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
Digiron [165]
3 years ago
6

Spring #1 has a force constant of k, and spring #2 has a force constant of 2k. Both springs are attached to the ceiling. Identic

al weights are hooked to their ends, and the weights are allowed to stretch the springs. The ratio of the energy stored by spring #1 to that stored by spring #2 is
1:1

1:4

2:1

1:2
Physics
1 answer:
Gre4nikov [31]3 years ago
4 0

Answer:

The ratio of the energy stored by spring #1 to that stored by spring #2 is 2:1

Explanation:

Let the weight that is hooked to two springs be w.

Spring#1:

Force constant= k

let x1 be the extension in spring#1

Therefore by balancing the forces, we get

Spring force= weight

⇒k·x1=w

⇒x1=w/k

Energy stored in a spring is given by \frac{1}{2}kx^{2} where k is the force constant and x is the extension in spring.

Therefore Energy stored in spring#1 is, \frac{1}{2}k(x1)^{2}

                                                              ⇒\frac{1}{2}k(\frac{w}{k})^{2}

                                                              ⇒\frac{w^{2}}{2k}

Spring #2:

Force constant= 2k

let x2 be the extension in spring#2

Therefore by balancing the forces, we get

Spring force= weight

⇒2k·x2=w

⇒x2=w/2k

Therefore Energy stored in spring#2 is, \frac{1}{2}2k(x2)^{2}

                                                              ⇒\frac{1}{2}2k(\frac{w}{2k})^{2}

                                                              ⇒\frac{w^{2}}{4k}

∴The ratio of the energy stored by spring #1 to that stored by spring #2 is \frac{\frac{w^{2}}{2k}}{\frac{w^{2}}{4k}}=2:1

You might be interested in
All objects are either ___________ or ___________. Charged objects can have a ____________ or _____________ charge. Uncharged ob
KATRIN_1 [288]

All objects are either <u>charged</u> or <u>uncharged</u><u>.</u> Charged objects can have a <u>positive</u> or <u>negative</u> charge. Uncharged objects will have a <u>no</u> charge. Charged objects that have the same charges will <u>repel</u> each other. Charged objects with opposite charges will <u>a</u><u>t</u><u>t</u><u>r</u><u>a</u><u>c</u><u>t</u> each other. Uncharged objects can become <u>charged</u><u>.</u> The charge an object has gives it <u>electric</u> energy. The charged object’s ability to attract (pull) or repel (push) other objects is called <u>electrostatic</u> force.

5 0
2 years ago
A student pushes a 21-kg box initially at rest, horizontally along a frictionless surface for 10.0 m and then releases the box t
Marrrta [24]

Answer:v=3.08 m/s

Explanation:

Given

mass of student m=21 kg

distance moved d=10 m

Force applied F=10 N

acceleration of system during application of force is a

a=\frac{F}{m}=\frac{10}{21}=0.476 m/s^2

using v^2-u^2=2 as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

v^2-0=2\times 0.476\times 10

v=\sqrt{9.52}

v=3.08 m/s

6 0
3 years ago
The smallest possible part of an element that can still be identified as the element is called
andreyandreev [35.5K]

Answer:

BLM

Explanation:

ACAB

4 0
3 years ago
Can someone see if it's right? thank you. ​
bearhunter [10]

Answer:

it's right you did a great job

8 0
2 years ago
A bowling ball is launched from the top of a building at an angle of 35° above the horizontal with an initial speed of 15 m/s. T
Mamont248 [21]

Let y_0 be the height of the building and thus the initial height of the ball. The ball's altitude at time t is given by

y=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ\,t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity.

The ball reaches the ground when y=0 after t=2.9\,\mathrm s. Solve for y_0:

0=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ(2.9\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(2.9\,\mathrm s)^2

\implies y_0\approx16.258\,\mathrm m

so the building is about 16 m tall (keeping track of significant digits).

3 0
3 years ago
Other questions:
  • A proton is at the origin. One electron is at the point (2m, 4m)
    8·1 answer
  • Consider the motion of a bullet that is fired from a rifle 6 m above the ground in a northeast direction. The initial velocity o
    8·1 answer
  • Under what conditions is the conservation of momentum applicable
    7·1 answer
  • The passing of the Moon directly between Earth and the Sun is a/an
    6·2 answers
  • Eric has a mass of 80 kg. He is standing on a scale in an elevator that is accelerating downward at 1.7 m/s2. What is the approx
    11·1 answer
  • I need help with this question id really appreciate it thanks !
    10·1 answer
  • Convert 550 cm into m. Please show your work
    14·1 answer
  • A(n) _____ wave best describes a wave whose direction of travel is perpendicular to the displacement of the medium.
    14·2 answers
  • How is the brightness of the star related to how quickly it pulses?​
    12·1 answer
  • 7. A loaded tractor-trailer with a total mass of 5000 kg traveling at hits a loading dock and comes to a stop in 0.64 s. What is
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!