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spin [16.1K]
3 years ago
10

A person wants to determine the spring constant of an exercise stretch cord. He pulls the cord with a force probe that exerts a

35-N force on the cord, causing it to stretch 20 cm. Using the Hooke's law find the spring constant of the cord. Assume that the stretch does not exceed the elastic range of the material. Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
amid [387]3 years ago
8 0

Answer: 180N/m(to 2 significant figures)

Explanation:

According to hooked law which states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically, F = ke where;

F is the applied force in newtons

k is the elastic/spring constant in N/m

e is the extension in meters

Given applied force = 35N

extension = 20cm = 0.2m

Since F = ke,

k = F/e = 35/0.2

k = 175N/m

The spring constant is 175N/m

= 180N/m (to 2significant figures)

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Answer:

6 m/s is the missing final velocity

Explanation:

From the data table we extract that there were two objects (X and Y) that underwent an inelastic collision, moving together after the collision as a new object with mass equal the addition of the two original masses, and a new velocity which is the unknown in the problem).

Object X had a mass of 300 kg, while object Y had a mass of 100 kg.

Object's X initial velocity was positive (let's imagine it on a horizontal axis pointing to the right) of 10 m/s. Object Y had a negative velocity (imagine it as pointing to the left on the horizontal axis) of -6 m/s.

We can solve for the unknown, using conservation of momentum in the collision: Initial total momentum = Final total momentum (where momentum is defined as the product of the mass of the object times its velocity.

In numbers, and calling P_{xi} the initial momentum of object X and P_{yi} the initial momentum of object Y, we can derive the total initial momentum of the system: P_{total}_i=P_{xi}+P_{yi}= 300*10 \frac{kg*m}{s} -100*6\frac{kg*m}{s} =\\=(3000-600 )\frac{kg*m}{s} =2400 \frac{kg*m}{s}

Since in the collision there is conservation of the total momentum, this initial quantity should equal the quantity for the final mometum of the stack together system (that has a total mass of 400 kg):

Final momentum of the system: M * v_f=400kg * v_f

We then set the equality of the momenta (total initial equals final) and proceed to solve the equation for the unknown(final velocity of the system):

2400 \frac{kg*m}{s} =400kg*v_f\\\frac{2400}{400} \frac{m}{s} =v_f\\v_f=6 \frac{m}{s}

7 0
3 years ago
Read 2 more answers
→Fo
irakobra [83]

Answer:

The mass of the block, M =T/(3a +g)  Kg

Explanation:

Given,

The upward acceleration of the block a = 3a

The constant force acting on the block, F₀ = Ma = 3Ma

The mass of the block, M = ?

In an Atwood's machine, the upward force of the block is given by the relation

                                     Ma = T - Mg

                                      M x 3a = T - Ma    

                                    3Ma + Mg = T

                                       M = T/(3a +g)  Kg

Where 'T' is the tension of the string.

Hence, the mass of the block in Atwood's machine is, M = T/(3a +g)  Kg

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= 2464.21 Joules
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