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3 years ago

A _______ is a rhythmic movement that carries energy through space or matter.

1 answer:

7 0

The answer is wave.

A wave can be defined as a rhythmic flow that moves over a medium from one place to the different area.

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URGENT HELP IT IS DUE BY TONIGHT PLEASE HELP

Law Incorporation [45]

Answer:

45.8

Explanation:

becuse 9.8+36=45.8 simple

5 0

3 years ago

PLEASE HELP!!!!!!!!!!!

Alexxandr [17]

I am 100% sure its A.

5 0

3 years ago

Read 2 more answers

A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl

zysi [14]

Answer: $\alpha =9.302\ rad/s^2$

Explanation:

Given

mass of sphere $m=250\ gm$

diameter of sphere $d=4.30\ cm$

radius $r=\frac{4.30}{2}\ cm$

$f=0.0200\ N$

friction will provide resisting torque so

$f\times r=I\times \alpha$

where $I=\text{moment of Inertia}$

$f=\text{friction force}$

$\alpha =\text{angular acceleration}$

$I=\frac{2}{5}mr^2$

$0.02\times r=\frac{2}{5}mr^2\times \alpha$

$\alpha =\frac{5}{2r}\times f$

$\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02$

$\alpha =9.302\ rad/s^2$

(b)time taken to decrease its rotational speed by $21\ rad/s$

$t=\dfrac{\Delta \omega }{\alpha }$

$t=\dfrac{21}{9.302}$

$t=2.25\ s$

6 0

3 years ago

A block is at rest. The coefficients of static and kinetic friction are s = 0.81 and k = 0.69, respectively. The acceleration of

lys-0071 [83]

Answer:

Explanation:

a ) The angle required = angle of repose = θ

Tanθ = .81

θ = 39⁰

b ) when angle of incline θ = 44

Net force on the block = mg sinθ - μ mg cosθ where μ is coefficient of kinetic friction

acceleration = gsinθ - μ g cosθ

= 9.8 ( sin44 - μ cos44 )

= 9.8 ( .695 - .69 x .72 )

= 9.8 ( .695 - .497 )

= 1.94 m /s²

7 0

3 years ago

The decomposition of sulfuryl chloride (so2cl2) is a first-order process. the rate constant for the decomposition at 660 k is 4.

Natali5045456 [20]

<span>THIS IS A GAS PHASE REACTION AND WE ARE GIVE PARTIAL PRESSURES . I WRITE IN TERMS OF P RATHER THAN CONCENTRATION : lnPso2cl12=-kt+lnPso2cl1 initial partial pressure Pso2cl12 the rate constant k and the time t lnPso2cl12=(4.5*10-2*s-1)*65*s+ln (375) so lnPso2cl12=3.002 we take the base e antilog: lnPso2cl12=e3.002 Pso2cl12=20 torr we use the integrated first order rate lnPso2cl12=3.002=k*t+ lnPso2cl12=3.002 we use the same rate constant and initial pressure k=4.5*10-2*s-1 Pso2cl12=375 Pso2cl12=1* so2cl12 Pso2cl12=37.5 torr subtract in Pso2cl12 grom both side lnPso2cl12- lnPso2cl12=-kt ln(x)-ln(y)=ln (x/y) ln (Pso2cl12/Pso2cl20)=-kt we get t -1/k*ln(Pso2cl12/Pso2cl20)=t t=51 s</span>

6 0

4 years ago