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Bond [772]
2 years ago
6

What causes a roller coaster to "lose" energy and stop? What force does it lose energy to? Give a detailed explanation that is e

asy to understand.
Physics
1 answer:
Zarrin [17]2 years ago
5 0

Roller coasters work by utilizing its potential energy. Potential energy is "lost" as the cars lose height, subsequently gaining kinetic energy as revealed by increased speeds (and loop the loop). As the cars of the roller coaster course through the changing heights, it constantly swaps between potential and kinetic energy. Theoretically, this process could be endless. However, energy is continually "lost" because of dissipative forces such as friction and air resistance.

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A block of 200 g is attached to a light spring with a force constant of 5 N / m and freely in a horizontal plane vibrates. The m
Finger [1]

Answer:

m  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 mm  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 m

Explanation:

That is a reason

8 0
2 years ago
A tiger travels 3m/s^2 for 4.1s,what was its initial speed if it's final speed was 55k/h?
zaharov [31]
Acceleration a=3m/s^2
time t= 4.1seconds
Final velocity V= 55km/h
initial velocity U= ?
First convert V to m/s
36km/h=10m/s
55km/h= 55*10/36=15.28m/s
Using the formula V= U+at
U= V-at
U= 15.28-3*4.1=15.28-12.3=2.98m/s
Initial velocity U= 2.98m/s or 10.73km/h (Using the conversion rate 36km/h=10m/s)
8 0
3 years ago
Ninais ba ni floranteng mapalapit ang kanyang loob kay adolfo
luda_lava [24]

Answer:

?

Explanation:

8 0
1 year ago
A long, thin solenoid has 450 turns per meter and a radius of 1.06 . The current in the solenoid is increasing at a uniform rate
kirill115 [55]

Answer:9.34 A/s

Explanation:

Given

radius of solenoid R=1.06 m

Emf induced E=8.50\times 10^{-6} V/m

no of turns per meter n=450

we know Induced EMF is given by

\int Edl=-\frac{\mathrm{d} \phi}{\mathrm{d} t}=-\frac{\mathrm{d} B}{\mathrm{d} t}A

Magnetic Field is given by

B=\mu _0ni

thus \frac{\mathrm{d} B}{\mathrm{d} t}=-\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}

Area of cross-section

A=\pi R^2 where

solving integration we get

E.\cdot 2\pi r=\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}\pi R^2

where r=distance from axis

R=radius of Solenoid

\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{Er}{\mu _0nR^2}

\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{8.50\times 10^{-6}\times 3.49\times 10^{-2}}{4\pi \times 10^{-7}\times 450\times 1.06^2}

\frac{\mathrm{d} i}{\mathrm{d} t}=9.34 A/s

4 0
3 years ago
A student performs an experiment in measuring the period of a simple pendulum of known length 49.0 cm.He performed five trials a
Levart [38]
Correct Answer is Bb
4 0
3 years ago
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