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alexdok [17]
3 years ago
14

When are the displacement and acceleration equal to zero for the motion of a mass on a spring

Physics
2 answers:
sveta [45]3 years ago
7 0
When the spring is compressed all the way down.hope this helps
emmasim [6.3K]3 years ago
7 0
At minimum displacement 
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The Zero Gravity Research Facility at the NASA Glenn Research Center includes a
denis23 [38]

Answer: (a) t = 5.44 sec

(b) vf = 53.31 m/s

(c) s = 5.0m

Explanation: from the question, given data

the Height of the tower, h = 145m

from question

(a)

the initial velocity, v₁ = 0 m/s

s = v₁t + 1/2 gt²

-145 m = 0(t) + 1/2 (-9.8t²)

  t² = 145/4.9

   t² = 29.59

    t = 5.44 sec

(b)

the speed of the sphere at the bottom of the tower is

vf² = vi² +2as

vf² = 0 + 2(-9.8 × -145)

vf² = 2842

vf = 53.31 m/s

(c)

when caught, the sphere experiences a deceleration of;

   a = -29.0g

the time it would take to decelerate becomes;

vf = vi + at

0 = (53.31) + (-29 ×9.8)t

where t = 53.31 / 284.2

t = 0.1876 sec

∴ the distance travelled during the deceleration becomes;

vf² = vi² + 2as

s = (vf² - vi²) / 2a

s = (0 - 53.31²) / 2×-29×9.8

s = -2841.9561 / -568.4

s = 4.99 ≈ 5.0m

i hope this helps, cheers

4 0
3 years ago
The moon orbits the earth once every 27 days at a distance of 384400 km. The international space station orbits the earth at an
Leno4ka [110]

Answer:

ive never done this befor but i think its 36 times aroud earth a day

Explanation:

384400/400=961  

961% of 27 days is 40minuets

1440/40=36

7 0
3 years ago
A 45.2-kg person is on a barrel ride at an amusement park. She stands on a platform with her back to the barrel wall. The 3.74-m
elena-14-01-66 [18.8K]

Answer:

  • <u><em>1,230N</em></u>

Explanation:

<u>1. Name of the variables:</u>

   f:frequency\\\\ \omega:angular\text{ }speed\\\\ a_c:centripetal\text{ }acceleration\\\\ F_c:centripetal\text{ }force\\ \\ m:mass\\ \\ d:diameter\\ \\ r:radius\\ \\ g:gravitational\text{ }acceleration

<u>2. Formulae:</u>

         f=\dfrac{number\text{ }of\text{ }revolutions}{time}

          \omega=2\pi f

          a_c=\omega^2 r

           F_c=m\times a_c

<u>3. Solution (calculations)</u>

       f=\dfrac{1}{1.65s}=0.\overline{60}s^{-1}

       \omega=2\pi\times0.\overline{60}\approx 3.808rad/s

      a_c=(3.808rad/s)^2\times (3.74/2m)=27.12m/s^2

      F_c=45.2kg\times27.12m/s^2=1,225.67N\approx 1,230N

3 0
3 years ago
A block of mass m is pushed a horizontal distance D from position A to position B, along a horizontal plane with friction coeffi
Wewaii [24]

Answer:

The total work done by friction is -2 · μ · m · g · D

Explanation:

Hi there!

The work done by a force is calculated as follows:

W = F · d · cos θ

Where:

W = work.

F = force that does the work.

d = displacement.

θ = angle between the displacement and the force.

If the force is horizontal, as in this case, cos θ = 1

The friction force is calculated as follows:

Ffr = μ · m · g

Where:

μ = friction coefficient.

m = mass of the object.

g = acceleration due to gravity.

Then, in this case, the work done by friction when pushing the block from A to B will be:

W AB = -Ffr · D

W AB = - μ · m · g · D

Notice that the friction force is negative because it is opposite to the pushing force P.

When the block is pushed from B to A, the work done by friction will be:

W BA = Ffr · (-D)

W BA = -μ · m · g · D

Now, the displacement is negative and the friction force is positive (in the opposite direction to -P).

The total work done by friction will be:

W AB + W BA = - μ · m · g · D  - μ · m · g · D  = -2 μ · m · g · D

5 0
3 years ago
If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular
erastova [34]

Answer;

D. The car would begin to move in the direction it was headed in a straight line.

Explanation;

-Centripetal force is any net force causing uniform circular motion. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration.

-The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.

-Therefore,If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular path were reduced then the car would begin to move in the direction it was headed in a straight line.

6 0
4 years ago
Read 2 more answers
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