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sukhopar [10]
4 years ago
14

A 60-kg skier starts at the top of a 10-meter high slope. At the bottom, she is traveling 10 m/s. How much energy does she lose

to friction?
Physics
1 answer:
Karo-lina-s [1.5K]4 years ago
4 0
The potential energy of the skier at the top is
60 kg (9.81 m/s2) (10 m) = 5886 J

Her kinetic energy at the bottom is
(1/2) (60 kg) (10 m/s)^2 = 3000 J

The energy lost to friction is
5886 J - 3000 J = 2886 J
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Answer:

the decay of half of the nuclei only a half-life has passed ,  b) in rock time it is 1 108 years

Explanation:

The radioactive decay is given by

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If half of the atoms have decayed

       ½ N₀ = N₀ e^{\lambda t}

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The definition of average life time is

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We replace

       t = -ln 0.5 / 0.693 10⁻⁸

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b) in rock time it is 1 108 years

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Approximately how many atoms are there along a 9.5 cm line
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First convert 5.5 cm to meters.

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The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti
Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

5 0
3 years ago
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