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saveliy_v [14]
4 years ago
6

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

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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 
<span>gain of Pot energy = .25*9.81*1.2 </span>
<span>work input = (1/4)(2 pi *.6)*30 </span>

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

8 0
3 years ago
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