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Vinvika [58]
3 years ago
15

A slinky is traveling down the stairs, like in the video clip below. What is the total KINETIC ENERGY of the slinky at the botto

m of the stairs (just before it stops moving) IF the Height of the stairs is 2 meters, the weight of the slinky is 4.41 Newtons, its spring constant is 0.84 N/m, and the distance the slinky is initially stretched (to get it going) is 0.25 meters??
Physics
1 answer:
yuradex [85]3 years ago
7 0

Answer:

8.79 J

Explanation:

Given that a slinky is traveling down the stairs, like in the video clip below. What is the total KINETIC ENERGY of the slinky at the bottom of the stairs (just before it stops moving) IF the Height of the stairs is 2 meters, the weight of the slinky is 4.41 Newtons, its spring constant is 0.84 N/m, and the distance the slinky is initially stretched (to get it going) is 0.25 meters??

Total energy = mgh

Total energy = 4.41 × 2 = 8.82 J

Elastic potential energy = 1/2 × Ke^2

Elastic potential energy = 1/2 × 0.84 × 0.25^2

Elastic potential energy = 0.02625

Also,

Total energy = P.E + K.E

Substitute them into the formula above

8.82 = 0.02625 + K.E

K.E = 8.82 - 0.02625

K.E = 8.79375

K.E = 8.79 J

Therefore, the KINETIC ENERGY of the slinky at the bottom of the stairs is 8.79 Joules approximately

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#1

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3 years ago
Approximately how much air is in a column 1-cm2 in cross section that extends from sea level to the top of the atmosphere?
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Approximately 101 N air is in a column 1-cm2 in cross-section that extends from sea level to the top of the atmosphere

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At the equator, the radius of the Earth at sea level is 6378.137 km (3963.191 mi). At the poles, it is 6,356.752 km (3,949.903 km), and on average, it is 6,371.001 km (3,958.756 mi). The elevation of the shoreline—the boundary between the ocean and the land—is referred to as sea level. Land that is higher than this altitude is above sea level, and land that is lower is below sea level.

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2 years ago
Two simple pendulum of slightly different length , are set off oscillating in step is a time of 20s has elasped , during which t
adell [148]

Answer:

Length of longer pendulum = 99.3 cm

Length of shorter pendulum = 82.2 cm

Explanation:

Since the longer pendulum undergoes 10 oscillations in 20 s, its period T = 20 s/10 = 2 s.

From T = 2π√(l/g), the length of the pendulum. l = T²g/4π²

substituting T = 2s and g = 9.8 m/s² we have

l = T²g/4π²

= (2 s)² × 9.8 m/s² ÷ 4π²

= 39.2 m ÷ 4π²

= 0.993 m

= 99.3 cm

Now, for the shorter pendulum to be in step with the longer pendulum, it must have completed some oscillations more than the longer pendulum. Let x be the number of oscillations more in t = 20 s. Let n₁ = number of oscillations of longer pendulum and n₂ = number of oscillations of longer pendulum.

So, n₂ = n₁ + x. Also n₁ = t/T₁ and n₂ = t/T₂ where T₂ = period of shorter pendulum.

t/T₂ = t/T₁ + x

1/T₂ = 1/T₁ + x  (1)

Also, the T₂ = t/n₂ = t/(n₁ + x)  (2)

From (1) T₂ = T₁/(T₁ + x) (3)

equating (2) and (3) we have

t/(n₁ + x) = T₁/(T₁ + x)

substituting t = 20 s and n₁ = 10 and T₁ = 2s, we have

20 s/(10 + x) = 2/(2 + x)

10/(10 + x) = 1/(2 + x)

(10 + x)/10 = (2 + x)

(10 + x) = 10(2 + x)

10 + x = 20 + 10x

collecting like terms

10x - x = 20 - 10

9x = 10

x = 10/9

x = 1.11

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substituting x into (2)

T₂ = t/n₂ = t/(n₁ + x)

= 20/(10 + 1)

= 20/11

= 1.82 s

Since length l = T²g/4π²

substituting T = 1.82 s and g = 9.8 m/s² we have

l = T²g/4π²

= (1.82 s)² × 9.8 m/s² ÷ 4π²

= 32.46 m ÷ 4π²

= 0.822 m

= 82.2 cm

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