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4 years ago

A dotted half note has beats when the time signature is 4/4

Physics

2 answers:

levacccp [35]4 years ago

8 0

A dotted half note has 3 beats in 4/4

AfilCa [17]4 years ago

3 0

There is 2 beats on a half note

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A uniform meter stick (with a length of 1.00 meter) has a mass of 108 g. It is supported at its midpoint by a vertical rigid blu

alexandr1967 [171]

Answer:4

Explanation:

Given

$m_1=21\ gm$

$m_2=27\ gm$

Mass of stick is $m=108\ gm$

Let T be the tension in the red string

Now if the red string is cut , suppose T is the tension in the blue rod immediately after cut

Therefore

$T=m_1g+m_2g$

$T=(0.021+0.027)\times 10$

$T=0.48\ N$

4 0

4 years ago

A pitcher hurls a 0.25kg ball around a vertical circular path of radius 0.6 m, applying a tangential force of 30N, before releas

Sever21 [200]

Thank you for posting your question here at brainly. Below is the solution:

Ke up top = 1/2*.25 *225
gain of Pot energy = .25*9.81*1.2 
work input = (1/4)(2 pi *.6)*30 

so 
sum of those 3 energies = 
(1/2)(.25)v^2

3 0

4 years ago

A charged particle is accelerated in a uniform electric field. When its velocity is 2 m/s, its electric potential energy is 100

zavuch27 [327]

Answer:

particle's potential energy = 70J

Explanation:

From conservation of energy; K1 + Ue1 = K2 + Ue2

where K1 and K2 are the kinetic energies at two positions and Ue1 and Uue2 are the electrical potential energies at two positions.

k1 = 10J, Ue1 = 100J

K2 = 40J

substitute into K1 + Ue1 = K2 + Ue2

Ue2 = K1 + Ue1 - K2

= 10 +100 - 40

Ue2 = 70J

7 0

3 years ago

What term is used to describe the 4 in the expression 4 Ca(NO³)²

NISA [10]

Hello,

The 4 describes the amount of calcium in that compound.

Have A Good Day.

3 0

3 years ago

At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constan

Jet001 [13]

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time $t_1$ , the acceleration vector has direction $\theta_1$ such that

$\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ$

which indicates the particle is situated at a point on the lower left half of the circle, while at time $t_2$ the acceleration has direction $\theta_2$ such that

$\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ$

which indicates the particle lies on the upper left half of the circle.

Notice that $\theta_1-\theta_2=90^\circ$ . That is, the measure of the major arc between the particle's positions at $t_1$ and $t_2$ is 270 degrees, which means that $t_2-t_1$ is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

$\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}$

where $R$ is the radius of the circle and $T$ is the period. We have

$t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s$

and the magnitude of the particle's acceleration toward the center of the circle is

$\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}$

So we find that the path has a radius $R$ of

$7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}$

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3 years ago