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AleksandrR [38]
3 years ago
11

A turtle and a rabbit are in a 150 meter race. The rabbit decides to give the turtle a 1 minute head start. The turtle moves at

a constant speed of 0.500 m/s through the whole race (in fact the turtle even starts at a velocity of 0.500 m/s as while he was still approaching the starting line he was allowed to continue to keep going without stopping). The rabbit starts the race from rest and accelerates at a rate or 1.50 m/s2 until she reaches her top speed of 10 m/s. She then finishes the race running at a constant speed of 10 m/s. a) What is the turtle’s position when the rabbit starts to run (1 minute into the race)? b) How long does it take the turtle to finish the race? c) How long does it take the rabbit to reach max speed? d) What is the rabbit’s position when she reaches max speed? e) How long does it take the rabbit to finish the race? f) Who won?
Physics
1 answer:
yan [13]3 years ago
7 0

Answer:

a) s_{T} = 30\,m, b) t = 5\,min, c) \Delta t = 6.667\,s, d) \Delta s_{R} = 33.333\,m, e) t' = 11.667\,s, f) The rabbit won the race.

Explanation:

a) As turtle moves at constant speed, its position is determined by the following formula:

s_{T} = v_{T}\cdot t

Where:

t - Time, measured in seconds.

v_{T} - Velocity of the turtle, measured in meters per second.

s_{T} - Position of the turtle, measured in meters.

Then, the position of the turtle when the rabbit starts to run is:

s_{T} = \left(0.5\,\frac{m}{s} \right)\cdot (60\,s)

s_{T} = 30\,m

The position of the turtle when the rabbit starts to run is 30 meters.

b) The time needed for the turtle to finish the race is:

t = \frac{s_{T}}{v_{T}}

t = \frac{150\,m}{0.5\,\frac{m}{s} }

t = 300\,s

t = 5\,min

The time needed for the turtle to finish the race is 5 minutes.

c) As rabbit experiments a constant acceleration until maximum velocity is reached and moves at constant speed afterwards, the time required to reach such speed is:

v_{R} = v_{o,R} + a_{R}\cdot \Delta t

Where:

v_{R} - Final velocity of the rabbit, measured in meters per second.

v_{o,R} - Initial velocity of the rabbit, measured in meters per second.

a_{R} - Acceleration of the rabbit, measured in \frac{m}{s^{2}}.

\Delta t - Running time, measured in second.

\Delta t = \frac{v_{R}-v_{o,R}}{a_{R}}

\Delta t = \frac{10\,\frac{m}{s}-0\,\frac{m}{s}}{1.50\,\frac{m}{s^{2}} }

\Delta t = 6.667\,s

The time taken by the rabbit to reach maximum speed is 6.667 s.

d) On the other hand, the position reached by the rabbit when maximum speed is reached is determined by the following equation of motion:

v_{R}^{2} = v_{o,R}^{2} + 2\cdot a_{R}\cdot \Delta s_{R}

\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}

\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}

Where \Delta s_{R} is the travelled distance of the rabbit from rest to maximum speed.

\Delta s_{R} = \frac{\left(10\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(1.50\,\frac{m}{s^{2}} \right)}

\Delta s_{R} = 33.333\,m

The distance travelled by the rabbit from rest to maximum speed is 33.333 meters.

e) The time required for the rabbit to finish the race can be determined by the following expression:

t' = \frac{\Delta s_{R}}{v_{R}}

t' = \frac{150\,m-33.333\,m}{10\,\frac{m}{s} }

t' = 11.667\,s

The time required for the rabbit from rest to maximum speed is 11.667 seconds.

f) The animal with the lowest time wins the race. Now, each running time is determined:

Turtle:

t_{T} = 300\,s

Rabbit:

t_{R} = 60\,s + 6.667\,s + 11.667\,s

t_{R} = 78.334\,s

The rabbit won the race as t_{R} < t_{T}.

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Nepal has high potential for producing hydroelectricity however it is difficult too.
Stels [109]

Answer:

I'm not a scholar of hydroelectric power in Nepal, so consider my answers carefully, below.

Explanation:

High Potential:  Hydroelectric power comes from the potential energy stored in a mass that is above Earth's surface.  As the word "hydro" implies, the mass in this case is water.  Water from snow and glacier melt, and from normal precipitation (rain) in mountainous regions eventually cascades down the mountains in fast-flowing rivers or waterfalls.  Often, there are lakes or man-made reservoirs to collect and store the water before it flows down.  Mt. Everest is 8848 meters tall (about 29,000 feet).    If a lake forms at just 2,000 meters, one can calculate the amount of energy in each kilogram of water stored in the lake that represents the potential energy available at that altitude.  1 kg of water at 2,000 meters has potential energy, PE, according to the equation:  PE = mgh, where m is the mass in kg, g is Earth's acceleration due to gravity (9.8 m/sec^2), and h is height, in meters.  

PE = mGH

PE = (1 kg)*(9.8 m/sec^2)*(2,000 meters) = 19,000 kg*m/sec^2

1 kg*m/sec^2 is the SI unit for 1 Joule, a measure of energy.

This potential energy can be converted into electrical energy by releasing the water so that it can flow down to a water-powered turbine that spins magnets and coils of wire that produce electricity.  The 19,000 Joules of water potential energy can be converted to electrical power, less any inefficiency in the system, such as friction.

Nepal has the natural advantage in that it has many high mountain ranges with water flows that can be used for generating electrical power.  The result is low operating costs (the fuel is the flowing water) and no greenhouse gas emissions

The difficulty in developing hydroelectric power in Nepal is due to the same factor that gives it an advantage:  it is difficult constructing large hydroelectric plants in such rough terrain, and the power lines that are needed to transport the power to its destination are expensive and difficult to maintain and repair.

8 0
2 years ago
Mark and David are loading identical cement blocks onto David’s pickup truck. Mark lifts his block straight up from the ground t
Pepsi [2]

Answer:

b) true. The jobs are equal

Explanation:

The work on a body is the scalar product of the force applied by the distance traveled.

    W = F. d

Work is a scalar, the work equation can be developed

    W = F d cos θ

Where θ is the angle between force and displacement

Let's apply these conditions to the exercise

a) False, if we see the expression d cosT is the projection of the displacement in the direction of the force, so there may be several displacement, but its projection is always the same

b) true. The jobs are equal dx = d cosθ

c) False, because the force is equal and the projection of displacement is the same

d) False, knowledge of T is not necessary because the projection of displacement is always the same

e) False mass is not in the definition of work

5 0
3 years ago
A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t
Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The  Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6×10^{24} be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67×10^{-11} N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v=\sqrt{\frac{Gmm}{R+h} }         [(R+h) is the distance of the satellite   from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = h_{i} = 98 km = 98000 m

∴Initial Energy (E_{i})  = \frac{1}{2}mv^{2} + \frac{-GMm}{(R+h_{i} )}

Substituting v=\sqrt{\frac{GMm}{R+h_{i} } } in the above equation and simplifying we get,

E_{i} = \frac{-GMm}{2(R+h_{i}) }

Similarly for final condition,

h=h_{f} = 198km = 198000 m

∴Final Energy(E_{f}) = \frac{-GMm}{2(R+h_{f}) }

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = E_{f} - E_{i}

        = \frac{GMm}{2}(\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} })

Substituting ,

M = 6 × 10^{24} kg

m = 1036 kg

G = 6.67 × 10^{-11}

R = 6400000 m

h_{i} = 98000 m

h_{f} = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = \frac{1}{2}m[v_{f} ^{2} - v_{i} ^{2}]

                                                          = \frac{GMm}{2}[\frac{1} {R+h_{f} } - \frac{1} {R+h_{i} }]

                                                          = -ΔE                                                            

                                                          = - 484.438 MJ

c)  Change in Potential Energy (ΔPE) = GMm[\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} }]

                                                             = 2ΔE

                                                             = 968.907 MJ

3 0
3 years ago
A small steel roulette ball rolls around the inside of a 30 cm diameter roulette wheel. It is spun at 150 rpm, but is slows to 6
liraira [26]

Solution :

Given

Diameter of the roulette ball = 30 cm

The speed ball spun at the beginning = 150 rpm

The speed of the ball during a period of 5 seconds = 60 rpm

Therefore, change of speed in 5 seconds = 150 - 60

                                                                      = 90 rpm

Therefore,

90 revolutions in 1 minute

or In 1 minute the ball revolves 90 times

i.e. 1 min = 90 rev

     60 sec = 90 rev

        1 sec = 90/ 60 rec

         5 sec = $\frac{90}{60}\times 5$

                   = 75 rev

Therefore, the ball made 75 revolutions during the 5 seconds.

7 0
2 years ago
Two forces that are not equal in size are
Bumek [7]
They are unbalanced forces ..... Hope this helps :3
4 0
2 years ago
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