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vova2212 [387]
3 years ago
6

When hung from an ideal spring with spring constant k = 1.5 N/m, it bounces up and down with some frequency ω, if you stop the b

ouncing and let it swing from side to side through a small angle, the frequency of the pendulum is half the bounce frequency ω/2. What is the length of the un–stretched spring l =? Note ω should not be part of your answer.
Physics
1 answer:
Umnica [9.8K]3 years ago
7 0

Answer:

L = ¼ k g / m

Explanation:

This is an interesting exercise, in the first case the spring bounces under its own weight and in the second it oscillates under its own weight.

The first case angular velocity, spring mass system is

    w₁² = k / m

The second case the angular velocity is

    w₂² = L / g

They tell us

    w₂ = ½ w₁

Let's replace and calculate

     √ (L / g) = ½ √ (k / m)

      L / g = ¼ k / m

       

      L = ¼ k g / m

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Consider a uniform horizontal wooden board that acts as a pedestrian bridge. The bridge has a mass of 300 kg and a length of 10
gayaneshka [121]

Answer:

F = 2123.33N

Explanation:

In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.

\Sigma \tau=0

Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.

Then, you obtain the following formula:

-\tau_l+\tau_p+\tau_{cm}=0          (1)

τl: torque produced by the left support

τp: torque produced by the person

τcm: torque produced by the center of mass of the wooden

The torque is given by:

\tau=Fd           (2)

F: force applied

d: distance to the pivot of the torque, in this case, distance to the right support.

You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:

-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N

Where d1, d2 and d3 are distance to the right support.

You solve the equation for F and replace the values of the other parameters:

F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N

The force applied by the left support is 2123.33 N

8 0
3 years ago
helpppp I just saw my crush when I was walking out of a supermarket and like I was looking a mess so now I'm just wondering how
UkoKoshka [18]

Answer:

okay

Explanation:

(wish i could help but im just answering for points)

3 0
3 years ago
What type of radioactive decay is shown in this equation?
svp [43]
There is no <span>radioactive decay</span>
6 0
3 years ago
The force of Earth's gravity on a capsule in space will lessen as it moves farther away. If the capsule moves to twice its dista
Bess [88]

Answer: One quarter of the force

Explanation:

According to Newton's law of Gravitation, the force F exerted between two bodies of masses m1 and m2  and separated by a distance r  is equal to the product of their masses and inversely proportional to the square of the distance:

F=G\frac{(m1)(m2)}{r^2}    (1)

Where Gis the gravitational constant

This means that the gravity force decreases when the distance between these two bodies increases.

In this context, if the distance between the capsule and the Earth increases twice, the new distance will be 2r.

Substituting this distance in (1):

F=G\frac{(m1)(m2)}{(2r)^2}    (2)

F=G\frac{(m1)(m2)}{4r^2}    

<u>Finally:</u>

F=\frac{1}{4}G\frac{(m1)(m2)}{r^2} >>>This means the force toward Earth becomes one quarter "weaker"

3 0
2 years ago
Austin performed an experiment. He put 100mL of vegetable oil (density 0.9 g/ml) in a graduated cylinder. He put a 100g mass of
Grace [21]

Answer:

Explanation:

mass of displaced oil = 11 x  .9

= 9.9 gm

9.9 x 10⁻³ kg

weight of displaced oil = 9.9 x 9.81 x 10⁻³ N

= .097 N .

buoyant force by oil = .097 N

weight of unknown metal = .1 x 9.8

= .98 N .

weight of metal in oil = .98 - .097

= .883 N .

=

6 0
3 years ago
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