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Murljashka [212]
3 years ago
12

What 3 things can you learn about energy in a roller coaster?

Physics
1 answer:
Alika [10]3 years ago
4 0

<u>Answer:</u>

The following are the things we can learn about energy in roller coaster.

  • All roller coasters work using the principle of gravitational energy.
  • All roller coaster utilise the Potential energy.
  • Roller coasters convert the potential energy into kinetic energy.
  • Friction is reduced in roller coasters as it decreases the speed of roller  coaster.

Roller coasters do not have engines whereas Gravity pulls them and potential energy gets converted into kinetic energy providing its motion. Due to the friction and air resistance, the energy gets reduced as the ride proceeds.

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Diferencia entre transferencia y transformación de energía. Da ejemplos de cada caso.
Firdavs [7]

Answer:

La transformación de energía es un proceso en el que la energía se intercambia entre un sistema y el medio ambiente en al menos dos formas de energía diferentes entre sí. Por ejemplo, un panel solar convierte la energía lumínica en energía eléctrica.

En cambio, en la transferencia de energía, esta no cambia su forma sino que es transmitida de un cuerpo a otro. El ejemplo más claro es el de la fogata, que transmite calor al medio ambiente a través de radiación.

6 0
3 years ago
. An object whose mass is 375 lb falls freely under the influence of gravity from an initial elevation of 253 ft above the surfa
lozanna [386]

Answer:

(a) Vf = 128 ft/s

(b) K.E = 122.8 Btu

Explanation:

(a)

In order to find the velocity of the object just before striking the surface of earth or the final velocity, we use 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = 32.2 ft/s²

h = height = 253 ft

Vf = Final Velocity = ?

Vi = Initial Velocity = 10 ft/s

Therefore,

(2)(32.2 ft/s²)(253 ft) = Vf² - (10 ft/s)²

16293.2 ft²/s² + 100 ft²/s² = Vf²

Vf = √(16393.2 ft²/s²)

<u>Vf = 128 ft/s</u>

<u></u>

(b)

The kinetic energy of the object before it hits the surface of earth is given by:

K.E = (0.5)(m)(Vf)²

where,

m = mass of object = 375 lb

K.E = Kinetic energy of object before it strikes the surface of earth = ?

Therefore,

K.E = (0.5)(375 lb)(128 ft/s)²

K.E = 3073725 lb.ft²/s²

Now, converting this to Btu:

K.E = (3073725 lb.ft²/s²)(1 Btu/25037 lb.ft²/s²)

<u>K.E = 122.8 Btu</u>

3 0
3 years ago
A dog sledding team is composed of a 82.8 kg musher (the person driving the sled), a 21.4 kg sled, and four dogs. Assume that ea
vlabodo [156]

Answer: 340.8W

Explanation: Please see the attachments below

6 0
3 years ago
Read 2 more answers
A basketball player shoots toward a basket 5.3 m away and 3.0 m above the floor. If the ball is released 1.9 m above the floor a
Snezhnost [94]

Answer:

Vi = 8.28 m/s

Explanation:

This problem is related to the projectile motion.

As we know there are two components of motion associated with this, the horizontal component and vertical component.

The horizontal distance covered by the ball is

Vx*t = x

Vx*t = 5.3

Vx = 5.3/t  eq. 1

Also we know that

Vx = Vicos(60)

Vx = Vi*0.5  eq. 2

equate eq. 1 and eq. 2

5.3/t = Vi*0.5

5.3/0.5 = Vi*t

Vi*t = 10.6  eq. 3

The vertical distance is

Vy = y1 + Vyi*t - 0.5gt²

also we know that

Vyi = Visin(60)

Vyi = Vi*0.866

It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance

3 = 1.9 + Vi*0.866*t - 0.5gt²

3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²

3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²

3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²

1.1 = 0.866(Vi*t) - 4.9t²

0.866(Vi*t) = 4.9t² + 1.1

substitute Vi*t = 10.6 in above equation

0.866(10.6) = 4.9t² + 1.1

9.18 = 4.9t² + 1.1

4.9t² = 8.08

t² = 8.08/4.9

t² = 1.648

t = 1.28 sec

Finally, initial speed can be found by substituting the value of t into eq. 3

Vi*t = 10.6

Vi = 10.6/t

Vi = 10.6/1.28

Vi = 8.28 m/s

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Answer:

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Explanation:

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3 years ago
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