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Irina-Kira [14]
2 years ago
15

Eating 2500 Cal every day a friend of mine maintains a stable weight of 70 kg. One day, after eating 3500 Cal, he decided to do

extra exercise to avoid gaining weight. He was doing jumps: he leaves the ground with a speed of 3.3m/s at every jump. Assuming that his body turns energy to mechanical work with a 25 % efficiency, how many jumps he will have to make
Physics
1 answer:
Ne4ueva [31]2 years ago
4 0

Answer:

2721

Explanation:

First, we find the change in calories

E = E(taken) - E(daily)

E = 3500 - 2500

E = 1000 cal

Next, we use the efficiency to find the available metabolic energy

E' = n.E, where n = efficiency

E' = 0.25 * 1000

E' = 250 cal

Converting this cal to J, we have

1 cal = 4.18 J

250 cal = 250 * 4.18 J

250 cal = 1045 kJ

Given that the final velocity is 0, we use equation of motion to find the height, h attained

u² = 2gh

h = u²/2g

h = 3.3² / 2 * 9.81

h = 10.89 / 19.62

h = 0.56 m

Potential energy, E = mgh

E = 70 * 9.8 * 0.56

E = 384 J

Total number of jumps, n =

n = 1045 / 384

n = 2721 jumps

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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
2 years ago
A boy pulls a 28.0-kg box with a 230-N force at 35° above a horizontal surface. If the coefficient of kinetic friction between t
ddd [48]

Answer:

1977.696 J

Explanation:

Given;

Weight of the box = 28.0 kg

Force applied by the boy = 230 N

angle between the horizontal and the force = 35°

Therefore,

the horizontal component of the force = 230 × cosθ

= 230 × cos 35°

= 188.405 N

Coefficient of kinetic friction, μ = 0.24

Force by friction, f = μN

here,

N = Normal force = Mass × acceleration due to gravity

or

N = 28 × 9.81 = 274.68 N

therefore,

f = 0.24 × 274.68

or

f = 65.9232 N

Now,

work done by the boy, W₁ = 188.405 N × Displacement  

= 188.405 N × 30

= 5652.15 J

and,

the

work done by the friction, W₂ = - 65.9232 N × Displacement  

= - 65.9232 N × 30 m

= - 1977.696 J

[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]

8 0
3 years ago
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