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
xxMikexx [17]
3 years ago
12

A force is applied to an ideal spring (initially in its equilibrium position) and does 1.9 JJ of work stretching it 2.2 cmcm . H

ow much work is required to stretch the spring by 6.5 cmcm from its equilibrium position?
Physics
2 answers:
coldgirl [10]3 years ago
7 0

Answer:

16.586 J

Explanation:

The energy stored in an elastic material is given as,

E = 1/2ke².................. Equation 1

Where E = work done by the  ideal spring, k = force constant of the spring, e = extension.

make k the subject of the equation

k = 2E/e²..................... Equation 2

Given: E = 1.9 J, e = 6.5 cm = 0.022 m

Substitute into equation 2

k = 2(1.9)/0.022²

k = 3.8/(0.022²)

k = 7851.24 N/m

When the spring is stretched by 6.5 cm

E = 1/2ke'²................. Equation 3

Where e = 6.5 cm = 0.065

Substitute into equation 3

E = 1/2(7851.24)(0.065²)

E = 16.586 J.

Verdich [7]3 years ago
5 0

Answer:

W=16.58J

Explanation:

initial information we have

work: W=1.9J

stretched distance: x=2.2cm=0.022m

from this, we can find the value of the constant of the spring k, with the equation for work in a spring:

W=\frac{1}{2} kx^2

substituting known values:

1.9J=\frac{1}{2}k(0.022)^2\\

and clearing for k:

k=\frac{2(1.9J)}{0.022^2} \\k=7,851.24

and now we want to know how much work is done when we stretch the spring a distance of 6.5cm from equilibrium, so now x is:

x=6.5cm=0.065m

and using the same formula for work, with the value of k that we just found:

W=\frac{1}{2} kx^2

W=\frac{1}{2}(7851.24)(0.065)^2\\W=16.58J

You might be interested in
A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carouse
Sidana [21]

Answer:

m = 35.98 Kg ≈ 36 Kg

Explanation:

I₀ = 125 kg·m²

R₁ = 1.50 m

ωi = 0.600 rad/s

R₂ = 0.905 m

ωf = 0.800 rad/s

m = ?

We can apply The law of conservation of angular momentum as follows:

Linitial = Lfinal

⇒    Ii*ωi = If*ωf   <em>(I)</em>

where    

Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m

If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m

Now, we using the equation <em>(I) </em>we have

(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800

⇒  m = 35.98 Kg ≈ 36 Kg

5 0
3 years ago
PLEASE HURRY
VLD [36.1K]

Answer:

Wilhelm Conrad Roentgen (1845-1923)

Antoine Henri Becquerel (1852-1908)

Pierre (1859-1906) and Marie (1867-1934) Curie

Explanation:

Wilhelm Conrad Roentgen (1845-1923)

Contribution:  Discovery of x-rays in 1901.

Antoine Henri Becquerel (1852-1908)

Contribution: He discovered that radioactivity is the separation of x-rays and document and the difference between two.

Pierre (1859-1906) and Marie (1867-1934) Curie

Contribution: She discovered Polonium and Radium in1911

5 0
3 years ago
A 5-m steel beam is lowered by means of two cables unwinding at the same speed from overhead cranes. As the beam approaches the
ArbitrLikvidat [17]

Answer:

a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)

b) The acceleration of point A is 3.25 m/s²

The acceleration of point E is 0.75 m/s²

Explanation:

a) The relative acceleration of B with respect to D is equal:

a_{B} =a_{D} +(a_{B/D} )_{n} +(a_{B/D} )_{t}

Where

aB = absolute acceleration of point B = 2.5 j (m/s²)

aD = absolute acceleration of point D = 1.5 j (m/s²)

(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam

(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)

a_{B} =a_{D}  +(a_{B/D} )_{t}

2.5j=1.5j  +(a_{B/D} )_{t}\\(a_{B/D} )_{t}=j=1m/s^{2}

We have that

(aB/D)t = BDα

Where α = acceleration of the beam

BDα = 1 m/s²

Where

BD = 2

2\alpha =1\\\alpha =0.5rad/s^{2}CW

b) The acceleration of point A is:

a_{A} =a_{D}  +(a_{A/D} )_{t}

(aA/D)t = ADαj

a_{A} =a_{D}  +AD\alpha j\\a_{A}=1.5j+(3.5*0.5)j\\a_{A}=3.25jm/s^{2}

The acceleration of point E is:

(aE/D)t = -EDαj

a_{E} =a_{D}  -ED\alpha j\\a_{E}=1.5j-(1.5*0.5)j\\a_{E}=0.75jm/s^{2}

7 0
3 years ago
Electric fields up to 2.00 × 10 5 N/C have been measured inside of clouds during electrical storms. Neglect the drag force due t
katrin2010 [14]

Answer:

1.9161676647\times 10^{13}\ m/s^2 or 1.9532799844\times 10^{12}g

23.4843749996 m

Yes

Explanation:

E = Electric field = 2\times 10^5\ N/C

c = Speed of light = 3\times 10^8\ m/s

m = Mass of proton= 1.67\times 10^{-27}\ kg

q = Charge of electron = 1.6\times 10^{-19}\ C

Acceleration is given by

a=\dfrac{Eq}{m}\\\Rightarrow a=\dfrac{2\times 10^5\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}\\\Rightarrow a=1.9161676647\times 10^{13}\ m/s^2

Dividing by g

\dfrac{a}{g}=\dfrac{1.9161676647\times 10^{13}}{9.81}\\\Rightarrow a=1.9532799844\times 10^{12}g

The acceleration is 1.9161676647\times 10^{13}\ m/s^2 or 1.9532799844\times 10^{12}g

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{(0.1\times 3\times 10^8)^2-0^2}{2\times 1.9161676647\times 10^{13}}\\\Rightarrow s=23.4843749996\ m

The distance is 23.4843749996 m

The gravitational field is very small compared to the electric field so the effects of gravity can be ignored.

5 0
3 years ago
An object of mass 10.0kg is released at point A, slidesto the bottom of the 30° incline, then collides with ahorizontal massless
Elenna [48]

Answer:

Explanation:

Potential energy lost by mass = mgh

= 10 x 9.8 x 2 = 196 J

a ) If v be velocity of mass at the bottom , its kinetic energy will be stored in spring as elastic energy

= 1/2 m v² = 1/2 k x² , k is spring constant , x is compression , m is mass falling down

.5 x 10 v² = .5 x 500 x .75²

v = 5.3 m /s

b ) kinetic energy of mass at the bottom

= /2 m v²

= .5 x 10 x 5.3²

= 140.45 J

energy lost by mass while coming down

=potential energy at the top - kinetic energy at bottom

=  196 - 140.45

= 55.55 J .

This is equal to negative work done by friction

work done by friction = - 55.55 J

c ) Since there will be no loss of energy in compression and extension of spring so , no loss of kinetic energy will take place of mass . So it wil have same velocity that is 5.3 m /s while on its return journey.

d ) kinetic energy at the bottom = 140.45

loss of energy by friction again

= 140.45  - 55.55

= 84.9 J

If h be the height attained

mgh = 84.9

10 x 9.8 x h = 84.9

h = .866 m

( We have assumed that loss of energy in return journey will be same due to friction . )

6 0
3 years ago
Other questions:
  • The synthesis of nitrogen trihydride from nitrogen gas and hydrogen gas is shown by which balanced chemical equation?
    6·2 answers
  • You are pulling your little sister on her sled across an icy (frictionless) surface. When you exert a constant horizontal force
    6·1 answer
  • 3. Pedro threw a penny into a fountain and it sank to the bottom. Which of the following would explain
    5·2 answers
  • What order shows decreasing wavelength
    7·1 answer
  • Two identical hard spheres, each of mass m and radius r, are released from rest in otherwise empty space with their centers sepa
    8·1 answer
  • There are stars located in the center bulge of the Milky Way and the spiral arms of the Milky Way. What is the difference betwee
    6·1 answer
  • Why do the compass needles change direction when the electric current is Flowing
    10·2 answers
  • Which countries have high productivity, but are not among the top 10 for human development? Check all that apply.
    9·1 answer
  • How long will it take you to drive the 256 miles to get to the Mackinac Island Ferry if you average 70 mi/hr? If you leave at 3:
    14·1 answer
  • Two large parallel plates are 0.00630 m apart and the voltage across them is 10.0 bolts. What is the electric field strength bet
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!