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marta [7]
4 years ago
8

It is estimated that 1kg of body fat will provide 3.8 * 10^7 J of energy. A 67kg mountain climber decides to climb a mountain 35

00 m high how much work is done by gravity on the climber for the trip up to the top of the mountain
Physics
1 answer:
xz_007 [3.2K]4 years ago
4 0
During a climb UP the mountain, gravity does NO work on the climber.
Actually, it's more correct to say that gravity does NEGATIVE work
on him.  The climber has to DO the positive work to haul himself up.
  
                     Work = (mass) x (gravity) x (height) .

For the guy in this problem:

                     Work = (67 kg) x (9.8 m/s²) x (3,500 meters)

                             =  2,298,100 joules.

If he eats no candy bars on the way, and completely depends on
his stored body fat for the energy, then he'll burn off

                       (2,298,100 joules) / (3.8 x 10⁷ joules/kg)

                   =          0.06 kg of fat.

That's only about 2.1 ounces.  We KNOW he'll lose more weight than that,
climbing 11,000 feet.  That's because climbing is pretty inefficient. 
In addition to the potential energy you have to give your body weight,
you also have to expend energy breathing, digesting, metabolizing,
and sweating.
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A car traveled north 25 km for 1 hour. The car’s SPEED is 25 km/hr north.
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6 0
3 years ago
Given that on Earth, gravity causes an acceleration of 9.8 m/s2, what is an acceleration of 7 g?
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68.6 m/s^2

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3 years ago
Mike is standing on the roof of a building looking at the roof of the neighboring building that is 15 meters away and 10 meters
amid [387]

Answer:

Part a)

t = 1.65 s

Part b)

x = 40.4 m

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

v = 27.3 m/s

direction of velocity is given as

[tex]\theta = 26.35 degree

Explanation:

Part a)

acceleration due to gravity on this planet is 3/4 times the gravity on earth

So the acceleration due to gravity on this new planet is given as

a = \frac{3}{4}(9.81)

a = 7.36 m/s^2

now the vertical displacement covered by the canister is given as

y = 10 m

now by kinematics we have

y = \frac{1}{2}gt^2

10 = \frac{1}{2}(7.36)t^2

t = 1.65 s

Part b)

Horizontal speed of the canister is given as

v_x = 24.5 m/s

now the distance moved by it

x = v_x t

x = 24.5 (1.65)

x = 40.4 m

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

Final velocity in X direction will remains the same

v_x = 24.5 m/s

final velocity in Y direction

v_y = v_i + at

v_y = 0 + (7.36)(1.65)

v_y = 12.14 m/s

now magnitude of velocity is given as

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{24.5^2 + 12.14^2}

v = 27.3 m/s

direction of velocity is given as

\theta = tan^{-1}\frac{v_y}{v_x}

\theta = tan^{-1}\frac{12.14}{24.5}

[tex]\theta = 26.35 degree

6 0
3 years ago
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