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alekssr [168]
3 years ago
6

An object–spring system undergoes simple harmonic motion. If the amplitude increases but the mass of the object is not changed,

the total energy of the system:______.a. undergoes a sinusoidal change. b. decreases exponentially. c. decreases. d. doesn't change. e. increases.
Physics
1 answer:
cricket20 [7]3 years ago
6 0

Answer:

The correct answer is:

doesn't change (d)

Explanation:

The total energy in a system is the sum of Kinetic and Potential energies in a system, assuming that energy is not lost to an external procedure. Now, let us define what potential and kinetic energies are:

Potential Energy: this is energy at rest or stored energy

Kinetic Energy: this is energy in motion

In a simple harmonic motion of a mass-spring system, there is no dissipative force, hence the total energy is equal to the potential and kinetic energies. The total energy is not changed rather, it varies between potential and kinetic energies depending on the point at which the mass is. The kinetic energy is greatest at the point of lowest amplitude (highest velocity) and lowest at the point of greatest amplitude (lowest velocity), while potential energy is greatest at the point of highest amplitude (lowest velocity) and lowest at the point of smallest amplitude ( highest velocity). However, at every point, the sum of kinetic and potential energies equals total energy.

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The velocity of a particle is given by v=20t² - 100t + 50, where v is in meters per second and t is in seconds. Evaluate the vel
allochka39001 [22]

Explanation:

It is given that,

The velocity of a particle is given by :

v=20t^2-100t+50

Where

v is in m/s and t is in seconds

Let a is the acceleration of the object at time t. So,

a=\dfrac{dv}{dt}

a=\dfrac{d(20t^2-100t+50)}{dt}

a=40t-100

When a = 0

40t-100=0

t = 2.5 s

a is zero at t = 2.5 s. Velocity, v=20(2.5)^2-100(2.5)+50

v = -75 m/s

Since, v=\dfrac{ds}{dt}, s is the distance travelled

s=\int\limits{vdt}

s=\int\limits{(20t^2-100t+50)dt}

s=\dfrac{20t^3}{3}-50t^2+50t

At t = 2.5 s, s=\dfrac{20(2.5)^3}{3}-50(2.5)^2+50(2.5)

s = −83.34 m

Hence, this is the required solution.

5 0
3 years ago
Read 2 more answers
Given: Saturated air changes temperature by 0.5°C/100 m. The air is completely saturated at the dew point. The dew point has bee
erma4kov [3.2K]

Answer:

Explanation:

Given

saturated air temperature by 0.5^{\circ}C/100 m

Dew point temperature is given by t=2^{\circ}C

Dew point is defined as the temperature after which air no longer to uphold the water vapor fuse with it and some water vapor may condense to a liquid.

air continues to rise for 1400 m

i.e. change in temperature would be \Delta t =\frac{0.5}{100}\times 1400=7^{\circ}C

Final temperature t_f

t_f+\Delta t=t

t_f=2-7=-5^{\circ}C

3 0
3 years ago
Water is flowing through a channel that is 12m wide with a
Alexus [3.1K]

Answer:

Velocity from second channel will be 1.6875 m/sec

Explanation:

We have given width of the channel , that is diameter of the channel 1 d_1 = 12 m

So radius r_1=\frac{d_1}{2}=\frac{12}{2}=6m

Speed through the channel 1 v_1=0.75m/sec

Width , that is diameter of the channel 2 d_2=4m

So r_2=2m

From continuity equation

A_1v_1=A_2v_2

\pi \times 12^2\times 0.75=4\times \pi\times  4^2\times v_2

v_2=1.6875m/sec

So velocity from smaller channel will be 1.6875 m /sec

6 0
3 years ago
List five items that you think would generate an electric field in your house
goldenfox [79]
Lamps 
electric wiring
extension cords 
electrical appliances 
power outlets
4 0
3 years ago
1. How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b.
jolli1 [7]

Answer:

a) The distance between the ineas doubles, intensity decreases with distance

b) The distance between the ineas doubles

Explanation:

The diffraction pattern of a grid is given as a percentage

                d sin θ = m λ

where d give the distance between two consecutive lines, θ it is at an angle, λ the wavelength and m is an integer that determines the order of diffraction, let's not forget that the entire spectrum is at a value of m and then it is repeated.

Let's apply this to our case

a) distance from grid to observation screen doubles

Here we have two effect:

* the energy of the source is constant, it must be distributed over a surface, therefore the intensity decreases with distance

* The other factor can be found using trigonometry

          tan θ= y / L

where y is the distance from the central maximum to the line under study and L is the distance to the screen

           

In general, diffraction experiments cover very small angles

             tan θ = sin θ/ cos θ = sin θ

we substitute

          sinθ = y / L

we subtitle into the diffraction equation

          d y / L = m λ

          y = L / d m λ

          L = 2 L₀

          y = 2 L₀ m λ / d

we see that by doubling the distance to the screen the lines we are seeing are separated by double

b) When the density of lines doubles, it means that in the same distance I have twice as many lines, therefore the distance between two consecutive lines is reduced by half

          d = d₀o / 2

          y = (L m λ) / d

          y = (L m λ/ d₀) 2

we see that The distance between the ineas doubles

5 0
3 years ago
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