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kozerog [31]
3 years ago
12

A 45.0 kg skier testing a new competitive ski wax drops off a ledge, goes into a crouch and goes straight down a slope of 10.0◦

with respect to the horizontal. The coefficient of kinetic friction between the snow and skis is 0.114. The skier’s speed after traveling 300 m along this slope is (in m/s):
Physics
1 answer:
Neko [114]3 years ago
6 0

Answer:

v = 21.03 m/s

Explanation:

given,

mass of skier = 45 kg        

the slope of the snow = 10.0◦      

coefficient of friction = 0.114  

distance traveled = 300 m      

speed = ?              

Acceleration = g sin θ - µ g Cos θ        

= 9.8 × Sin (10°) - 0.10 × 9.8 × Cos(10°)        

= 0.737 m/s²      

using equation of motion        

v² = u² + 2 a s        

v² = 0 + 2 × 0.737 × 300            

v = 21.03 m/s                  

Speed of skier's after travelling 300 m speed is  equal to 21.03 m/s

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A roller coaster starts at the top of a hill of height h, goes down the hill, and does a circular loop of radius r before contin
jeka94

a) See free-body diagram in attachment

b) Net force in the y-direction: F_y=mg+N[/tex]

c) The velocity at which the roller coaster will fall is [tex]v=\sqrt{gr}[/tex]

d) The speed of the roller coaster must be 17.1 m/s

e) The roller coaster should start from a height of 90 m

f) The roller coaster should start from a height of 100 m

Explanation:

a)

See the free-body diagram in attachment. There are only two forces acting on the roller coaster at the top of the loop:

  • The weight of the roller coaster, acting downward, indicated by mg (where m is the mass of the roller coaster and g is the acceleration of gravity)
  • The normal reaction exerted by the track on the roller coaster, acting downward, and indicated with N

The two forces are represented in the diagram as two downward arrows (the length is not proportional to their magnitude, in this case)

b)

Since there are only two forces acting on the roller coaster at the top of the loop, and both forces are acting downward, then we can write the vertical net force as follows (we take downward as positive direction):

F_y = mg + N

where

mg is the weight

N is the normal reaction

Since the roller coaster is in circular motion, this net force must be equal to the centripetal force, therefore

m\frac{v^2}{r}=mg+N

where v is the speed of the car at the top of the loop and r is the radius of the loop.

c)

For this part of the problem, we start from the equation written in part b)

m\frac{v^2}{r}=mg+N

where the term on the left represents the centripetal force, and the terms on the right are the weight and the normal reaction.

We now re-arrange the equation making v (the speed) as the subject:

v=\sqrt{gr+\frac{Nr}{m}}

However, the velocity at which the roller coaster will fall is the velocity at which the normal reaction becomes zero (the roller coaster loses contact with the track), so when

N = 0

And as a result, the minimum velocity of the cart is

v=\sqrt{gr}

d)

In this part, we are told that the radius of the loop is

r = 30 m

And the mass of the cart is

m = 50 kg

Moreover, the acceleration of gravity is

g=9.8 m/s^2

We said that the minimum velocity that the cart must have in order not to fall at the top is

v=\sqrt{gr}

And substituting, we find

v=\sqrt{(9.8)(30)}=17.1 m/s

e)

According to the law of conservation of energy, the initial gravitational energy of the roller coaster at the starting point must be equal to the sum of the kinetic energy + gravitational potential energy at the top of the loop, therefore:

mgh = \frac{1}{2}mv^2 + mg(2r)

where

h is the initial height at the starting point

(2r) is the height of the roller coaster at the top of the loop

We can re-arrange the equation making h the subject,

h=\frac{v^2}{2g}+2r

And substituting the minimum speed of the cart,

v=\sqrt{gr}

this becomes

h=r+2r=3r

And since r = 30 m, we find

h=3(30)=90 m

f)

In this case, 10% of the initial energy is lost during the motion of the roller coaster. We can rewrite the equation of the previous part as

0.90mgh = \frac{1}{2}mv^2 + mg(2r)

Because only 90% (0.90) of the initial energy is converted into useful energy (kinetic+potential) when the cart reaches the top of the loop.

Re-arranging the equation, this time we get

h=\frac{\frac{v^2}{2g}+2r}{0.90}

Again, by substituting v=\sqrt{gr}, we get

h=\frac{3r}{0.90}

And therefore, the new initial height must be

h=\frac{3(30)}{0.9}=100 m

Learn more about kinetic and potential energy:

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(a) Frequency of sound wave is inversely proportional to string length.

(b) Pitch of a sound depends on the frequency while loudness depends on the amplitude of sound.

<h3>Relationship between string length and frequency</h3>

The relationship between string length and frequency is given as;

λ = 2L

where;

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v = fλ

f = v/λ

f = v/2L

Thus, frequency of sound wave is inversely proportional to string length.

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The pitch of a sound depends on the frequency while loudness of a sound depends on the amplitude of sound waves.

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The model that describes a Sun-centered solar system is called theA.laws of motion.B.geocentric model.C.heliocentric model.D.gen
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The answer is C the heliocentric model.
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3 years ago
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3 years ago
What type of force pulls in two opposite directions?
Mnenie [13.5K]

<u>Answer</u>

a. tension (associated with normal faults)

c. diagonal

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3 years ago
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