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Paraphin [41]
3 years ago
10

When a wave crashes on the beach, a child exclaims that the particles in the water wave may have traveled all the way across the

world. What is wrong with her reasoning?
A. waves cannot travel long distances
B. The particles in a wave tend to fall off as the wave travels.
C. Waves do not transmit matter.
D. waves do not have enough energy to transmit a water particle across the world.
Physics
1 answer:
Brilliant_brown [7]3 years ago
7 0

Answer:

C. Waves do not transmit matter.

Explanation:

A wave is a disturbance that transmits energy from one point to another.

Waves are simply means of transferring energy form places to places without moving the materials of the medium.

The particles of wave are not simply transported from one part to another, in water waves only the energy moves.

It will be a wrong assertion to claim that water waves transmits particles from one place to another round the world.

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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
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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

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The equation is garbled and the question is missing.

I found this equation for the same statement:

S = - 2.7t ^2 + 30t + 6.5

And one question is: after how many seconds is the ball 12 feet above the moon's surface?

Given that S is the height of the ball, you just have to replace S with 12 and solve for t.

=> 12 = - 2.7 t^2 + 30t + 6.5

=> 2.7t^2 - 30t - 6.5 + 12 = 0

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Now you can use the quadratic equation fo find t:

t = { 30 +/- √ [30^2) - 4(2.7)(5.5)] } / (2*2.7)

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