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Brut [27]
3 years ago
13

At least how many Calories does a mountain climber need in order to climb from sea level to the top of a 5.42 km tall peak assum

ing the muscles of the climber can convert chemical energy to mechanical energy with an efficiency of 16.0 percent. The total mass of the climber and the equipment is 78.4 kg. (Enter your answer as a number without units.)
Physics
1 answer:
Verdich [7]3 years ago
4 0

Answer:

Ec = 6220.56 kcal

Explanation:

In order to calculate the amount of Calories needed by the climber, you first have to calculate the work done by the climber against the gravitational force.

You use the following formula:

W_c=Mgh        (1)

Wc: work done by the climber

g: gravitational constant = 9.8 m/s^2

M: mass of the climber = 78.4 kg

h: height reached by the climber = 5.42km = 5420 m

You replace in the equation (1):

W_c=(78.4kg)(9.8m/s^2)(5420m)=4,164,294.4\ J     (2)

Next, you use the fact that only 16.0% of the chemical energy is convert to mechanical energy. The energy calculated in the equation (2) is equivalent to the mechanical energy of the climber. Then, you have the following relation for the Calories needed:

0.16(E_c)=4,164,294.4J

Ec: Calories

You solve for Ec and convert the result to Cal:

E_c=\frac{4,164,294.4}{016}=26,026,840J*\frac{1kcal}{4184J}\\\\E_c=6220.56\ kcal

The amount of Calories needed by the climber was 6220.56 kcal

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Light from a laser forms a 1.31-mm diameter spot on a wall. If the light intensity in the spot is what is the power output of th
Wewaii [24]

Answer:

0.04973 W

Explanation:

I = Intensity of laser = 3.69\times 10^4\ W/m^2 (assumed, as it is not given)

d = Diameter of spot = 1.31 mm

r = Radius = \dfrac{d}{2}=\dfrac{1.31}{2}\ mm

A = Area = \pi r^2

Power is given by

P=IA\\\Rightarrow P=3.69\times 10^4\times \pi \dfrac{(1.31\times 10^{-3})^2}{4}\\\Rightarrow P=0.04973\ W

The power output of the laser is 0.04973 W

5 0
3 years ago
A spherical wave with a wavelength of 2.0 mm is emitted from the origin. At one instant of time, the phase at rrr = 4.0 mm is πr
max2010maxim [7]

Complete Question

A spherical wave with a wavelength of 2.0 mm is emitted from the origin. At one instant of time, the phase at r_1 = 4.0 mm is π rad. At that instant, what is the phase at r_2 = 3.5 mm ? Express your answer to two significant figures and include the appropriate units.

Answer:

The phase at the second point is  \phi _2  = 1.57 \  rad

Explanation:

From the question we are told that

    The wavelength of the spherical wave is  \lambda =  2.0 \ mm =  \frac{2}{1000} = 0.002 \ m

    The first radius  is  r_1  = 4.0 \ mm  = \frac{4}{1000}  = 0.004 \ m

     The phase at that instant is  \phi _1 = \pi \ rad

     The second radius is  r_2  = 3.5 \ mm  = \frac{3.5}{1000}  = 0.0035 \ m

Generally the phase difference is mathematically represented as

          \Delta  \phi =  \phi _2 -  \phi _1

this can also be expressed as

         \Delta \phi =  \frac{2 \pi }{\lambda } (r_2 - r_1 )

So we have that

   \phi _2 -  \phi _1 =   \frac{2 \pi }{\lambda } (r_2 - r_1 )

substituting values

     \phi _2 -  \pi =   \frac{2 \pi }{0.002 } ( 0.0035 - 0.004 )

    \phi _2  =   \frac{2 \pi }{0.002 } ( 0.0035 - 0.004 ) +   3.142

   \phi _2  = 1.57 \  rad

7 0
3 years ago
A current-carrying wire oriented north-south and laid over a compass deflects the compass 8° to the east. What is the magnitude
zhenek [66]

2.8 \times 10^{-6}\ T is the magnitude of the magnetic field made by the current

<u>Explanation:</u>

Given data:

\theta=8^{\circ}

Magnetic field of earth, B_{\text {earth}}=2 \times 10^{-5}\ \text {tesla}

We need to find the magnetic field of wire, B_{\text {wire }}

The compass needle moves toward a direction of magnetic field. The current in wire makes a magnetic field in available space where the compass is on the ground. The vector sum of the Earth's magnetic field and the wire's magnetic field represents the net magnetic field, as shown in the attached drawing, expressing the angle:

                   \tan \theta=\frac{B_{\text {wire}}}{B_{\text {earth}}}

By substituting the given values, we get

                  B_{\text {wire}}=\tan \theta \times B_{\text {earth}}=\tan 8^{\circ} \times 2 \times 10^{-5}=0.1405 \times 2 \times 10^{-5}

                  B_{\text {wire}}=0.28 \times 10^{-5}=2.8 \times 10^{-6}\ T

4 0
3 years ago
Two-out-of-tune flutes play the same note. One produces a tone that has a frequency of 250 Hz, while the other produces 266 Hz.
Andru [333]

Answer:

 x = 259 Hz

Explanation:

given,

frequency of one tuning fork = 250 Hz

frequency of another tuning fork = 266 Hz

when a tuning fork is sounded together beat frequency heard = 9

 let x be the frequency of unknown

 x - 250 = 9 Hz..............(1)

 x = 259 Hz

when a another tuning fork is sounded together beat frequency heard = 7

266 - x  = 7 Hz..............(2)

 x = 259 Hz

now, on solving both the equation the frequency comes out to be 259 Hz.

so, The frequency of the tuning fork is equal to 259 Hz

6 0
4 years ago
A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia
Sloan [31]

Answer:

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Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0
3 years ago
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