Question 1

Rhythms that occur faster and slower than the beat are. Select one:. a. incorrect. b. not synchronized with the time signature.

zlopas [31]

Rhythms that occur faster and slower than the beat are b.<span>not synchronized with the time signature. The synchronization follows the same beat or rhythm. If the time signature say is lower than the original, then the rhythm should be faster. Otherwise, the rhythm is slower than the original one.</span>

4 0

2 years ago

At resonance, what is impedance of a series RLC circuit? less than R It depends on many other considerations, such as the values

denis23 [38]

Answer:

at resonance impedence is equal to resistance and quality factor is dependent on R L AND C all

Explanation:

we know that for series RLC circuit impedance is given by

$Z=\sqrt{R^2+\left ( X_L-X_C \}right )^2$

but we know that at resonance $X_L=X_C$

putting $X_L=X_C$ in impedance formula , impedance will become

Z=R so at resonance impedance of series RLC is equal to resistance only

now quality factor of series resonance is given by

$Q=\frac{\omega L}{R}=\frac{1}{\omega CR}=\frac{1}{R}\sqrt{\frac{L}{C}}$ so from given expression it is clear that quality factor depends on R L and C

3 0

2 years ago

Can someone help me pls

Dmitry_Shevchenko [17]

First question: 800J
Second question: 20.4m

4 0

2 years ago

An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get

Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

$mV_i = (m-m_O)V_f - m_OV_O$

where,

m=Total mass

$m_O =$ Mass of Object

$V_i =$ Velocity before throwing

$V_f =$ Final Velocity

$V_O =$ Velocity of Object

Our values are:

$m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg$

Solving to find the final speed, after throwing the object we have

$V_f=\frac{mV_0+m_TV_O}{m-m_O}$

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$

$V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}$

$V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}$

$V_{f1}=0.8511m/s$

B) $7.9Kg\rightarrow 11.2m/s$

$V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}$

$V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}$

$V_{f2} = 2.0173m/s$

C) $10.5Kg\rightarrow 7m/s$

$V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}$

$V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}$

$V_{f3} = 3.63478m/s$

Therefore the final velocity of astronaut is 3.63m/s

7 0

2 years ago

At high elevations or high latitudes, some of the water that falls on land does not immediately soak in , run off, evaporate or

LUCKY_DIMON [66]

Answer:

The water is stored in ice sheets and as snow

Explanation:

Temperature reduces with an increase in altitudes. The standard laps rate is 6.5°C per 1,000 m gained in elevation

At very high elevations, therefore, the air is usually very cold such that when an elevation of 4,500 meters is reached at the equator, it is possible to observe snowfall and the water remain temporarily stored on the surface of the mountain as ice and snow

7 0

2 years ago