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
Bond [772]
3 years ago
6

If it requires 2.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re

quired to stretch it an additional 4.0 cm?
Physics
1 answer:
valina [46]3 years ago
7 0

Answer:

16 J

Explanation:

It is given that,

Work done, W = 2 J

A spring is stretched by 2.0 cm from its equilibrium length

We need to find how much more work will be required to stretch it an additional 4.0 cm.

Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :

U=\dfrac{1}{2}kx^2\\\\k=\dfrac{2U}{x^2}\\\\k=\dfrac{2(2)}{(0.02)^2}\\\\k=10000\ N/m

The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.

W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 10000\times ((0.06)^2-(0.02)^2)\\\\W=16\ J

So, the required work done is 16 J.

You might be interested in
In an experiment to measure the acceleration due to gravity, g two values, 9.96 m/s2 and 9.72 m/s2 , are determined. Find (1) th
alexira [117]

Answer:

(1) Percent Difference = 2.47%

(2) Percent Error (9.96 m/s²) = 1.63 %

    Percent Error (9.72 m/s²) = 0.82 %

(3) Percent Error (Mean) = 0.41 %

Explanation:

(1)

Percent Difference = [(9.96 m/s² - 9.72 m/s²)/(9.72 m/s²)]*100 %

<u>Percent Difference = 2.47%</u>

<u></u>

<u>(2)</u>

Percent Error = (|Measured Value - Original Value|/Original Value)*100%

Therefore,

Percent Error (9.96 m/s²) = (|9.96 m/s² - 9.8 m/s²|/9.8 m/s²)*100%

<u>Percent Error (9.96 m/s²) = 1.63 %</u>

Now,

Percent Error (9.72 m/s²) = (|9.72 m/s² - 9.8 m/s²|/9.8 m/s²)*100%

<u>Percent Error (9.72 m/s²) = 0.82 %</u>

<u></u>

<u>(</u>3<u>)</u>

First we need to find the mean of values:

Mean = (9.96 m/s² + 9.72 m/s²)/2

Mean = 9.84 m/s²

Therefore,

Percent Error (Mean) = (|9.84 m/s² - 9.8 m/s²|/9.8 m/s²)*100%

<u>Percent Error (Mean) = 0.41 %</u>

8 0
3 years ago
A rectangular sharp-crested weir is contracted on both sides, and the opening is 1.2 m wide. At what height (Hw) should it be pl
Alex

Answer:

H_w = 2.129 m

Explanation:

given,

Width of the weir, B = 1.2 m

Depth of the upstream weir, y = 2.5 m

Discharge, Q = 0.5 m³/s

Weir coefficient, C_w = 1.84 m

Now, calculating the water head over the weir

Q = C_w BH^{3/2}

H = (\dfrac{Q}{C_wB})^{2/3}

H = (\dfrac{0.5}{1.84\times 1.2})^{2/3}

H = 0.371\ m

now, level of weir on the channel

H_w = y - H

H_w = 2.5 - 0.371

H_w = 2.129 m

Height at which weir should place is equal to 2.129 m.

7 0
3 years ago
Astronomers define the __________ as all of space and everything in it. It is enormous, almost beyond imagination. Question 2 op
Genrish500 [490]

Answer:

Universe

Explanation:

I took the quiz.

3 0
3 years ago
Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.9 m tall window in 0.134 sec. From how high above th
timama [110]

Answer:

Explanation:

Given

length of window h=2.9\ m

time Frame for which rock can be seen is \Delta t=0.134\ s

Suppose h is height above which rock is dropped

Time taken to cover h+2.9 is t_1

so using equation of motion

y=ut+\frac{1}{2}at^2

where  y=displacement

u=initial velocity

a=acceleration

t=time

time taken to travel h  is

h=0+0.5\times g\times (t_2)^2---2

Subtract 1 and 2 we get

2.9=0.5g(t_1^2-t_2^2)

5.8=g(t_1+t_2)(t_1-t_2))

and from equation t_1-t_2=0.134\ s

so t_1+t_2=\frac{5.8}{9.8\times 0.134}

t_1+t_2=4.416\ s

and t_1=t_2+\Delta t

so t_2+\Delta t+t_2=4.416

2t_2+0.134=4.416

t_2=0.5\times 4.282

t_2=2.141\ s

substitute the value of t_2 in equation 2

h=0.5\times 9.8\times (2.141)^2

h=22.46\ m

                                                     

8 0
3 years ago
An aircraft has a liftoff speed of 33 m/s. What is the minimum constant acceleration an
JulsSmile [24]

Answer: The minimum acceleration for the air plane is 2.269m/s2.

Explanation: To solve such problem the equation of motion are applicable.

The initial velocity is 0 since the airplane was initially standing. We are going to use this equation

V^2=U^2+2as

33^2=0+2a (240)

a= 2.269m/s2

5 0
3 years ago
Other questions:
  • What kind of image does the lens in a camera produce?
    14·1 answer
  • 4 meters and the frequency is 3 hz what’s the wave speed
    5·2 answers
  • A photon has momentum of magnitude 8.30×10−28 kg⋅m/s . Part APart complete What is the energy of this photon? Give your answer i
    9·1 answer
  • Which factors are used to calculate the kinetic energy of an object? Check all that apply.
    8·1 answer
  • Fluids flow and exert forces on objects.
    7·1 answer
  • An archer fires and arrow while standing atop a 5.15 m tall wall. The arrow is fired at an angle of 55 degrees and has a launch
    8·1 answer
  • The high-speed police chase ends at an intersection as a 2,150-kg Ford Explorer (driven by Robin) traveling south at 35 m/s coll
    12·1 answer
  • A vector has components x=6 m and y=8 m. what is its magnitude and direction?
    9·1 answer
  • Pls help me with this one
    13·1 answer
  • Running water has materials such as dirt, sand, and dead plants and animals in it. When this water ends up in a lake, the materi
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!