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3 years ago

Helpppp!!!ASAP!! THANK YOUUUU!

Physics

1 answer:

mario62 [17]3 years ago

4 0

Answer:

F = 9.675Hz

Explanation:

pls for certain reasons let us make

wavelength = $
frequency = F
V = velocity

3 loops : 6$/4 = L

6$/4 = 2

$ = 4/3 = 1.333

V = F x $

F = V/$

F = 12.9/1.333 = 9.675Hz

F = 9.675Hz

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Is displacement a fundermetal unit or derived unit.

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Displacement is fundamental unit

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Which of the following accurately describes the way in which a muscle moves?

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3 years ago

Force F = (-6.0 N)ihat + (2.0 N) jhatacts on a particle with position vector r =(4.0 m) ihat+ (4.0 m) jhat.

finlep [7]

Answer:

$\tau=(-32k)\ N-m$

$\theta=116.55^{\circ}$

Explanation:

Given that.

Force acting on the particle, $F=(-6i + 2.0j)\ N$

Position of the particle, $r=(4i+4j)\ m$

To find,

(a) Torque on the particle about the origin.

(b) The angle between the directions of r and F

Solution,

(a) Torque acting on the particle is a scalar quantity. It is given by the cross product of force and position. It is given by :

$\tau=F\times r$

$\tau=(-6i + 2.0j)\times (4i+4j)$

$\tau=\begin{pmatrix}0&0&-32\end{pmatrix}$

$\tau=(-32k)\ N-m$

So, the torque on the particle about the origin is (32 N-m).

(b) Magnitude of r, $|r|=\sqrt{4^2+4^2}=5.65\ m$

Magnitude of F, $|F|=\sqrt{(-6)^2+2^2}=6.324\ m$

Using dot product formula,

$F{\circ}\ r=|F|.|r|\ cos\theta$

$cos\theta=\dfrac{F{\circ} r}{|F|.|r|}$

$cos\theta=\dfrac{-24+8}{6.324\times 5.65}$

$\theta=116.55^{\circ}$

Therefore, this is the required solution.

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3 years ago

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The uniform 100-kg beam is freely hinged about its upper end A and is initially at rest in the vertical position with theta = 0.

max2010maxim [7]

Answer:

ac = 2.86 m / s²

Explanation:

Image can detail the system to determine the force in the FA to understand the system into the applicated force

m = 100 kg , L = 3 m

∑ F = 0 ⇒ Ay - 100 kg + P * cos (45) = 0

Ay = 768.86 N

∑ Mₐ = α * I ₐ

I ₐ = m * L² / 3 ⇒ I ₐ = 100 kg * 4² m / 3

Replacing

P * sin (45) * 3 = α * 100 kg * 4² m / 3

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4 0

3 years ago

The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate

ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

$\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}$ ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

$\theta=\frac{3.7m}{0.162m}=22.83rad$

$22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}$

to calculate the angular speed w we can use $\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}$

Thus, wf=12.68rad/s

C) We can use our result in B)

$\alpha=3.52\frac{rad}{s^2}$

I hope this is useful for you

regards

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