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IRISSAK [1]
3 years ago
15

A mass attached to a spring is displaced from its equilibrium position by 5cm and released. The system then oscillates in simple

harmonic motion with a period of 1s. If that same mass–spring system is displaced from equilibrium by 10cm instead, what will its period be in this case?(A) 0.5s(B) 2s(C) 1s(D) 1.4s
Physics
1 answer:
aliina [53]3 years ago
5 0

Answer:

T= 1 s

Explanation:

Given that

When x=  cm ,T= 1

we know that time period of spring mas system given as

T=2\pi \sqrt{\dfrac{m}{k}}

T= Time period

m= mass

k=spring constant

So from above equation we can say that time period of system does not depends on the value of x.

So when x= 10 cm ,still time period will be 1 s.

T= 1 s

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Consider an elevator with a table and a book on top of the table. The mass of the table is 10kg and the mass of the book is 2kg.
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Newton's second law allows us to find the force of the block on the table is 126 N

Newton's second law says that the net force is proportional to the product of the mass and the acceleration of the body

                Σ F = m a

Where the bold letters indicate vectors, m is the mass and the acceleration of the body

A free body diagram is a diagram where the forces are represented without the details of the bodies, in the attached we can see a free body diagram of the system.

Let's start by finding the acceleration of the elevator with kinematics  

                 v = v₀ + a t

                 a = \frac{v-v_o}{a}  

Where v and v₀ are the current and initial velocity, respectively, at acceleration and t is the time

                 a = \frac{8-1}{2}

                 a = 3.5 m / s²

Let's write Newton's second law for each body

The book

                N₂ - W₂ + N₁ = m a

               

Table

                N₁ - W₁ - W₂ = M a

                W₁ = Mg

                W₂ = mg

                N₁ = (M + m) g + M a

                N₁ = (10 + 2) 9.8 + 10 3.5

                N₁ = 152.6 N

This is the reaction of the earth to the support of the block and the table

               N₂ = ma + m g  - N₁

               N₂ = m ( a +g) - N₁  

               N₂ = 2 (3.5 + 9.8) - 152.6

               N₂ = 26.6 - 152.6

               N₂ = -126 N

The negative sign indicates that the direction is opposite to the one assigned, this is the action of the block on the table.

In conclusion using Newton's second law we can find the forces of the block on the table is 126 N

Learn more here: brainly.com/question/19860811

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To demonstrate the tremendous acceleration of a top fuel dragracer, you attempt to run your car into the back of a dragster that
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Answer:

a. 2v₀/a   b. 2v₀/a  

Explanation:

a. Since you are moving with a constant velocity v₀, the distance, s you cover in time = t max is s = v₀t.

Since the dragster starts from rest with an acceleration, a, using

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Since the distance moved by me and the dragster must be the same,

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v₀t. - 1/2at² = 0

t(v₀ - 1/2at) = 0

t= 0 or v₀ - 1/2at = 0

t= 0 or v₀ = 1/2at

t= 0 or t = 2v₀/a  

So the maximum time tmax = 2v₀/a

b. Since the distance covered by me to meet the dragster is s = v₀t in time, t = tmax which is also my distance from the dragster when it started. So, my distance from the dragster when it started is s =  v₀(2v₀/a)

= 2v₀/a  

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