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12345 [234]
3 years ago
11

block is attached to an oscillating spring. The function below shows its position (cm) vs. time (s). What is the angular frequen

cy ( ω ω ) of oscillation? x ( t ) = 1.5 cos ( 20 t ) x(t)=1.5cos⁡(20t)
Physics
1 answer:
faltersainse [42]3 years ago
7 0

Answer:

Angular frequency is 20 rad/s.      

Explanation:

Given that,

A block is attached to an oscillating spring. The function below shows its position (cm) vs. time (s) is given by :

x(t)=1.5\cos(20\ t).....(1)

The general equation of oscillating particle is given by :

x(t)=A\cos(\omega t).......(2)

Compare equation (1) and (2) we get :

\omega=20\ rad/s

So, the angular frequency of the oscillation is 20 rad/s.

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(TCO 4) A signal consists of only two sinusoids, one of 65 Hz and one of 95 Hz. This signal is sampled at a rate of 245 Hz. Find
storchak [24]

Answer:

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

Explanation:

Given :

Frequencies of the sinusoids,

$f_{m_1}= 65 \ Hz$ ,  and

$f_{m_2}= 95 \ Hz$

Sampling rate f_s = \ 245 \ Hz

The positive frequencies at the output of the sampling system are :

$f_{o_1}=\pm f_{m_1} \pm nf_s, f_{o_2}=\pm f_{m_2} \pm nf_s $

When n = 0,

$f_{o_1}= f_{m_1} = 65 \ Hz,\ \  f_{o_2}= f_{m_2} = 95 \ Hz $

when n  = 1,

$f_{o_1}=\pm f_{m_1} \pm f_s, \ \ f_{o_2}=\pm f_{m_2} \pm  f_s $

$f_{o_1}= \pm 65 \pm 245,\ \  f_{o_2}=\pm 95 \pm 245$

$f_{o_1}= 180 \ Hz, 310 \ Hz,\ \  f_{o_2}= 150 \ Hz,340 \ Hz$

When n = 2,

$f_{o_1}= \pm 65 \pm 2(245),\ \  f_{o_2}=\pm 95 \pm 2(245)$

$f_{o_1}= 555 \ Hz, 425 \ Hz,\ \  f_{o_2}= 395 \ Hz,585 \ Hz$

Therefore, the first six positive frequencies present in the replicated spectrum are :

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

8 0
3 years ago
Which range is the approximate audible range for humans?
allsm [11]

Answer:

a

Explanation:

bcz mre than that they will be affected

8 0
3 years ago
Read 2 more answers
Determine the critical crack length for a through crack contained within a thick plate of 7150-T651 aluminum alloy that is in un
Mila [183]

Explanation:

Formula to determine the critical crack is as follows.

          K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}

  \gamma = 1,     K_{IC} = 24.1

  [/tex]\sigma_{y}[/tex] = 570

and,   \sigma_{f} = 570 \times \frac{3}{4}

                       = 427.5

Hence, we will calculate the critical crack length as follows.

      a = \frac{1}{\pi} \times (\frac{K_{IC}}{\sigma_{f}})^{2}

        = \frac{1}{3.14} \times (\frac{24.1}{427.5})^{2}

       = 10.13 \times 10^{-4}

Therefore, largest size is as follows.

            Largest size = 2a

                                 = 2 \times 10.13 \times 10^{-4}

                                 = 20.26 \times 10^{-4}

Thus, we can conclude that the critical crack length for a through crack contained within the given plate is 20.26 \times 10^{-4}.

4 0
3 years ago
Determine the automobile’s braking distance from 90 km/h when it is going up a 5° incline. (Round the final value to one decimal
NeX [460]

Answer:

366 m

Explanation:

u = 90 km/h = 25 m/s,

theta = 5 degree

acceleration, a = g Sin theta = 9.8 x Sin 5 = 0.854 m/s^2

The final velocity os zero and let the braking distance be s.

Use third equation of motion

v^2 = u^2 - 2 a s

0 = 25 x 25 - 2 x 0.854 x s

s = 366 m

8 0
3 years ago
Ok sooo I need help I don’t understand this lol
Burka [1]

Answer:

I dont either but I'm sorry and hopefully you figured it out

3 0
3 years ago
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