A 70kg student, happy to see snow in early
November, rushes out to sled on a hill with a 10kg
sled. They start h=12m above a horizontal plane.
How fast are they moving by the time they reach
the bottom of the hill? (assume the coefficient of
friction is zero here) After a short distance the
reach a point where there are some dry leaves are
scattered producing an effective coefficient of
friction uk=0.4. How far do they travel through the leaves before they
come to rest?
Use the fundamental principle

Answers

Answer 1

(a) The speed of the student at the bottom of the hill is 15.34 m/s.

(b) The speed of the student at the presence of friction force is 11.88 m/s.

What is the speed of the student at the bottom of the hill?

The speed of the student at the bottom of the hill is calculated by applying the principle of conservation of energy as shown below.

Kinetic energy at the bottom hill = potential energy at maximum height

¹/₂mv² = mgh

where;

m is mass of the studentv is the speed of the speed of the student at bottom hillh is the height of the hill

v² = 2gh

v = √2gh

v = √(2 x 9.8 x 12)

v = 15.34 m/s

The speed of the student at the presence of friction force is calculated as;

Kinetic energy at the bottom hill + work done against friction  = potential energy at maximum height

¹/₂mv² + μmgh = mgh

¹/₂v² + μgh = gh

v²  +  2μgh = 2gh

v² = 2gh - 2μgh

v² = (2 x 9.8 x 12) - (2 x 0.4 x 9.8 x 12)

v² = 141.12

v = √141.12

v = 11.88 m/s

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Related Questions

If a 62 kg panther sits in a tree 1.3 meters above the ground, how much gravitational potential energy does it have?

If the air resistance cause the panther to lose 200 J of energy as it falls, what is the KE of the panther just before it hits the ground?

What was the velocity of the panther just before it hit the ground?

Answers

Answer:

below

Explanation:

PE = mgh = 62 * 9.81 * 1.3 = 791 J

  now lose 200 J    to air resistance  = 591 J   that is converted to kinetic energy

KE = 1/2 m v^2

591 = 1/2 (62)(v^2)     shows v = 4.4 m/s

URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!

Answers

Answer: B

Explanation:

You will have air resistance as a type of friction, and since there isn't a normal force opposing it, gravity will bring the ball toward the ground.

7. A child of mass m starts from rest and slides without friction from a height h along a curved waterslide (Fig. P5.46). She is launched from a height into the pool. Figure P5.46 4/5 Fmax (a) Is mechanical energy conserved? Why?​

Answers

The mechanical energy of the girl will be conserved because the system is isolated and the initial potential energy will be equal to final kinetic energy.

What is the law of conservation of energy?

The law of conservation of energy states that energy can neither be created nor destroyed but can be transformed from one form to another.

The change in the potential energy of the  launched from a height into the pool without friction from the given height h is calculated by applying the following kinematic equation.

ΔP.E = ΔK.E

where;

ΔP.E is change in potential energy of the childΔK.E is change in the kinetic energy of the child

mghf - mghi = ¹/₂mv²  - ¹/₂mu²

where;

m is the mass of the girlg is acceleration due to gravityhi is the initial height of the girlhf is the final height when she is launched into the poolu is the initial velocityv is the final velocity of the girl

Thus, for every closed or isolated system such as this case, mechanical energy is always conserved because the initial potential energy of the girl will be converted into her final kinetic energy.

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Will give Brainliest!
A thin 2.50 kg box rests on a 5.50 kg board that hangs over the end of a table, as shown in (Figure 1).

How far can the center of the box be from the end of the table before the board begins to tilt?

Answers

Based on the principle of moments. the distance from the end of the table should the box be placed before the board begins to tilt is 8.6 cm.

What distance from the end of the table should the box be placed before the board begins to tilt?

The distance from the end of the table should the box be placed before the board begins to tilt is determined from the principle of moments as follows:

sum of clockwise moments = sum of anticlockwise moments

The block is 30 cm on the table and 20 cm outside it.

The downward force acting on the left-hand side of the box = 3/5 x 5.50  = 3.3kg.

This force acts at the center of gravity, 15 cm or 0.15 m away.

Therefore, anticlockwise moments o the left side = 3.3 x 0.15 = 0.495 J

Also, the clockwise moment on the right side = Force * distance

Force = 2/5 x 5.5 = 2.2 N

Distance from the center of gravity = 10 cm or 0.10 m away.

the clockwise moment on the right side due to the board = 2.2 x 0.1 = 0.22.

The moment due to the box with a weight of 2.5 kg at a distance of x meters will be:

the total clockwise moment on the right side = 0.28 + 2.5 * x.

When the board is just about to tilt:

0.495 = 0.28 + 2.5 * x

2.5x = 0.495 - 0.28

x = 0.086 m or 8.6 cm

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A race car leaves the starting line and travels 36,000 m in the first 600 seconds of the race. They are then forced to take a pit stop and don't go anywhere for 250 seconds. After the pit stop, they finish the race, going 24,500 m in 350 seconds.

a. What is the car's average speed during the first part of the race? (before the pit stop)

b. What is the car's average speed during the pit stop?

c. What is the car's average speed after the pit stop?

d. What is the car's average speed for the whole trip?

Answers

Answer:s= A. 60 m/s

B. s=0 m/s

c. s=70 m/s

D. s=50.4 m/s

Explanation:

Sorry if its wrong

Question 3 (2 points)
Of the following exercise stimuli, ordered from the lightest to the heaviest, select the
intensity that has been shown to significantly improve affect for most people (albeit
perhaps for a relatively short period of time)
a 10-minute walk at a self-selected intensity
a 20-minute session of a cycle ergometer at 60% VO2max
a 30-minute treadmill run at 75% VO2max
a 60-minute session of aerobics at 75% VO2max
multiple 30 second intervals at 150% VO2max

Answers

A 20-minute cycle ergometer exercise at 60% VO2max has been demonstrated to dramatically improve impact for the majority of persons.

Intensity and an example are what?

the property of being strongly felt or having a powerful impact: The explosion was so loud that it could be heard from five kilometres distant. Measures of luminosity. [C or U] the strength of being something that may be measured, such as sunlight, sound, etc.

The measurement of intensity:

A measure that is produced from a randomised measure is known as an intensity measure in probability theory. Since the randomised measure of a set's expected value is the intensity measure, which is a non-random measure, it equates to the average volume that the random measure assigns.

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b) Robert pulls at a steady speed for
2m/s how far will she travel in 10 minutes
10mins?
I need help someone pleaseeeee
Tysm

Answers

Robert pulls at a steady speed for 2m/s. She travelled 1200m in 10 minutes.

Distance : Path travelled by an object is the total distance.speed : distance travelled in an interval of time.Time : It is inversely proportional to the time.

Speed = 2m/s

time taken = 10 minutes

(1 minute = 60 seconds )

then,

10 mint = 10 x 60 = 600 seconds

To calculate how far she travel we use :

distance travelled = speed x time

distance = 2 x 600

distance = 1200m

Robert pulls at a steady speed for 2m/s. She travelled 1200m in 10 minutes.

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Units and magnitudes 1. Make the following transformations 500 cm to m:​

Answers

Answer:

5 m

500 cm to m is 5 meters

To convert one length to another, a conversion is applied, but first we will propose some data to serve us when applying the conversation.

Data:

100 cm is the same as a meter.& 1 meter is the same as 100 cm.

Now, we convert centimeters to meters.

Conversion:

500cm • (1m / 100cm)5m

Explanation:

They are 5 meters because multiplying 500 by 1 will give you the same result of 500, then we divide the 500 with the last amount that we have left, which is 100, thus giving us a result of 5 meters.

[100 points]a force of 330 newtons is required to lift a crate up a ramp a distance of 5 meters in 2 seconds. how much power is used?

Answers

Answer:

17,482.36 J

Explanation:

Ans. W = F •d cos  = 120 N • 165 m cos 28.0 = 17,482.36 J

Answer:

Explanation:

Given:

F = 330 N

h = 5 m

t = 2 s

_________

N - ?

N = A / t = F·h / t = 330·5 / 2 = 825 W

how can an object have the same speed and mass but change its momentum?​

Answers

Answer: It can change its direction

Explanation:

Momentum is a vector, which has magnitude and direction.  An object can have the same speed and mass (magnitude) but if it changes direction, it just changed its momentum

2. What is the force on a 1 kg bal that is falling freely due to the pull of gravity?

Answers

Answer: 9.8 N

Explanation:

A
where light does not strike.
A) mirror
C) wave
is an area
B) transparent
D) shadow
q

Answers

Answer:

d

Explanation:

because when an object stand in front of light, light bends around the other and doesn't go through it

a bolder of mass 45kg is pushed on a surface with a coefficient of kinetic friction of 0.85. what force has to be applied to produce an acceleration of 2 m/s^2

Answers

Answer: The answer is 386 N

Explanation:

a baby carriage is sitting at the top of the hill that is 21 m high. the carriage with the baby has a mass or 1.5kg. the carriage has ___ energy. calculate it.

Answers

Answer:
309 Joules

Steps:
We can use the gravitational potential energy formula.
GPE = mgh
We are given the mass, the height, and we know gravity. Let’s solve for GPE

1.5 * 9.81 * 21 = 309 Joules

The unbalanced force on the miners elevator is 60.0 N and the mass of the loaded elevator is 150 kg what is the acceleration of the elevator down the shaft

Answers

Answer:

Explanation:

F = m*(g - a)

g - a = F / m

a = g - F / m = 10 - 60.0 / 150 = 9.6 m/s²


A man walking through a field travelled 30 m south and then 20 m west.
Calculate the distance he travelled and his displacement

Answers

The net distance and displacement traversed is 50m and  36.0555 m respectively.

Given - A man has walked 30m and 20m south and west respectively.

To find distance and displacement .

principle- Distance means path covered and a vector quantity. Displacement means shortest distance covered and state function. It is also scalar quantity. Displacement can be zero but distance can't be zero.

Total distance traversed= 30 + 20 = 50m

Total displacement traversed (s) ⇒√(x₁² + x₂²)

⇒√(30² + 20²)    ⇒√(900 + 400)

=√1300

=36.0555 m

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Derrick goes to a party and has a bad time, which decreases his party going behavior. This is an
example of
A) positive reinforcement
B) negative reinforcement
C positive punishment
D) negative punishment

Answers

Answer:

B) negative reinforcement

Explanation:

Derrick goes to a party and has a bad time, which decreases his party going behavior. This is an example of negative reinforcement. Hence option B is correct.

What is reinforcement ?

According to the reinforcement hypothesis, "contingent consequences" of human activities lead to human behaviour. This means that when employees get the appropriate reinforcers, their conduct can change for the better, and bad behaviour can be eliminated.

Self-awareness, self-reflection, and self-regulation are the three key tenets of the self-regulation paradigm of human behaviour. Historically, rewards have correlated with self-regulation. Although the result may have an impact on the conduct, behaviour also need antecedents. Positive reinforcement, negative reinforcement, extinction, and punishment are the four different forms of reinforcement. The use of a positive reinforcer is known as positive reinforcement. In order to encourage the antecedent behavior from that, negative reinforcement is the technique of eliminating anything undesirable from the subject's environment.

Hence option B is correct.

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A rocket blasts off from rest and attains a speed of 33.9 m/s in 12.7 s. An astronaut has a mass of 65.1 kg. What is the astronaut's
apparent weight during takeoff?

Answers

The apparent weight of the astronaut of mass 65.1 kg moving with a speed of 33.9 m/s in 2.7 s is 811.797 N.

What is weight?

Weight is the force with which a body is attracted toward the earth or a celestial body by gravitation and which is equal to the product of the mass and the local gravitational acceleration.

To calculate the astronaut's apparent weight, we use the formula below.

Formula:

W = m{[(v-u)/t]+g}............ Equation 1

Where:

W = The apparent weight of the astronautm = Mass of the astronautv = Final speedu = Initial speedt = Timeg = Acceleration due to gravity

From the question,

Given:

m = 65.1 kgv = 33.9 m/su = 0 m/s (from rest)t = 12.7 sg = 9.8 m/s²

Substitute these values into equation 1

W = 65.1{[(33.9-0)/12.7]+9.8]W = 65.1×12.47W = 811.797 N

Hence, the apparent weight of the astronaut is 811.797 N.

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what is the answer? is it equilibrium or not?​

Answers

No, the system is not in equilibrium because the forces are not balanced which caused acceleration In system.

In mechanics, a force is any action that seeks to preserve, modify, or deform a body's motion. Isaac Newton's three principles of motion, which are outlined in his Principia Mathematica, are frequently used to illustrate the idea of force (1687).

Newton's first law states that unless a force is applied to a body, it will stay in either its resting or uniformly moving condition along a straight path. According to the second law, when an external force applies on a body, the body accelerates (changes velocity) in the force's direction.

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Which term describes light passing straight through an object?
O transmission
O reflection
O refraction
O absorption

Answers

Answer:

transmission

Explanation:

Answer: transmission

Explanation: took test!

Can you help? It give a example and I need help seeing what type of boundary it is.

Answers

Answer:

Explanation:


4. Convergent boundary- when two plates move closer together.


5. Divergent boundary- when two plates move apart


6. Transform boundary- when two plates move past each other

A 0.80-m-tall barrel is filled with water (with a weight density of 9800 N/m³). Find the water
pressure on the bottom of the barrel.
Express your answer to two significant figures and include the appropriate units.

Answers

Pressure on the bottom of the barrel = 1.78 * 10⁵Pₐ

a) The force applied perpendicular to the object's surface and dispersed over the area is referred to as the pressure.

Given,

height of barrel = 0.8mP

density = 9800N/m²

atmospheric pressure = Pₙ = 1.013 * 10⁵Pₐ

From the relation,

Pressure at the bottom

Pₓ = Pₙ + pgh

Pₓ = 1.013 * 10⁵ + 9800 * 9.8 * 0.8

Pₓ = 1.78 * 10⁵Pₙ

b)

Given,

mass = 1kg

area, a =1.2cm² =1.2 * 10⁻⁴ m²

Gravitational acceleration g = 9.8m/s²

We know that

Pressure on the finger,

Pf = force/area

Pf = (mass*g) / a

Pf = (1 * 9.8)/ (1.2 * 10⁻⁴)

Pf = 8.17 * 10⁴ N/m²

a) pressure on the bottom = 1.78 * 10⁵Pₐ

b) pressure on the finger = 8.17 * 10⁴ N/m²

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Which quantities (initial horizontal speed, initial vertical speed, range, flight time, or maximum height) varies linearly with the initial velocity?

Answers

Initial horizontal speed and  flight time of a projectile are proportional to of the initial velocity.

What is projectile motion?

A projectile is any object that is sent into space with only gravity acting on it. The primary force affecting a projectile is gravity.

This doesn't mean that other forces don't have an impact; it just means that they have a much smaller one compared to gravity. The trajectory of a projectile is the path it takes after being fired. The projectile is something that is batted or hurled, similar to a baseball.

Initial horizontal speed  projectile= u

flight time of a projectile = 2usinθ/g.

So, these two quantities of projectile motion are  proportional to the initial velocity.

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Questions
1. a. Explain why you need to use half the time taken for an echo to be
detected when you are doing calculations of distance.
b. Sound travels at 330 m/s in air. It takes 1.5s to detect the echo
from a building. Calculate the distance to the building.
(Hint: Remember that the time is for the journey there and back).
2. Explain why dolphins can find food at greater distances than bats.
3. A woman is having an ultrasound scan. Ultrasound travels at
1500 m/s inside the body. The ultrasound transmitter sends out a
pulse of ultrasound waves, which are reflected from a tissue surface
7.5 cm away. Calculate the length of time before the echo is received
(Hint: Be careful with units!)

Answers

1.a. The reason of using half time is  an echo refers to the Reflection of sound after hitting the target just like a basket ball bouncing back as it hits its target i.e. earth.

So parameters like total energy , distance, amplitude of waves,.etc..  refers to the both action and reaction journey.

Like for example if a rubber ball travels 5m to hit the wall and returns back . so the net distance = incident distance + reflection distance⇒ 10m

But in Physics to compute the accurate result of incoming signals or energy perceived we just have to take either of the 2 journeys i.e. incidence or reflection. Second perception is the concept of taking average of multiple data. When there is 2 times i.e. one time to take incidence and other one is reflection , so in such cases average of the time for close vicinity of result.

1.b . The net distance propagated is 247.5 m.

Given - Sound travels 330m/s taking time 1.5s

To find the distance travelled -

distance travelled in 1 sec= 330 m

distance for 0.75 sec= 330 x 0.75 = 247.5m

Since 1.5 sec is the total time of the process of echo.

[tex]echo = incidence + reflection[/tex]

so taking time of one journey = 1.5 sec/ 2 = 0.75 sec

Hence the distance traversed is 247.5 m .

2. Dolphins uses echolocation as an ability or tool to perceive the incoming waves in atmosphere that includes sound waves.

They generate a special kind of sound that makes its target i.e. food to come to it.

Now talking about bats , although they also have similar ability of echolocation , but its mechanism to capture a food is time consuming comparatively and dolphins use higher frequencies compared to bats.

3.The length of time before the echo is received is 0.0001 sec.

Given - Speed = 1500m/s , distance =7.5cm ⇒0.075m

To find how long it will take before the receptance of incoming sound signal :-

[tex]2d=speed * time \\2d=vt[/tex]

[2 x 0.075]/ 1500= t

t= 0.0001 sec

so it takes 0.0001 sec to traverse before the echo is achieved.

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During a medieval siege of a castle, the attacking army uses a trebuchet to hurl heavy stones at the castle walls. If the trebuchet launches the stones with a velocity of +30.0m/s at an angle of 50.0°, how long does it take the stone to hit the ground? What is the maximum distance that the trebuchet can be from the castle wall to be in range? How high will the stones go? Show all your work.

Answers

(a) The time of motion of the stone is 4.69 seconds.

(b) The maximum distance the trebuchet can be from the castle wall to be in range is 90.44 m.

(c) The maximum height reached by the stone is 26.95 m.

What is the time of motion of the stone?

The time of motion of the stone is calculated as follows;

t = 2u sinθ / g

where;

u is the initial velocity of the stoneθ is the angle of projection of the stoneg is acceleration due to gravity

t = (2 x 30 x sin50) / (9.8)

t = 4.69 seconds

The maximum distance the trebuchet can be from the castle wall to be in range is calculated as;

R = u²sin(2θ) / g

R = (30² sin(2 x 50))/(9.8)

R = 90.44 m

The maximum height reached by the stone is calculated as follows;

H = u² sin²θ / 2g

H = (30²  x (sin 50)²) / (2 x 9.8)

H = 26.95 m

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Calculate the change in kinetic energy of a ball of mass 200 g when it bounces. Assume that it hits the ground with a speed of 15.8 m/s and leaves it at 12.2 m/s.

Answers

Answer: So, the kinetic energy of the ball is 0.02 J

Explanation:

An ideal massless spring, inclined plane, that makes an angle theta with the horizontal, and mass arrangement is given. A block of mass m is released from rest at the top of a frictionless incline. The block comes to rest momentarily after it has compressed this spring by ∆x. Initially, distance between the block and the spring is d.
Find ∆x.

Answers

Compression is  Δx = √(2mgd·sinθ/k).

Given parameters:

Mass of the block = m.

Distance between the spring and the block is = d

The spring constant = k.

And, angle of inclination = θ.

And,  The block comes to rest momentarily after it has compressed this spring by ∆x.

Now, loss of potential energy of the block = mgd sinθ.

And, gain in potential energy of the spring due to compression = 1/2k(Δx)²

From principle of conservation of energy,

1/2k(Δx)² =  mgd sinθ.

⇒ Δx = √(2mgd·sinθ/k)

So, amount of compression is  Δx = √(2mgd·sinθ/k).

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The half-life of chromium-51 is 28 days. If the sample contained 510 grams, how much chromium would remain after 56 days?

Answers

The quantity of 128 grams of chromium would be remaining.

The time required for half the reaction to change is called half life period.

This concept is useful for understanding carbon dating in Nuclear Physics. The term is used more to characterize any type of exponential decay. Nuclei that decay easily have shorter half life.

Given :- Half Life of Cr = 28 days

m= 510 grams

To find :- quantity of Cr after 56 days.

Formula : Nₙ= N₀/(2)ⁿ

Nₙ= 510/2²

Nₙ= 128g

Therefore 128 grams will be available after 56 days.

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my thoughts about earthwake

Answers

An earthquake is a weak to violent shaking of the ground produced by the sudden movement of rock materials below the earth's surface. The earthquakes originate in tectonic plate boundary.
Earthquake is a sudden violent movement of the earth's surface, causing great damages.



2. A runner whose initial speed is 8.06 m/s increases her speed to 8.61 m/s in order to win a race. If the
runner takes 5.00 seconds to complete this increase in speed, what is her acceleration?

Answers

Answer:

Explanation:

Given:

V₀ = 8.06 m/s

V = 8.61 m/s

t = 5.00

________

a - ?

Acceleration:

a = (V - V₀) / t

a = (8.61 - 8.06) / 5.00 ≈ 0,11 m/s²

Answer:0.11

Explanation:

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