m/s^2)\\\\\mu=\frac{25\ N - 4.98\ N}{34.33\ N}\\\\\mu = 0.58[/tex]

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The attached picture shows Newton's Second Law of Motion.

Explanation:

(a)   Formula to calculate the force of kinetic friction is as follows.

             f = [tex]\mu N[/tex]

                = [tex]\mu mg Cos (\theta)[/tex]

Putting the given values into the above formula as follows.

          f = [tex]\mu mg Cos (\theta)[/tex]

           = [tex]0.118 \times 57.4 kg \times 9.8 \times Cos (28.1^{o})[/tex]

           = [tex]0.118 \times 57.4 kg \times 9.8 \times 0.882[/tex]

           = 58.54 N

Hence, the force of kinetic friction is 58.54 N.

(b)    Net force experienced by the block will be as follows.

            F = [tex]mg Sin (\theta) - f[/tex]

         ma = [tex]mg Sin (\theta) - \mu mg Cos (\theta)[/tex]

or,         a = [tex]g[Sin (\theta) - \mu Cos (\theta)][/tex]                  

                = [tex]9.8[Sin(28.1) - Cos(28.1)][/tex]

                = [tex]9.8 \times (0.471 - 0.882)[/tex]

                = 4.03 [tex]m/s^{2}[/tex]

Therefore, the acceleration is 4.03 [tex]m/s^{2}[/tex].

(c)    According to the third equation of motion,

          [tex]v^{2} = u^{2} + 2as[/tex]

                    = [tex]0 + 2 \times 4.03 \times 17.2[/tex]        

                    = 138.63 m/s

Hence, the speed she is traveling when she reaches the bottom of the slide is 138.63 m/s.

Answer:

Explanation:

mass, m = 57.4 kg

distance, d = 17.2 m

angle of inclination, θ = 28.1°

initial velocity, u = 0 m/s

coefficient of kinetic friction, μk = 0.108

(a) N is the normal reaction acting on the student.

N = mg Cosθ

N = 57.4 x 9.8 x Cos 28.1

N = 496.2 N

Friction force = μk x N

Friction force = 0.108 x 496.2 = 53.6 N

Let a is the acceleration

ma = mg Sinθ - friction force

ma = 57.4 x 9.8 x Sin 28.1 - 53.6

a = 3.7 m/s²

Let the speed is v.

v² = u² + 2ad

v² = 0 + 2 x 3.7 x 17.2

v = 11.3 m/s