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2 years ago

Describing Energy Transformations

Physics

1 answer:

UNO [17]2 years ago

4 0

Answer:

chemical energy to electrical energy to sound energy to heat energy

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a) A Kuiper Belt Object orbits the Sun with an average orbital distance of 40 AU. What is its orbital period? Round your answer

denpristay [2]

Answer:

253 years

Explanation:

a = Distance in Astronomical units from the sun = 40 AU

P = Orbital period in years

From Kepler's law we have

$P^2=a^3$

$\Rightarrow P^2=40^3$

$\Rightarrow P^2=64000$

$\Rightarrow P=\sqrt{64000}$

$\Rightarrow P=252.982212813$

$\Rightarrow P\approx 253\ years$

The orbital period of the Kuiper belt object is 253 years.

7 0

2 years ago

Classical conditioning requires _____.

LekaFEV [45]

I would say A classical conditioning definitely demands a pairing of two stimuli
hope i helped

5 0

2 years ago

Read 2 more answers

if you were floating in the solar system at equal distances between mars (small mass) and jupiter (large mass) which planet woul

tester [92]

You would gravitate towards Jupiter because if it’s large mass it has a stronger gravitational pull

3 0

2 years ago

Electrical wire with a diameter of .5 cm is wound on a spool with a radius of 30 cm and a height of 24 cm.

kow [346]

Answer:

a) # lap = 301.59 rad , b) L = 90.48 m

Explanation:

a) Let's use a direct proportions rule (rule of three). If one turn of the wire covers 0.05 cm, how many turns do you need to cover 24 cm

# turns = 1 turn (24 cm / 0.5 cm)

# laps = 48 laps

Let's reduce to radians

# laps = 48 laps (2 round / 1 round)

# lap = 301.59 rad

b) Each lap gives a length equal to the length of the circle

L₀ = 2π R

L = # turns L₀

L = # turns 2π R

L = 48 2π 30

L = 9047.79 cm

L = 90.48 m

6 0

3 years ago

Find the fundamental frequency and the next three frequencies that could cause standing-wave patterns on a string that is 30.0 m

maksim [4K]

Answer:

0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz

Explanation:

The fundamental frequency of a standing wave on a string is given by

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

L is the length of the string

T is the tension in the string

$\mu$ is the mass per unit length

For the string in the problem,

L = 30.0 m

$\mu=9.00\cdot 10^{-3} kg/m$

T = 20.0 N

Substituting into the equation, we find the fundamental frequency:

$f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz$

The next frequencies (harmonics) are given by

$f_n = nf$

with n being an integer number and f being the fundamental frequency.

So we get:

$f_2 = 2 (0.786 Hz)=1.572 Hz$

$f_3 = 3 (0.786 Hz)=2.358 Hz$

$f_4 = 4 (0.786 Hz)=3.144 Hz$

6 0

3 years ago