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natka813 [3]
3 years ago
9

A 2.00 kg block is placed against a spring on a frictionless 36° incline. the spring, whose spring constant is 19.8n/cm, is com

pressed 22.8 cm and then released. what is the elastic potential energy of the compressed spring?
Physics
1 answer:
alekssr [168]3 years ago
5 0
A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs. 
Read more on Brainly.com - brainly.com/question/1581851#readmore
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lisov135 [29]

Answer:

10m/s

Explanation:

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An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi
mel-nik [20]

Answer:

Explanation:

All the displacement will be converted into vector, considering east as x axis and north as y axis.

5.3 km north

D = 5.3 j

8.3 km at 50 degree north of east

D₁= 8.3 cos 50 i + 8.3 sin 50 j.

= 5.33 i + 6.36 j

Let D₂ be the displacement which when added to D₁ gives the required displacement D

D₁ + D₂ = D

5.33 i + 6.36 j + D₂ = 5.3 j

D₂ = 5.3 j - 5.33i - 6.36j

= - 5.33i - 1.06 j

magnitude of D₂

D₂²= 5.33² + 1.06²

D₂ = 5.43 km

Angle θ

Tanθ = 1.06 / 5.33

= 0.1988

θ =11.25 ° south of due west.

4 0
3 years ago
An RL circuit contains a resistor with R = 6800 Ω and an inductor with L = 2300 µH. If the impedance of this circuit is 160,000
Rainbow [258]

| Impedance | = √ [R² +(ωL)²]

R² = 6800² = 4.624 x 10⁷
 
(ωL)² = (2 · π · f · 2.3 · 10⁻³)²

          = 2.0884 x 10⁻⁴  f²

| Z | =  √[ (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²) ]  =  1.6 x 10⁵

     (1.6 x 10⁵)²  =  (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²)

     (2.56 x 10¹⁰) - (4.624 x 10⁷)  =  2.0884 x 10⁻⁴ f²


Frequency² =   (2.56 x 10¹⁰ - 4.624 x 10⁷)  /  2.0884 x 10⁻⁴

                    =       2.555 x 10¹⁰ / 2.0884 x 10⁻⁴

                    =          1.224 x 10¹⁴ 

                    =          122,400 GHz          <== my calculation

                                      11.1 MHz           <== online impedance calculator

Obviously, I must have picked up some rounding errors
in the course of my calculation. 
  











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