1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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.

You might be interested in
When the mass of the cylinder increased by a factor of 3, from 1.0 kg to 3.0 kg, what happened to the cylinder’s gravitational p
const2013 [10]

Answer: Fourth option. It increased by a factor of 3.

Solution:

m1=1.0 kg

Cylinder's gravitational potential energy: Ep=m*g*h

Ep1=(1.0 kg)*g*h

Ep1=g*h

m2=3.0 kg

Ep2=(3.0 kg)*g*h

Ep2=3*g*h

Replacing g*h by Ep1 in the equation above:

Ep2=3*Ep1

Then, the cylinder's gravitational potential energy increased by a factor of 3.

3 0
3 years ago
Read 2 more answers
A parallel RLC resonant circuit has a resistance of 200 Ω. If it is known that the bandwidth is 80 rad/s and the lower half-powe
d1i1m1o1n [39]

Answer:

L= 3.6mH

C =9.9 microfarad

Explanation:

Resonant frequency fr

fr = fl + 1/2 BW

fr = 800+ 1/2×80

=800+40

=840 rad/s

Bandwidth (BW)

BW = fr/Q

Q = quality factor

Q= fr/ BW

Q = 840/80

Q= 10.5

Quality factor = R/Xl

Xl = inductive reactance

Xl = R/Q

Xl = 200/10.5

Xl = 19.05 ohms

Xl =2πfL

L= Inductance

L = Xl /2πf

L =19.05/5278.56

L= 3.6mH

Capacitor C

1= fr^2 × 4π^2LC

C = 1/fr^24π^2L

C = 1/100307.5

C= 9.9microfarad

5 0
3 years ago
What is the magnitude of the impulse that would cause the 2-kg box to accelerate from 2 m/s to 5 m/s?
Law Incorporation [45]

6 J is the impulse caused by the change in velocity of 2 kg box from 2 m/s to 5 m/s.

Answer:

The magnitude of impulse is 6 J.

Explanation:

Impulse is the force acting on any object for a given time interval. As force is equal to the product of mass and acceleration and acceleration is the rate of change of velocity with time. Then the product of force with time interval will be equal to the product of mass with change in velocity.

F = m a = \frac{m(v-u)}{(t_{2}-t_{1}  )}

FΔt = mΔv

Impulse = FΔt=mΔv

As the mass of box is given as 2 kg and the velocity changes from 2 m/s to 5 m/s, then the impulse = 2 × (5-2) = 2 ×3 =6 J

So 6 J is the impulse caused by the change in velocity of 2 kg box from 2 m/s to 5 m/s.

3 0
3 years ago
What are ten different uses for solar energy?
Yuki888 [10]

Answer:

Top 10 Residential Uses for Solar Energy.

01. Solar Powered Ventilation Fans.

02. Solar Heating for Your Swimming Pool.

03. Solar Water Heater.

04. Solar House Heating.

05.Solar Powered Pumps.

06. Charging Batteries With Solar Power.

07. Power Your Home With Photo-Electric.

08. Solar Energy For Cooking.

09. Solar energy for outdoor lighting.

10. Solar transportation.

8 0
3 years ago
A spring with spring constant of 34 N/m is stretched 0.12 m from its equilibrium position. How much work must be done to stretch
Nesterboy [21]

Answer:0.253Joules

Explanation:

First, we will calculate the force required to stretch the string. According to Hooke's law, the force applied to an elastic material or string is directly proportional to its extension.

F = ke where;

F is the force

k is spring constant = 34N/m

e is the extension = 0.12m

F = 34× 0.12 = 4.08N

To get work done,

Work is said to be done if the force applied to an object cause the body to move a distance from its initial position.

Work done = Force × Distance

Since F = 4.08m, distance = 0.062m

Work done = 4.08 × 0.062

Work done = 0.253Joules

Therefore, work done to stretch the string to an additional 0.062 m distance is 0.253Joules

8 0
3 years ago
Other questions:
  • A magnesium surface has a work function of 3.60 eV. Electromagnetic waves with a wavelength of 320 nm strike the surface and eje
    14·1 answer
  • An x-ray has a wavelength of 4.18 Å. Calculate the energy (in J) of one photon of this radiation. Enter your answer in scientifi
    12·1 answer
  • What is the difference in the charges on a ballon rubbed in your hair and a glass rod rubbed
    5·1 answer
  • Write a paragraph explaining how the scientific method exemplified the new emphasis on reason
    13·1 answer
  • Identify whether the features and functions belong to the Hubble Space Telescope or the ALMA radio telescope. 1.installed on lan
    11·2 answers
  • Constants Canada geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, trav
    10·1 answer
  • An amusement park ride moves a rider at a constant speed of 14 meters per second in a horizontal circular path of radius 10. met
    9·1 answer
  • A block of mass m is attached to the end of a spring (spring stiffness constant k ). The mass is given an initial displacement x
    12·1 answer
  • The graph shows the amplitude of a passing wave over time in seconds (s) What is the approximate frequency of the wave shown? A.
    15·1 answer
  • A vibrating mass of 300 kg mounted on a massless support by a spring of stiffness 40,000 N>m and a damper of unknown damping
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!