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inessss [21]
3 years ago
12

If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it oscillates. Suppose

it is displaced 0.122 m from its equilibrium position and released with zero initial speed. After 0.810 s, its displacement is found to be 0.122 m 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:
o-na [289]3 years ago
5 0

Answer:

a. Amplitude = 0.244 meters

b. Period = 1.62 seconds

c. Frequency = 0.6173 Hz

Explanation:

a.

With the position goes from 0.122 meters to -0.122 meters (negative because it is in the opposite side of the equilibrium point), the amplitude is the maximum value minus the minimum value:

Amplitude = 0.122 - (-0.122) = 0.122 + 0.122 = 0.244 meters

b.

The period is the amount of time the object takes to arrive in the same position again. So, if it takes 0.81 seconds to go to -0.122 m, it will take another 0.81 seconds to come back to 0.122 m, so the period is the sum of these two times:

Period = 0.81 + 0.81 + 1.62 seconds

c.

The frequency of the movement is the inverse of the period:

Frequency  = 1 / Period

So if the period is 1.62 seconds, the frequency is:

Frequency = 1 / 1.62 = 0.6173 Hertz

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zubka84 [21]

50m

Explanation:

Displacement is the length of path traveled which is measured from start to the finishing of the path.

    Analysis of the journey;

 Starts from:

 0                                    30m                                              from right

                  15m                                                                       to left

                                                                       50m              to right

                 

The displacement is 50m from the starting point.

Distance is total path traveled and for this problem it is 30+ 15 + 50 = 95m

learn more:

displacement brainly.com/question/5461768

#learnwithBrainly

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2 years ago
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Sever21 [200]
Genes. you inherit 50% from your mom and dad. I believe that is the answer....
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3 years ago
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4 0
3 years ago
A car with a mass of 850kg is moving at a speed of 72km/h when colliding with a concrete wall until it stops. After the collisio
Sergeeva-Olga [200]

Answer:

Explanation:

The vehicle is experiencing a large force created by the concrete wall.

Equation

vf^2 = vi^2 + 2*a * d

Givens

vf = 0   The car eventually does stop.

vi = 72 km/hr * [ 1000 m/  km] * [1 hour / 3600 seconds]

vi = 20 meters / second

a = ?

m = 850 kg

Solution

vf^2 = vi^2 + 2a*d

0 = 20 m/s + 2* 2 *a

-20 m/s = 4a

-20/4 =   a

a = - 5 m/s^2   The minus sign tells you the vehicle is slowing down. It sure should be.

Force = m * a

F = - 850 * (-5)

F =  - 4250 N

The car provides a 4250 N force on it going east to west and a 4250 N force going from west to east provided by the concrete wall.

8 0
3 years ago
A steel beam that is 5.50 m long weighs 332 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending
ElenaW [278]

Explanation:

The given data is as follows.

    Length of beam, (L) = 5.50 m

    Weight of the beam, (W_{b}) = 332 N

     Weight of the Suki, (W_{s}) = 505 N

After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

                 \sum M_{o} = 0

     -W_{s} \times x + W_{b} \times 1.5 = 0

                x = \frac{W_{b} \times 1.5}{W_{s}}

                   = \frac{332 N \times 1.5}{505 N}

                   = 0.986 m

Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.

Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.

8 0
3 years ago
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