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Maslowich
2 years ago
13

ME Project- Decomposer

Physics
1 answer:
Alenkasestr [34]2 years ago
3 0

Answer:

Explanation:

Decomposer.

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The amplitude of a lightly damped oscillator decreases by 3.42% during each cycle. What percentage of the mechanical energy of t
frozen [14]

Answer:

The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

Explanation:

Mechanical energy (Potential energy, PE) of the oscillator is calculated as;

PE = ¹/₂KA²

During the first oscillation;

PE₁ = ¹/₂KA₁²

During the second oscillation;

A₂ = A₁ - 0.0342A₁ = 0.9658A₁

PE₂ = ¹/₂KA₂²

PE₂ = ¹/₂K (0.9658A₁)²

PE₂ = (0.9658²)¹/₂KA₁²

PE₂ = (0.9328)¹/₂KA₁²

PE₂ = 0.9328PE₁

Percentage of the mechanical energy of the oscillator lost in each cycle;

Change \ in \ percent= \frac{PE_2 - PE_1}{PE_1} \\\\Change \ in \ percent= \frac{0.9328PE_1 -PE_1}{PE_1} *100\\\\Change \ in \ percent= \frac{-0.0672PE_1}{PE_1}*100 \\\\Change \ in \ percent= -0.0672*100\\\\Change \ in \ percent= -6.72 \ \% \\\\Loss \ in \ percent= 6.72 \ \%

Therefore, the percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

4 0
2 years ago
A delivery truck leaves a warehouse and travels 2.60 km north. The truck makes a left turn and travels 1.25 km west before makin
yulyashka [42]

Answer:

4.19 km and 107.35 degrees north of east

Explanation:

So in the end, the truck is (2.6 + 1.4 = 4km) north and 1.25 km west from the warehouse. We can use the Pythagorean formula to calculate the magnitude and direction α of the truck displacement from the warehouse:

s = \sqrt{s_n^2 + s_w^2} = \sqrt{4^2 + 1.25^2} = \sqrt{16 + 1.5625} = \sqrt{17.5625} = 4.19 km

tan\alpha = \frac{s_n}{s_w} = \frac{4}{1.25} = 3.2

\alpha = tan^{-1}3.2 = 1.27 rad \approx 72.65 degrees north or west or (180 - 72.65) = 107.35 degrees north of east

3 0
3 years ago
Which of the following is NOT a mental process?
sergij07 [2.7K]

Answer:

Its A. Walking on the beach

4 0
3 years ago
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Which body is in equilibrium?
sergey [27]
"(1) a satellite moving around Earth in a circular <span>orbit" is the only option from the list that describes an object in equilibrium, since velocity and gravity are working together to keep the orbit constant. </span>
6 0
2 years ago
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A 2 kg ball is thrown upward with a velocity of 15 m/s. What is the kinetic energy of the ball as it is being thrown?
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KE=1/2 m v^2
KE= .5 x 2kg x 15m/s to the 2nd power
KE=225 km/s
7 0
3 years ago
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