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3 years ago

What is a Rest in music?

SAT

2 answers:

Anuta_ua [19.1K]3 years ago

4 0

Answer:

A rest is a musical notation causing you to 'rest' for one beat

Explanation:

mafiozo [28]3 years ago

4 0

DescriptionA rest is a musical notation sign that indicates the absence of a sound.

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Which part of this chemical process requires energy?

Radda [10]

Answer:

Endothermic reactions.

Explanation:

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2 years ago

Panthers and leopards prey on deer in an ecosystem. For unknown reasons, the population of leopards increased drastically. What

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2 or 3 because, less food means more fighting to get the remaining

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How many days do I have to wait to know my FSA score?

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A bicycle wheel is mounted on a fixed, frictionless axle, with a light string wound around its rim. The wheel has moment of iner

xz_007 [3.2K]

The final rotational speed ω_final and the instantaneous power P delivered to the wheel are; ω_f = √((ω_i)² + 2(FL/(kmr²) and P = Frω_i

<h3>What is the Instantaneous Power?</h3>

A) From rotational kinematics, the formula for the final angular velocity is;

ω_f = √((ω_i)² + 2αθ)

where;

α is angular acceleration

θ = L/r. Thus;

ω_f = √((ω_i)² + 2α(L/r))

Now, α = T/I

Where;

I is moment of inertia = k*m*r²

T is t o r q u e = F * r

Thus;

α = (F * r)/(kmr²)

α = F/(kmr)

ω_f = √((ω_i)² + 2(F/(kmr))(L/r))

ω_f = √((ω_i)² + 2(FL/(kmr²)

B) Formula for instantaneous power is;

P = Fv

where at t = 0; v = rω_i

Thus;

P = Frω_i

Read more about Instantaneous Power at; brainly.com/question/14244672

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2 years ago

suppose you push a hockey puck of mass m across frictionless ice for a time 1.0 s, starting from rest, giving the puck speed v a

saul85 [17]

Answer: Twice the previous time would be taken to reach the same speed v with the puck of mass 2m.

Explanation:

Let a Force pushes the hockey puck of mass m.

Then acceleration, a= \frac{F}{m}a=mF

From the equation of motion,

\begin{gathered}\➪ v=u+at\\ v=0+\frac{F}{m}\Delta t\end{gathered}⇒v=u+atv=0+mFΔt ......(1)

In the second case, when mass is 2m, then acceleration,

a'=\frac{F}{2m}a′=2mF

and t' is the time taken.

The final speed is v,

\begin{gathered}\➪ v=0+ a't'\\ \➪ \frac {F}{m}\Delta t=\frac{F}{2m}t'\\ \➪ t'= 2\Delta t\end{gathered}⇒v=0+a′t′⇒mFΔt=2mFt′⇒t′=2Δt using equation (1)

Hence, it would take two times the previous amount of time to push the pluck of double mass.

 I hope this is understandable and if my answer helped you please can you give me a brainliest its will really help me and thank you have a nice day!!

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3 years ago

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