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Sedbober [7]
2 years ago
9

At the top of a hill a roller coaster has gravitational potential energy due to its position. What happend to this potential ene

rgy as the roller coaster speeds up on the way down the hill?
Physics
1 answer:
Anuta_ua [19.1K]2 years ago
3 0

As the roller coaster speeds up on the way down the hill, the potential energy of  roller coaster will be converted to kinetic energy.

<h3>What is Conservation of Energy ?</h3>

Conservation of energy state that energy is neither created nor destroy, they can only be transformed from one form to another. Energy of and object can transform from Potential energy to kinetic energy and vice versa

Given that at the top of a hill a roller coaster has gravitational potential energy due to its position. What will happen to this potential energy as the roller coaster speeds up on the way down the hill is that the potential energy to the roller coaster will start decreasing while the kinetic energy will start to increase.

The total energy of the roller coaster will be constant because of conservation of energy. As the roller coaster speeds up on the way down the hill, the potential energy will eventually reduce to zero where the total energy of the as the roller coaster will be equal to maximum kinetic energy.

Therefore, as the roller coaster speeds up on the way down the hill, the potential energy of  roller coaster will be converted to kinetic energy.

Learn more about Energy here: brainly.com/question/25959744

#SPJ1

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Which non-mineral nutrients are essential in photosynthesis? (Select all that apply.) oxygen calcium hydrogen carbon
Lunna [17]

Answer:

carbón, hydrogen and oxygen

4 0
3 years ago
A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 15 m/sm/s. A passenger accidentally drops his camera
jeka94

To solve this problem we will apply the linear motion kinematic equations. With the information provided we will calculate the time it takes for the object to fall. From that time, considering that the ascent rate is constant, we will take the reference distance and calculate the distance traveled while the object hit the ground, that is,

h = v_0 t -\frac{1}{2} gt^2

-18 = 15*t + \frac{1}{2} 9.8*t^2

t = 3.98s

Then the total distance traveled would be

h = h_0 +v_0t

h = 18+15*3.98

h = 77.7m

Therefore the railing will be at a height of 77.7m when it has touched the ground

5 0
3 years ago
2. Michelle bought 15.6 yards of fabric. She used 5.43 yards in her art installation. About what percent of the fabric did she u
photoshop1234 [79]
35%
5.43/15.6
This is how you calculate the percentage the answer is 35%
4 0
3 years ago
Could you please solve it with shiwing the full work
tia_tia [17]

Answer:

1.V= 640.48 m/s :total velocity in t= 5s

2. Y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away

3. v =25m/s

4. s= (-1.5t³+26t ) m

Explanation:

1. Parabolic movement in the x-y plane , t=5s

V₀=638.6 m/s=Vx  :Constant velocity in x

Vy=V₀y +gt= 0+9.8*5  = 49 m/s : variable velocity in y

v=\sqrt{v_{x} ^{2} +v_{y} ^{2} }

v=\sqrt{ 638.6^{2} +49 ^{2} }

V= 640.48 m/s : total velocity in t= 5s

2. v_{ox} =v_{o} cos33.2=20.9*cos33,2= 17.49 m/s

v_{oy}=v_{o}*sin33,2 =20.9*sin33,2=11.44 m/s

x=v₀x*t

13=v₀x*t

13=17.49*t

t=13/17.49=0.743s : time for 13.0 m away

th=v₀y/g=11.44/9.8= 1,17s :time for maximum height

at t=0.743 sthe ball is going up ,then g is negative

y=v₀y*t - 1/2 *g¨*t²

y=11.44*0.743 -1/2*9.8*0.743²

y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away

3. s = (1t3 + -5t2 + 3) m

v=3t²-10t=3*25-50=75-50=25m/s

at t=0, s=3 m

at t=5s s=5³-5*5²+3

4.  a = (-9t) m/s2

a=dv/dt=-9t

dv=-9tdt

v=∫ -9tdt

v=-9t²/2 + C1 equation (1)

in t=0  , v₀=26m/s ,in the equation (1) C1= 26

v=-9t²/2 + 26=ds/dt

ds=( -9t²/2 + 26)dt

s= ∫( -9t²/2 + 26)dt

s= -9t³/6+26t+C2 Equation 2

t = 0, s = 0 , C2=0

s= (-9t³/6+26t ) m

s= (-1.5t³+26t ) m

5 0
3 years ago
Professional baseball pitchers deliver pitches that can reach the blazing speed of 100 mph (miles per hour). A local team has dr
Wittaler [7]

Answer:

A. ) K =126. 7 J

B. ) h= 91.1 m.

Explanation:

A)

  • Assuming no air resistance, once released by the pitcher, the speed must keep constant through all the trajectory, so the kinetic energy of the ball can be expressed as follows:

       K = \frac{1}{2}*m*v^{2}  =  \frac{1}{2}*0.142 kg*(42.24m/s)^{2} = 126.7 J (1)

B)

  • Neglecting air resistance, total mechanical energy must be the same at any point, so, if we choose the ground level as the zero reference level for the gravitational potential energy, and assuming that the ball attains this kinetic energy just before striking ground, this value must be equal to the gravitational potential energy just before be dropped, so we can write the following equality:

        U_{o} = K_{f} = 126. 7 J (2)

        ⇒ m*g*h = 126. 7 J

  • Solving for h, we get:

       h = \frac{K_{f}}{m*g} = \frac{126.7J}{0.1420kg*9.8m/s2} = 91.1 m (3)

4 0
3 years ago
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