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Citrus2011 [14]
3 years ago
5

1. If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released. It will oscillate. I

f is displaced 0.120m from its equilibrium position and released with zero initial speed, then after 0.800s its displacement is found to be 0.120m on the opposite side, and it has passed the equilibrium position once during this interval. Find, a) the amplitude; b) the period; c) the frequency.
Physics
1 answer:
AURORKA [14]3 years ago
3 0

Answer: a) 0.12m; b) 1,6 s; c) 0.625 1/s

Explanation: The simple harmonic movement can be described by a sin or cosine function  in time.

This can be in the form:

X(t)= A Sin/Cos (wt+φ) where φ is initial phase o position at t=0

w the angular frequency are related to the frequency (f) as 2Pif

and f=1/T period of oscillating

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A driver in a 2000 kg Porsche wishes to pass to pass a slow-moving school bus on a four-lane road. What is the average power in
creativ13 [48]

Answer:

The power require to accelerate the car is, P = 299700 watts

Explanation:

Give data,

The mass of the car, m = 2000 kg

The initial velocity of the sports car, u = 30 m/s

The final velocity of the sports car, v = 60 m/s

The time period of acceleration, t = 9 s

The acceleration of the car, a =  (v-u) / t

                                                  = (60 - 30) / 9

                                                  = 3.33 m/s²

The displacement of the car,

                                               S = ut + ½ at²

                                                  = 30 x 9 + ½ x 3.33 x 9²

                                                  = 405 m

The force acting on the car, F = m x a

                                                  = 2000 x 3.33

                                                  = 6660 N

The work done by the car, W = F  S

                                                  = 6660 x 405

                                                  = 2697300 J

The power of the car,           P = W / t

                                                  = 2697300 / 9

                                                  = 299700 watts

Hence, the power require to accelerate the car is, P = 299700 watts

4 0
4 years ago
The manometer shown in fig. 2 contains water and kerosene. with both tubes open to the atmosphere, the free-surface elevations d
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The solution for this problem is: In the figure, you now know that total length of the kerosene column
So at x – xPatm + Pkg(H0 th) = Pa + Pwgh
Now H0 + h = 20 + 91.1 mm = 111.1 mm
Therefore = Pkg 0.1111 – P2g= h = 56 x 0.111 – 98 / 1000 x 9.81= 0.081 m or 81 mn
Therefore H0 = 111.1 - 81= 30.1 mm
6 0
3 years ago
A 2000 kg truck is traveling at 25 m/s. Determine the impulse needed
Vitek1552 [10]

Answer:

A 7kg impulse

Explanation:

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3 0
3 years ago
3. If a pipe with flowing water has a cross-sectional area nine times greater at point 2 than at point 1, what would be the rela
shusha [124]

Answer:

The velocity of the flow at point 1 is nine times greater than the velocity at point 2.

Explanation:

We know that the same volume of water entering through point 1 must exit through point 2.

The volume of water per unit of time or the volumetric flow is defined by:

V_{flow}=v*A\\ where:\\V_{flow}= flow [m^3/s]\\v=velocity[m/s]\\A=area [m^2]

If  A_{2}=9*A_{1}  \\

therefore

v_{1} *A_{1} = v_{2} *A_{2} \\v_{1} *A_{1} = v_{2} *9A_{1} \\\\v_{1} = 9v_{2}  \\

4 0
4 years ago
To understand the relationships between the parameters that characterize a wave. It is of fundamental importance in many areas o
Monica [59]

Answer:

1) c.

2) a.)

Explanation:

1)

  • At any wave, if its waveform repeats itself every time interval T, it is said that the wave is periodic, with a period T, which is the time needed to complete an entire cycle. The other options refer at the way in the waves propagates (longitudinal or transversal) and to the type of waveform (sinusoidal), so the right answer is c).

2)

  • At any wave that propagates at a constant speed, there exists a fixed relationship between the velocity v, the frequency  f and the wavelength λ, as follows:

       v = \lambda * f   (1)

  • So in order to v keep constant, if the frequency is increased, the wavelength will decrease in the same proportion, so a) is the right answer.
7 0
3 years ago
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