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rusak2 [61]
3 years ago
6

Which song form has alternating repetitions of a main theme which two or more contrasting sections?

Physics
2 answers:
dlinn [17]3 years ago
5 0

Your answer would be A, Rando. (I did the quiz on K12 and thats what it said) hope it helps.

scoundrel [369]3 years ago
3 0
The answer to this item is rondo, letter A. As already mentioned in the above's definition, this has a recurring or repetitive lead them two contrasting sections. The contrasting sections are more commonly referred to as "episodes" and occasionally as "digressions" or "couplets". 
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The diagram is being used to illustrate the second law of thermodynamics, where Qh represents a hot object and Qc represents a c
katrin [286]

Answer:

The answer is A. on edgen.

Explanation:

A. adding in the boxes an arrow that points from Qh to Qc

6 0
3 years ago
Read 2 more answers
The tent meteorite, found in 1897 near cape York, on the west coast of Greenland, is the largest meteorite exhibited by any muse
sweet-ann [11.9K]
Answer:
3.0883 x 10^10mg

Explanation:
1 kilogram = 1000 000 milligrams
So, 30 883 x 1000 000 = 30 883 000 000mg
6 0
3 years ago
A cat jumps from 2.5-meter-tall bookshelf to a 1.3-meter-tall countertop. If the cat
UkoKoshka [18]

The change in Potential energy of the cat is 176.4 J.

<h3 /><h3>Potential Energy:</h3>

This is the energy due to the position of a body. The S.I unit is Joules (J)

The formula for change in potential energy.

<h3 /><h3>Formula:</h3>
  • ΔP.E = mg(H-h).............. Equation 1

<h3>Where:</h3>
  • ΔP.E = Change in potential energy
  • m = mass of the cat
  • g = acceleration due to gravity
  • H = First height
  • h = second height.

From the question,

<h3>Given:</h3>
  • m = 15 kg
  • H = 2.5 m
  • h = 1.3 m
  • g = 9.8 m/s²

Substitute these values into equation 1

  • ΔP.E = 15×9.8(2.5-1.3)
  • ΔP.E = 15×9.8×1.2
  • ΔP.E = 176.4 J.

Hence, The change in Potential energy of the cat is 176.4 J

Learn more about Potential energy here: brainly.com/question/1242059

5 0
2 years ago
Given: G = 6.672 × 10−11 N · m2 /kg2 Io, a satellite of Jupiter, has an orbital period of 1.24 days and an orbital radius of 4.1
Dahasolnce [82]

Answer:

Mass of Jupiter = 4.173×10^15kg

Explanation:

Using Kepler's 3rd law, it states that the orbital period T is related to the distance,r as:

T^2 = GM/4 pi × r^3

Where G = universal gravitational constant

r = radius

M = masd of jupiter

Rearranging the formular to make M the subject of formular

T^2 × 4 pi = G M × r^3

(T^2 × 4 pi) / (G× r^3) = M

(1.24^2 × 4 × 3.142) /(6.672×10^-11)(4.11×10^8)^3

M = 19.32 /6.672×10^-11)(4.11×10^8)^3

M = 19.32 / 4.63 ×10^15

M = 4.173×10^15kg

6 0
3 years ago
A spring stretches 3.5 cm when a 9 g object is hung from it. The object is replaced with a block of mass 26 g which oscillates i
evablogger [386]

Answer:

The period of motion  of new mass T = 0.637 sec

Explanation:

Given data

Mass of object (m) = 9 gm = 0.009 kg

Δx = 3.5 cm = 0.035 m

We know that spring force is given by

F = m g

F = 0.009 × 9.81 = 0.08829 N

Spring constant

k = \frac{F}{x}

k = \frac{0.08829}{0.035}

k = 2.522 \frac{N}{m}

New mass(m_1) = 26 gm = 0.026 kg

Now the period of motion is given by

T = 2 \pi \sqrt{\frac{m}{k} }

T = 2 \pi \sqrt{\frac{0.026}{2.522} }

T = 0.637 sec

This is the period of motion  of new mass.

3 0
3 years ago
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