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marta [7]
3 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]3 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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What type of specialized cell in the eye is used for detecting low levels of light?
Hunter-Best [27]

Answer:

There are about six to seven million cones in a human eye and are most concentrated towards the macula. Cones are less sensitive to light than the rod cells in the retina (which support vision at low light levels), but allow the perception of color.

8 0
3 years ago
Read 2 more answers
Very large accelerations can injure the body, especially if they last for a considerable length of time. The severity index (SI)
Ludmilka [50]

Answer:

a) The severity index (SI) is 3047.749, b) The injured travels 0.345 meters during the collision.

Explanation:

a) The g-multiple of the acceleration, that is, a ratio of the person's acceleration to gravitational acceleration, is:

a' = \frac{35\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }

a' = 3.569

The time taken for the injured to accelerate to final speed is given by this formula under the assumption of constant acceleration:

v_{f} = v_{o} + a \cdot t

Where:

v_{o} - Initial speed, measured in meters per second.

v_{f} - Final speed, measured in meter per second.

a - Acceleration, measured in \frac{m}{s^{2}}.

t - Time, measured in seconds.

t = \frac{v_{f}-v_{o}}{a}

t = \frac{\left(12\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}  \right)}{35\,\frac{m}{s^{2}} }

t = 0.095\,s

Lastly, the severity index is now determined:

SI = \frac{a'^{5}}{2\cdot t}

SI = \frac{3.569^{5}}{2\cdot (0.095\,s)}

SI = 3047.749

b) The initial and final speed of the injured are 1.944\,\frac{m}{s} and 5.278\,\frac{m}{s}, respectively. The travelled distance can be determined from this equation of motion:

v_{f}^{2} = v_{o}^{2} + 2\cdot a \cdot \Delta s

Where \Delta s is the travelled distance, measured in meters.

\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a}

\Delta s = \frac{\left(5.278\,\frac{m}{s} \right)^{2}-\left(1.944\,\frac{m}{s} \right)^{2}}{2\cdot \left(35\,\frac{m}{s^{2}} \right)}

\Delta s = 0.345\,m.

8 0
3 years ago
Graphs help you see..
Diano4ka-milaya [45]
I think it's B because when you graph you do see the relationship between the dependent and independent variable
8 0
3 years ago
The surface tension of water was determined in a laboratory by using the drop weight method. 100 drops were released from a bure
Lapatulllka [165]

Answer:

γ = 0.06563 N / m

9.78% difference

Explanation:

Solution:-

- Surface tension is the ability of any fluid to resist any external force which causes a decreases in surface area of the impact area due to inward compressive forces. These compressive forces occur due to cohesive nature of the fluid molecules.

- Mathematically, surface tension ( γ ) is defined as the force felt per unit length by the fluid.

                           γ = F / L

Where,

              F: Force imparted

              L: The length over which force is felt

- We are given the mass ( M ) of ( n = 100 ) water droplets to e 3.78 g. The mass of a single droplet ( m ) can be evaluated as follows:

                         m = M / n

                         m = 3.78 / 100

                        m = 0.0378 g        

- The force ( F ) imparted by a single drop of water from the burette can be determined from the force balance on a single droplet. Assuming the distance over which the drop falls is negligible and resistive forces are negligible. Then the only force acting on the droplet is due to gravity:

                        F = m*g

                        F = 0.0378*9.81*10^-3

                        F = 0.000370818 N      

- The length over which the force is felt can be magnified into a circular area with diameter equal to that of a single droplet ( d ). The circumferential length ( L ) of the droplet would be as follows:

                        L = π*d

                        L = π*( 0.0018 )

                        L = 0.00565 m

- Then the surface tension would be:

                        γ = F / L

                        γ = 0.000370818 / 0.00565

                        γ = 0.06563 N / m

- The tabulated value of water's surface tension is given as follows:

                       γa = 0.07275 N/m

- We will determine the percentage difference between the value evaluated  and tabulated value as follows:

                     p.diff = \frac{gamma_a - gamma}{gamma_a} * 100\\\\p.diff = \frac{0.07275- 0.06563}{0.07275} * 100 \\\\p.diff = 9.78 %

- The %difference between is within the allowable practical limits of 10%. Hence, the evaluated value ( γ = 0.06563 N / m ) can be accepted with 9.78% error.

8 0
3 years ago
True or false cold air can hold more moisture than warm air
agasfer [191]
It is true
I hope this helps
5 0
3 years ago
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