What helps determine how easily magma flows?

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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4 0

3 years ago

What are generated at transform fault plate boundaries?

tankabanditka [31]

Answer: The Earth's layer, which has the covering and layer, is made of a progression of things, or structural plates, that creep after some time. Along these lines, at intersecting limits, mainland outside is made and maritime covering is devastated. 2 plates slippy past each other structures a redesign plate limit.

5 0

3 years ago

suppose the distance between the sun and the earth were quadrupled. what exactly (numerically) would happen to the gravitational

Wittaler [7]

Answer:

The gravitational force between the sun and the earth drops to (1/16) of its original value.

Explanation:

According to the gravitational law, gravitational force between two bodies is directly proportional the product of the masses of the two bodies and inversely proportional to the square of their distances apart.

If the two masses and the other constants of proportionality are held constant, the gravitational force between two bodies is inversely proportional to the square of their distances apart.

F ∝(1/R²)

If the initial gravitational force between the sun and the earth is given by F₀ and the distance between the sun and the earth is R₀.

F₀ = (k/R₀²)

where k is constant regardless of the values of corresponding gravitational forces and distances apart.

Then, let the new distance apart be R₁,

R₁ = 4 R₀

The new gravitational force between the sun and the earth, F₁ will be given by

F₁ = k/(R₁²)

Recall, R₁ = 4 R₀

F₁ = k/(4R₀)² = k/16(R₀²) = F₀/16

Hence, the gravitational force between the sun and the earth drops to (1/16) of its original value.

6 0

3 years ago

The periodic table includes___periods.

Brums [2.3K]

Answer:

Seven.

Explanation:

A period is a horizontal row of the periodic table. There are seven periods in the periodic table, with each one beginning at the far left. A new period begins when a new principal energy level begins filling with electrons.

8 0

4 years ago

The batter swings and misses the 40 m/s (90 mph) fastball, and the ball (mass 150 grams) ends up at rest in the catcher's mitt.

Nadusha1986 [10]

Answer:

The work done by the catcher is 120 J.

Explanation:

Given;

velocity of the fastball, v = 40 m/s

mass of the fastball, m = 150 g = 0.15 kg

Based on work-energy theorem, the work done by the catcher is equal to the kinetic energy of the fastball.

The kinetic energy of the fastball is given as;

K.E = ¹/₂mv²

K.E = ¹/₂ x 0.15 x 40²

K.E = 120 J

Therefore, the work done by the catcher is 120 J.

5 0

3 years ago