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4 years ago

9

How does a roller coaster convert potential energy to kinetic energy and then back to potential energy again and again?

1 answer:

andrey2020 [161]4 years ago

5 0

When the coaster ascends one of the smaller hills that follows the initial lift hill, its kinetic energy changes back to potential energy. In this way, the course of the track is constantly converting energy from kinetic to potential and back again.

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A rollercoaster accelerates from 10 m/s to 100 m/s2 for 25 seconds. What is the acceleration?

UkoKoshka [18]

Answer:

A roller coasters accelerates from an initial velocity of of 6.0 m/s to a final velocity of 70 m/s over 4 seconds. What's the acceleration? Q. Acceleration only takes place when things speed up. Q. A drag racer accelerated from 0 m/s to 200 m/s in 5 s.

Explanation:

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3 years ago

Answer the following. (a) What is the surface temperature of Betelgeuse, a red giant star in the constellation of Orion, which r

bagirrra123 [75]

Answer:

(a) T = 2987.6 k

(b) T = 19986.2 k

Explanation:

The temperature of a star in terms of peak wavelength can be given by Wein's Displacement Law, which is as follows:

$T = \frac{0.2898\ x\ 10^{-2}\ m.k}{\lambda_{max}}$

where,

T = Radiated surface temperature

$\lambda_{max}$ = peak wavelength

(a)

here,

$\lambda_{max}$ = 970 nm = 9.7 x 10⁻⁷ m

Therefore,

$T = \frac{0.2898\ x\ 10^{-2}\ m.k}{9.7\ x\ 10^{-7}\ m}$

<u>T = 2987.6 k</u>

(b)

here,

$\lambda_{max}$ = 145 nm = 1.45 x 10⁻⁷ m

Therefore,

$T = \frac{0.2898\ x\ 10^{-2}\ m.k}{1.45\ x\ 10^{-7}\ m}$

<u>T = 19986.2 k</u>

6 0

3 years ago

A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250

Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30$

coefficient of static friction $\mu =0.35$

coefficient of kinetic friction $\mu _k=0.25$

distance traveled $d=77.3 cm$

spring constant $k=35 N/m$

work done by gravity+work done by friction=Energy stored in Spring

$mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}$

$mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}$

$0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}$

$x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}$

$x=0.234 m$

$x=23.4 cm$

6 0

4 years ago

A uniform marble rolls down a symmetric bowl, starting from rest atthe top of the left side. The top of each side is a distanceh

Debora [2.8K]

Answer:

A. 5/7h

B. Same height

C. See attachment below

Explanation:

Find the attachment below for better understanding.

7 0

3 years ago

A person walks first at a constant speed of 5.50 m/s along a straight line from point A to point B and then back along the line

Sav [38]

Answer:

4.25 m/s

Explanation:

They walked the first distance at 5.50 m/s, then the same distance at 3 m/s.

Since the distances are equal, the average speed is simply the average of 5.50 and 3.

(5.50 + 3) / 2 = 4.25

Her average speed over the entire trip is 4.25 m/s.

8 0

4 years ago