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2 years ago

005 (part 1 of 2) 10.0 points

Physics

1 answer:

Anton [14]2 years ago

3 0

A constant speed motion is one in which equal distances are covered in equal times

The bird travels a cumulative distance of 7. $\overline 3$ km

The reason the above value is correct is as follows:

The known parameters are;

Speed of the runner, $v_r$ = 2.1 km/hr

The location where the bird begins to fly to the finish line = When the runner is 4.4 km from the finish line

The speed of the bird, $v_b$ = 10.5 km/hr = 5 times the runners speed

The bird reaches the finish line, turns, and returns back to the runner

Required:

The find cumulative distance traveled by the bird

Solution:

The distance the bird travels is five times the distance the runner travels, therefore,

Let x represent the distance the runner ravels before the bird returns, we have;

The distance the bird travels = 4.4 + 4.4 - x = 8.8 - x

The distance the runner travels = x

The time the runner runs x km = The time the bird flies (8.8 - 4) km

$From \ velocity = \dfrac{Distance }{Time}$ , we have;

$Time= \dfrac{Distance }{Velocity}$

Given the time taken by the runner is equal to the time taken by bird, while running, we have;

$Time= \dfrac{x}{2.1} = \dfrac{8.8 - x}{10.5}$

Therefore;

10.5·x = 2.1·(8.8 - x) = 2.1×8.8 - 2.1·x

10.5·x + 2.1·x = 2.1×8.8 = 18.84

12.6·x = 18.84

$x = \dfrac{18.84}{12.6} =\dfrac{22}{15} = 1.4 \overline 6$

$The \ distance \ the \ bird \ travels = 8.8 - \dfrac{22}{15} \approx \dfrac{22}{3} = 7. \overline 3$

The cumulative distance the bird travels is 7. $\overline 3$ km

Learn more about constant speed motion here:

brainly.com/question/12684433

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Directions: Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to the ground. Think about t

satela [25.4K]

1) At the top, the ball has more potential energy

2) Halfway through the fall, potential energy and kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) Potential energy at the top: 784 J

5) Potential energy halfway through the fall: 392 J

6) Kinetic energy halfway through the fall: 392 J

7) KInetic energy before hitting the ground: 784 J

Explanation:

The potential energy of an object is the energy possessed by the object due to its position in a gravitational field. It is given by

$PE=mgh$

where

m is the mass of the object

g is the acceleration of gravity

h is the height of the object above the ground

The kinetic energy of an object is the energy possessed by the object due to its motion, and it is given by

$KE=\frac{1}{2}mv^2$

where v is the speed of the object

For the bowling ball in the problem, when it sits on top of the building it has no kinetic energy (because its speed is zero, v = 0), therefore it has more potential energy than kinetic energy.

The total mechanical energy of the ball, which is the sum of the potential and the kinetic energy, is constant during the fall:

$E=PE+KE=const.$

When the ball is at the top, all its energy is potential energy, since the kinetic energy is zero:

$E=PE=mgH$

where H is the initial height.

When the ball is halfway through the fall, the height is H/2, so:

$PE=mg\frac{H}{2}$

which means that the potential energy is now half of the total mechanical energy: but since the total energy must be constant, this means that the kinetic energy is now also half of the total energy. Therefore, potential energy and kinetic energy are equal.

When the ball is just before hitting the ground, the height of the ball is now zero

h = 0

This also means that the potential energy is zero

PE = 0

Therefore, all the energy of the ball is now kinetic energy:

$KE=E$

which means that the kinetic energy is maximum, and therefore it is larger than the potential energy: this is because the ball accelerates during the fall, and therefore its speed is maximum just before hitting the ground.

The potential energy of the ball is given by

$PE=mgh$

where

m is the mass of the object

g is the acceleration of gravity

h is the height of the object above the ground

When the ball sits at the top, we have

m = 2 kg

$g=9.8 m/s^2$

h = 40 m

Therefore, the potential energy is

$PE=(2)(9.8)(40)=784 J$

The potential energy of the ball is given by

$PE=mgh$

where

m = 2 kg is the mass

$g=9.8 m/s^2$ is the acceleration due to gravity

When the ball is halfway through the fall, the height of the ball is

h = 20 m

Therefore, its potential energy is

$PE=(2)(9.8)(20)=392 J$

which is half of the initial potential energy.

The kinetic energy of the ball is given by

$KE=\frac{1}{2}mv^2$

where

m is the mass of the ball

v is its speed

When the ball is halfway through the fall, we have:

m = 2 kg (mass of the ball)

v = 19.8 m/s (speed of the ball)

Therefore, the kinetic energy is

$KE=\frac{1}{2}(2)(19.8)^2=392 J$

Which is equal to the potential energy.

The kinetic energy of the ball just before hitting the ground is

$KE=\frac{1}{2}mv^2$

where in this case,

m = 2 kg is the mass

v = 28 m/s is the speed of the ball

Therefore, kinetic energy is

$KE=\frac{1}{2}(2)(28)^2=784 J$

And we see that the kinetic energy of the ball just before hitting the ground is equal to the potential energy of the ball when it sits at the top: therefore, all the mechanical energy has converted from potential energy into kinetic energy.

Learn more about kinetic and potential energy:

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