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nekit [7.7K]
4 years ago
9

Finding the work done in pulling a stranded climber to safety. In an unfortunate accident, a rock climber finds herself stuck 17

m from the top of a 80 m rock face. Rescuers sent to help the climber lower a harness attached to a cable that will pull the climber to the top of the rock face. If the climber, secured in the harness, weighs 56 kg and the cable weighs 0.7kgm, how much work is done in raising the climber, harness, and cable to the top of the rock face? Note: the weight of the climber and cable is really a mass so you will need to multiply the mass in kg by the acceleration due to gravity (approximately 9.8ms2).
Physics
1 answer:
Gnesinka [82]4 years ago
4 0

Answer:

-9446.22 J

Explanation:

Parameters given:

Mass of climber and harness = 56kg

Mass of cable = 0.7kg

Distance between climber and top of rock face = 17m

The work done in pulling the climber is given as:

W = F * d

F is the force applied on the rope. It is opposite the force of gravity pulling the climber, hence, it is given as:

F = -Fg

Fg = force of gravity

Fg = m * g

g = acceleration due to gravity.

The mass of the climber, harness and cable = 56 + 0.7 = 56.7kg

=> Fg = 56.7 * 9.8

Fg = 555.66 N

Therefore, the work done will be:

W = - 555.66 * 17

W = -9446.22 J

The negative value of work means that the work done is opposite the value of the force acting on the climber.

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Given parameters:

Speed of car = 50m/s

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Unknown;

Distance of travel = ?

To solve this problem, we need to understand how speed relates with distance and time.

 Speed is a physical quantity. It is the distance traveled divided by time taken.

                Speed  = \frac{distance}{time}

   So,

                Distance  =  speed x time

Input the parameters and solve for the distance;

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The distance covered during this time interval and speed is 15000m

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What direction does centripetal acceleration point?
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3 0
3 years ago
A spaceship maneuvering near planet zeta is located at r⃗ =(600i^−400j^+200k^)×103km, relative to the planet, and traveling at v
slava [35]

<u>Answer:</u>

 The spaceship's position when the engine shuts off = (708.15 i - 444.1 j + 200 k)*10^3km

<u>Explanation:</u>

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  Initial velocity = 9500 i m/s

  Acceleration = (40 i - 20 k)10^3m/s^2

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 We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  Substituting

       s= (9500 i)*2100+\frac{1}{2}*(40 i - 20 k)*2100^2\\ \\ s=9500*2100 i+20*2100^2i-10*2100^2j\\ \\ s=19.95*10^6i+88.2*10^6i-44.1*10^6j\\ \\ \\s=(108.15i-44.1j)*10^6m

     So final position = ((600 i - 400 j + 200 k)+(108.15i-44.1j))*10^6=(708.15 i - 444.1 j + 200 k)*10^6m

                              =(708.15 i - 444.1 j + 200 k)*10^3km

    The spaceship's position when the engine shuts off = (708.15 i - 444.1 j + 200 k)*10^3km

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ella [17]

Answer:

Explanation:

Hi there,

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or 4.12 x 60

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or 247.2 X 60

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Hope this helps :-)

6 0
4 years ago
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