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AfilCa [17]
2 years ago
8

An object attached to an ideal massless spring is pulled across a frictionless surface. If the spring constant is 45 N/m and the

spring is stretched by 0.88 m when the object is accelerating at 0.8 m/s2, what is the mass of the object
Physics
2 answers:
Nataly [62]2 years ago
6 0

Answer:

The mass of the object is 49.5kg which is approximately 50kg

Explanation:

Given that

Spring constant (k)=45N/m

The extension (e)=0.88m

Also given that the acceleration is 0.8m/s²

Force by the spring is given as

Using hooke's law

According to Hooke's law which states that the extension of an elastic 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

k is the spring constant

e is the extension

From the formula k = F/e

F=ke

m is the mass of the block = ?

a is the acceleration = 0.8m/s²

e is the extension of the spring =  0.88m

k is the spring constant = 45N/m

F=45×0.88

F=39.6N

Now this force will set the object in motion, now using newton second law of motion

F=ma

Then, m=F/a

m=39.6/0.8

m=49.5kg

The mass of the object is 49.5kg which is approximately 50kg

otez555 [7]2 years ago
4 0

Answer:

Explanation:

Using Hooke's law,

F = -k × delta x

Where,

F = force

k = spring constant

= 45 N/m

Delta x = spring displacement

= 0.88 m

F = 45 × 0.88

= 39.6 N

a = 0.8 m/s^2

Force, F = Mass, M × acceleration, a

M = 39.6/0.8

= 49.5 kg

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A radioactive isotope has a half-life of 2 hours. If a sample of the element contains 600,000 radioactive nuclei at 12 noon, how
storchak [24]

Answer: There will be 75258 nuclei left at 6 pm.

Explanation:

a) half-life of the radioactive substance:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.69}{k}

k=\frac{0.69}{t_{\frac{1}{2}}}=\frac{0.693}{2hours}=0.346hours^{-1}

b) Expression for rate law for first order kinetics is given by:

A=A_0e^{-kt}

where,

k = rate constant  

t = time for decomposition = 6 hours ( from 12 noon to 6 pm)

A = activity at time t = ?

A_0 = initial activity  = 600, 000

A=600000\times e^{-0.346\times 6}

A=75258

Thus there will be 75258 nuclei left at 6 pm.

7 0
2 years ago
A damped harmonic oscillator consists of a block of mass 2.5 kg attached to a spring with spring constant 10 N/m to which is app
Cerrena [4.2K]

Answer:

0.5% per oscillation

Explanation:

The term 'damped oscillation' means an oscillation that fades away with time. For Example; a swinging pendulum.

Kinetic energy, KE= 1/2×mv^2-------------------------------------------------------------------------------------------------------------(1).

Where m= Mass, v= velocity.

Also, Elastic potential energy,PE=1/2×kX^2----------------------------------------------------------------------------------------------------------------------(2).

Where k= force constant, X= displacement.

Mechanical energy= potential energy (when a damped oscillator reaches maximum displacement).

Therefore, we use equation (3) to get the resonance frequency,

W^2= k/m--------------------------------------------------------------------------------------(3)

Slotting values into equation (3).

= 10/2.5.

= ✓4.

= 2 s^-1.

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F^2= (-0.1)^2

Potential energy,PE= 1/2 ×0.01

Potential energy= 0.05 ×100

= 0.5% per oscillation.

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4 0
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A force of 5000 n is applied outwardly to each end of a 5.0-m long rod with a radius of 34.0 mm and a young's modulus of 125 x 1
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