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olga55 [171]
3 years ago
13

A 40 kg-skier starts at the top of a 12-meter high slope. At the bottom, she is traveling 10 m/s. How much energy does she lose

to friction? A.She doesn't lose any because mechanical energy is always conserved.
B.4,704 J
C. 2,000 J
D.2,704 J
Physics
2 answers:
julia-pushkina [17]3 years ago
8 0

2704 j is my answer i did the math. good luck

Ad libitum [116K]3 years ago
8 0

Answer : Lost energy, E = 2704 J

Explanation :

Given that,

Mass of the skier, m = 40 kg

Height from bottom, h = 12 m

Velocity at the bottom, v = 10 m/s

At the top she will have potential energy and at the bottom she will have kinetic energy. Finding one by one :

Potential energy, P = m g h

P = 40 kg × 9.8 m/s² × 12 m

P = 4704 J

Kinetic energy, K = 0.5 × m v²

K = 0.5 × 40 kg × (10 m/s)²

K = 2000 J

Lost energy, E = P - K

E = 4704 J - 2000 J

E = 2704 J

The energy that is lost to friction is 2704 J.

So, the correct option is (D) " 2704 J "

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Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc
Sergeeva-Olga [200]

Answer:

a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.

b) Average power

P(w)= 1062.07 [w]

P(hp)=1.42 [hp]

c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.

Explanation:

First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)

Work:

W= F*d= m*g*d

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )

W= F*d= m*g*d=85* 9.8*12.75=10620.75 [J]

Converting from Joules to Kcals:

\frac{10620.75}{4186} =2.537 Kcal

Now lets take into account the efficiency of the human body (20%)

2.537 ---> 20%

 x       ---> 100%

x=\frac{2.537*100}{20} =12.685

So the student is consuming 12.685 KCals each time he runs up the stairs.

Now,

1 g --> 9 Kcals

1000 g --> 9000KCals

Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

\frac{9000Kcals}{12.685Kcals} =709.5 times

He must run up the stairs 709.5 times, to burn 1 Kg of fat.

********************

For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)

Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\

P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp]

*****

4 0
3 years ago
A large solar panel on a spacecraft in Earth orbit produces 1.0 kW of power when the panel is turned toward the sun. What power
Mandarinka [93]

Answer:

e*P_s = 11 W

Explanation:

Given:

- e*P = 1.0 KW

- r_s = 9.5*r_e

- e is the efficiency of the panels

Find:

What power would the solar cell produce if the spacecraft were in orbit around Saturn

Solution:

- We use the relation between the intensity I and distance of light:

                                  I_1 / I_2 = ( r_2 / r_1 ) ^2

- The intensity of sun light at Saturn's orbit can be expressed as:

                                  I_s = I_e * ( r_e / r_s ) ^2

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- We know that P = I*a, hence we have:

                                  P_s = I_s*a

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Hence,                       e*P_s = 11 W

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2 years ago
A 22.0 ohm and 75.0 ohm resistor are in parallel, connected to a 5.00 v battery. how much current flows out of the battery
Julli [10]

Answer:

294 mAmps

Explanation:

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5 0
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