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3 years ago

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 container contains

monitta

What is the question

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2 years ago

The gap between electrodes in a spark plug is 0.060 cm. Producing an electric spark in a gasoline-air mixture requires an electr

VladimirAG [237]

The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V

<u>Explanation:</u>

Given data,

E= 3 ×10 ⁶ Δx=0.06/100

We have to find the minimum potential difference

E= -ΔV/Δx

ΔV=- E × Δx

ΔV =-3 ×10 ⁶ . 0.06/100

ΔV=-1800 V

The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V

6 0

3 years ago

a certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigg

IrinaK [193]

Answer:

f = 1 m

Explanation:

The magnification of the lens is given by the formula:

$M = \frac{q}{p}$

where,

M = Magnification = 4

q = image distance = 5 m

p = object distance = ?

Therefore,

$4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m$

Now using thin lens formula:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\$

6 0

2 years ago

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of

wariber [46]

Answer:

For left = 0 N/C

For right = 0 N/C

At middle = $-7.6836 * 10^{-11} \vec{i}$ N/C

Explanation:

Given data :-

б = $6.8 * 10^{-22}$ C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

$E = \frac{\sigma }{2\varepsilon }$

1) To the left of the plate

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

2) To the right of them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

3) Between them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(-\vec{i})$ = $(\frac{\sigma }{\varepsilon })(-\vec{i})$ = $-\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} } \vec{i}$ = $-7.6836 * 10^{-11} \vec{i}$ N/C

5 0

3 years ago

Cole is riding a sled with initial speed of 5 m/s from west to east. the frictional force of 50 n exists due west. the mass of t

stepan [7]

We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
$F=ma$
where $F=-50 N$ is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
$a= \frac{F}{m}= \frac{-50 N}{100 kg}=-0.5 m/s^2$

Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
$a= \frac{v_f^2-v_i^2}{2d}$
where
$v_f=0$ is the final speed of the sled
$v_i=5 m/s$ is the initial speed
$d$ is the distance covered

By rearranging the equation, we find d:
$d= \frac{v_f^2-v_i^2}{2a}= \frac{-(5 m/s)^2}{2 \cdot (-0.5 m/s^2)}=25 m$

3 0

3 years ago