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Ludmilka [50]
3 years ago
9

Envision holding the end of a ruler with one hand and deforming it with the other. When you let go, you can see the oscillations

of the ruler. In what way could you modify this simple experiment to increase the rigidity of the system
Physics
1 answer:
lapo4ka [179]3 years ago
5 0

Answer: To increase the rigidity of the system you could hold the ruler at its midpoint so that the part of the ruler that oscillates is half as long as in the original experiment.

Explanation:

When a rule is displaced from its vertical position, it oscillates back and forth because of the restoring force opposing the displacement. That is, when the rule is on the left there is a force to the right.

By holding a ruler with one hand and deforming it with the other a force is generated in the opposite direction which is known as the restoring force. The restoring force causes the ruler to move back toward its stable equilibrium position, where the net force on it is zero. The momentum gained causes the ruler to move to the right leading to opposite deformation. This moves the ruler again to the left. The whole process is repeated until dissipative forces reduce the motion causing the ruler to come to rest.

The relationship between restoring force and displacement was described by Hooke's law. This states that displacement or deformation is directly proportional to the deforming force applied.

F= -kx, where,

F= restoring force

x= displacement or deformation

k= constant related to the rigidity of the system.

Therefore, the larger the force constant, the greater the restoring force, and the stiffer the system.

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002 10.0 points
snow_lady [41]

Answer:

-2040 m/s²

Explanation:

Taking toward the wall to be positive, the initial velocity is 10.1 m/s and the final velocity is -8.3426 m/s.

Average acceleration is the change in velocity over change in time.

a = Δv / Δt

a = (-8.3426 m/s − 10.1 m/s) / 0.00905 s

a = -2040 m/s²

3 0
3 years ago
The highest energy waves have the<br> {SHORTEST}wavelength
Likurg_2 [28]

Answer:

infared rays pls give me brainiest]

8 0
3 years ago
Some enterprising physics students working on a catapult decide to have a water balloon fight in the school hallway. The ceiling
sergejj [24]

Answer:

\alpha =54.7º

Explanation:

From the exercise we have our initial information

y=3.4m\\v_{o}=10m/s\\g=-9.8m/s^2

When the balloon gets to the ceiling its velocity at that moment is 0 m/s. Being said that we can calculate velocity at the vertical direction

v_{y}^2=v_{oy}^2+ag(y-y_{o})

Since v_{y}=0 and y_{o}=0

0=v_{oy}^2-2(9.8m/s^2)(3.4m)

v_{oy}=\sqrt{2(9.8m/s^2)(3.4m)}=8.16m/s

Knowing that

v_{oy}=v_{o}sin\alpha

sin\alpha =\frac{v_{oy} }{v_{o} }

\alpha =sin^{-1}(\frac{v_{oy}}{v_{o}})=sin^{-1}(\frac{8.16m/s}{10m/s})=54.7º

8 0
3 years ago
What’s Newton’s second law? Explain and mention some examples in daily life
Marina CMI [18]
Newton’s second law of motion is force equals mass times acceleration.

F = m•a

An example of this would be hitting a ball. If you hit the ball, it will move however fast you hit the ball. The harder you hit the ball, the faster it will move.

hope this helps and brainliest please
8 0
2 years ago
(a) Triply charged uranium-235 and uranium-238 ions are being separated in a mass spectrometer. (The much rarer uranium-235 is u
stiv31 [10]

Answer:

(a) 2.5 cm

(b) Yes

Solution:

As per the question:

Mass of Uranium-235 ion, m = 3.95\times 10^{- 25}\ kg

Mass of Uranium- 238, m' = 3.90\times 10^{- 25}\ kg

Velocity, v = 3.00\times 10^{5}\ m/s

Magnetic field, B = 0.250 T

q = 3e

Now,

To calculate the path separation while traversing a semi-circle:

\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})

The radius of the ion in a magnetic field is given by:

R = \frac{mv}{qB}

\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})

\Delta x = 2(\frac{mv}{qB} - \frac{m'v}{qB})

\Delta x = 2(\frac{m - m'}{qB}v)

Now,

By putting suitable values in the above eqn:

\Delta x = 2(\frac{3.95\times 10^{- 25} - 3.90\times 10^{- 25}}{3\times 1.6\times 10^{- 19}\times 0.250}\times 3.00\times 10^{5}) = 2.5\ cm

\Delta x = 1.25\ cm

(b) Since the order of the distance is in cm, thus clearly this distance is sufficiently large enough in practical for the separation of the two uranium isotopes.

3 0
3 years ago
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