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enot [183]
3 years ago
7

How much would a spring scale with k = 120 N/m stretch, if it had a 3.75 J of work done on it?

Physics
1 answer:
iren [92.7K]3 years ago
3 0

Answer:

0.25m

Explanation:

Given parameters:

Spring constant , K  = 120N/m

Work done  = 3.75J

Unknown:

magnitude of extension = ?

Solution:

To solve this problem;

           Work done  = \frac{1}{2}kx²  

K is the spring constant

x is the extension

               3.75  =  \frac{1}{2} x 120x²

               3.75  = 60x²

                x²  = 0.06

                x = √0.06  = 0.25m

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A spring requires 10 N of force to hold it 1 m past its natural length. How much work is required to stretch it from its natural
Ulleksa [173]

Answer:

5 J

Explanation:

F = spring force required to stretch the spring = 10 N

x = stretch in the spring = 1 m

k = spring constant of the spring

Spring force is given as

F = k x

Inserting the values

10 = k (1)

k = 10 N/m

W = work done in stretching the spring

Work done is given as

W = (0.5)k x²

Inserting the values

W =  (0.5) (10) (1)²

W = 5 J

5 0
4 years ago
True or false? Scientific laws can be expressed through a law that relates several variables.
Marina86 [1]

Answer:

True

Explanation:

Scientific laws are often written as expressions that contains variables and are laws that are binding themselves.

In science, laws are natural phenomenon that draws from careful observations that holds through following a series of detailed study. Within the range of assumed parameters, a law will always hold true.

Most laws in science are denoted using mathematical variables which helps to interpret them.

The variables shows the relationship between the different parts of the law.

For example, Newton's law of universal gravitation is expressed mathematically as shown below;

                          F  =  \frac{G m_{1} m_{2}  }{r^{2} }

where G, m and r are all variables.

  G is the universal gravitation constant

   m is mass

   r is the distance between them.

   F is the gravitational force.

Most scientific laws are often expressed in this format.

5 0
3 years ago
Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block
Oksana_A [137]

Answer:

av=0.333m/s, U=3.3466J

b.

v_{A2}=-1.333m/s,\\ v_{B2}=0.667m/s

Explanation:

a. let m_A be the mass of block A, andm_B=10.0kg be the mass of block B. The initial velocity of A,\rightarrow v_A_1=2.0m/s

-The initial momentum =Final momentum since there's no external net forces.

pA_1+pB_1=pA_2+pB_2\\\\P=mv\\\\\therefore m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}

Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):

v_A_1-v_B_1=v_{B2}-v_{A2}

-Applying the conservation of momentum. The blocks have the same velocity after collision:

v_{B2}=v_{A2}=v_2\\\\2\times 2+10\times 0=2v_2+10v_2\\\\v_2=0.3333m/s

#Total Mechanical energy before and after the elastic collision is equal:

K_1+U_{el,1}=K_2+U_{el,2}\\\\#Springs \ in \ equilibrium \ before \ collision\\\\U_{el,2}=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\\\\U_{el,2}=0.5\times 2\times 2^2-0.5(2+10)(0.333)^2\\\\U_{el,2}=3.3466J

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0

We plug these values in the equation:

m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}

2\times2+10\times0=2v_A_2+10v_B_2\\\\2=v_A_2+5v_B_2\\\\#Eqtn 2:\\v_A_1-v_B_1=v_{B2}-v_{A2}\\\\2-0=v_{B2}-v_{A2}\\\\2=v_{B2}-v_{A2}\\\\#Solve \ to \ eliminate \ v_{A2}\\\\6v_{B2}=2.0\\\\v_{B2}==0.667m/s\\\\#Substitute \ to \ get \ v_{A2}\\\\v_{A2}=\frac{4}{6}-2=1.333m/s

7 0
4 years ago
1. A 3.5 kg object experiences an acceleration of 0.5 m/s2. What net force does the object experience
Leviafan [203]

Answer:

<h2>1.75 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 3.5 × 0.5 = 1.75

We have the final answer as

<h3>1.75 N</h3>

Hope this helps you

7 0
3 years ago
Find the magnitude of the force at point x = 15cm if k=10500 N/m when the work done by this force is WD= 1/2k x2
Ann [662]

Answer:

sqdqk3

Explanation:

qdjqưbdkq ưqdhud j

6 0
3 years ago
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