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Airida [17]
2 years ago
14

The drawing shows an object attached to an ideal spring, which is hanging from the ceiling. The unstrained length of the spring

is indicated. For purposes of measuring the height h that determines the gravitational potential energy, the floor is taken as the position where h = 0 m. The equilibrium position at which the object hangs stationary is identified as position 2. The object is set into vertical simple harmonic motion between positions 1 and 3. Identify the positions where the kinetic energy KE, the elastic potential energy EPE, and the gravitational potential energy GPE each have their maximum values during an oscillation cycle.

Physics
1 answer:
Andrew [12]2 years ago
7 0

Image is missing so I have attached it.

Also, the options are missing and it is;

A) KE is has a maximum value at position 3. EPE has a maximum value at position 2. GPE has a maximum value at position 1.

B) KE is has a maximum value at position 1. EPE has a maximum value at position 2. GPE has a maximum value at position 3.

C) KE is has a maximum value at position 2. EPE has a maximum value at position 3. GPE has a maximum value at position 1.

D) KE is has a maximum value at position 1. EPE has a maximum value at position 3. GPE has a maximum value at position 2.

E) KE is has a maximum value at position 2. EPE has a maximum value at position 1. GPE has a maximum value at position 3.

Answer:

Option C is the correct answer which says; KE is has a maximum value at position 2. EPE has a maximum value at position 3. GPE has a maximum value at position 1.

Explanation:

If an object vibrates about its mean position, under the influence of a restoring force, such that restoring force is directly proportional to the displacement from the mean position, the motion of the object is called simple harmonic motion. During Simple harmonic motion, the sum of Kinetic and potential energy remains constant.

Now, Looking at the diagram, Kinetic Energy (KE) is maximum at position 2.

Looking at the options, only C and E agree with this.

Thus, our answer is either option C or E.

However, Option E is not going to be right because it says that GPE is at a maximum at position 3, which is not true as the maximum GPE will occur at position 1.

Thus,

Option C fulfills that and therefore will be the correct answer.

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Ksivusya [100]

Answer:

- 1 m/s, 20 m

Explanation:

u = 9 m/s, a = - 2 m/s^2, t = 5 sec

Let s be the displacement and v be the velocity after 5 seconds

Use first equation of motion.

v = u + a t

v = 9 - 2 x 5 = 9 - 10 = - 1 m/s

Use second equation of motion

s = u t + 1/2 a t^2

s = 9 x 5 - 1/2 x 2 x 5 x 5

s = 45 - 25 = 20 m

4 0
1 year ago
A certain factory whistle can be heard up to a distance of 2.5 km. Assuming that the acoustic output of the whistle is uniform i
enyata [817]

Answer:

Emitted power will be equal to 7.85\times 10^{-5}watt

Explanation:

It is given factory whistle can be heard up to a distance of R=2.5 km = 2500 m

Threshold of human hearing I=10^{-12}W/m^2

We have to find the emitted power

Emitted power is equal to P=I\times A

P=I\times 4\pi R^2

P=10^{-12}\times 4\times 3.14\times  2500^2=7.85\times 10^{-5}watt

So emitted power will be equal to 7.85\times 10^{-5}watt

4 0
2 years ago
What is the change in internal energy (in J) of a system that does 4.50 ✕ 105 J of work while 3.20 ✕ 106 J of heat transfer occu
Dmitrij [34]

Answer:

-3.25\times 10^6 J

Explanation:

We are given that

Work done by the system=4.5\times 10^5 J

Heat transfer into the system=U_1=3.2\times 10^6 J

Heat transfer to the environment=U_2=6\times 10^6 J

We have to find the change in internal energy

By first law of thermodynamics

\Delta Q=\Delta U+w

\Delta Q=U_1-U_2=3.2\times 10^6-6\times 10^6=-2.8\times 10^6J

Substitute the values then we get

-2.8\times 10^6=\Delta U+4.5\times 10^5

\Delta U=-2.8\times 10^6-4.5\times 10^5=-28\times 10^5-4.5\times 10^5=-32.5\times 10^5=-3.25\times 10^6 J

Hence, the change in internal energy =-3.25\times 10^6 J

7 0
2 years ago
The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 20
NNADVOKAT [17]

Answer:

The average angular acceleration of the Earth, α  = 6.152 X 10⁻²⁰ rad/s²

Explanation:

Given data,

The period of 365 rotation of Earth in 2006, T₁ = 365 days, 0.840 sec

                                                                                  = 3.1536 x 10⁷ +0.840

                                                                                 = 31536000.84 s

The period of 365 rotation of Earth in 2006, T₀ = 365 days

                                                                               = 31536000 s

Therefore, time period of one rotation on 2006, Tₐ = 31536000.84/365

                                                                                   = 86400.0023 s

The time period of rotation is given by the formula,

                                <em>Tₐ = 2π /ωₐ</em>

                                 ωₐ = 2π / Tₐ

Substituting the values,

                                  ωₐ = 2π /  365.046306        

                                      = 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation on 1906, Tₓ = 31536000/365

                                                                                    = 86400 s

Time period of rotation,

                                   Tₓ = 2π /ωₓ

                                    ωₓ = 2π / T

                                           =  2π /86400

                                          = 7.272205217  x 10⁻⁵ rad/s

The average angular acceleration

                                   α  = (ωₓ -   ωₐ) /  T₁

             = (7.272205217  x 10⁻⁵ - 7.272205023 x 10⁻⁵) / 31536000.84

                                    α  = 6.152 X 10⁻²⁰ rad/s²

Hence the average angular acceleration of the Earth, α  = 6.152 X 10⁻²⁰ rad/s²

3 0
2 years ago
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Degger [83]

Answer:

C) the Fahrenheit thermometer is incorrect

Explanation:

Since

1) K = °C + 273

2) °F = 9/5 °C + 32

for 0 °C

1) K = 0°C + 273 = 273 K

2) °F = 9/5 * 0°C + 32 = 32 °F

Thus the Kelvin thermometer measurement coincides with the Celsius measurement but not with the °F . On the other hand, if the Fahrenheit measurement is right, the Celsius thermometer and the Kelvin one should be wrong.

Therefore is more reasonable to assume that one thermometer failed (the one of Fahrenheit and both Kelvin and Celsius are right ) that 2 thermometers ( Celsius and Kelvin thermometers fail and the one of Fahrenheit is right)

3 0
2 years ago
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