Find an equation relating the rate of change of kinetic energy to the rate of change of velocity

How far away is the closest star?

sukhopar [10]

Answer: Proxima Centauri is the closet star about 40,208,000,000,000 km away.

Explanation:

8 0

2 years ago

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Birdman is flying horizontally at a

den301095 [7]

Answer:

68 m

Explanation:

Given that the horizontal velocity of the birdman = 17 m/s and

the height, h= 78 m.

The gravitational force is acting in the downward direction, so it will not change the horizontal speed.

The horizontal speed will remains be constant and will be equal to the initial horizontal speed of the turd.

Initially, the turd was also flying horizontally with the birdman, so the initial velocity of the turd is the same as the horizontal velocity of the birdman, i.e In the horizontal direction, $u_0=17$ m/s.

In the vertical direction, u = 0,

The distance to be traveled, in the direction of application of force, is equals to the height of the turd, i.e

s= 78 m

Let t be the time taken to cover a distance of 78 m.

Now, applying the equation of motion in the vertical direction,

$s=ut+\frac 12 at^2$

where u is the initial velocity and a is the acceleration due to gravity in the direction of displacement,s.

Here, $a=g=9.81 m/s^2$ , so

$78=0\times t +\frac 12 (9.81)t^2$

$\Rightarrow t^2=(78\times2)/9.81$

$\Rightarrow t = 4$ seconds.

Hence, the time taken to reach the ground is 4 seconds.

As the horizontal speed, $u_0=17$ m/s, is constant throughout the journey, so

the horizontal distance covered by turd

$= u\times t$

$= 17 \times 4 = 68$ m.

So, the distance of landing from the start of the field is 68 m as the birdman releases a turd directly above the start of the field.

Hence, the robot must hold the bucket at a distance of 68 m from the start of the field.

5 0

3 years ago

Why are there so few eclipses in 2014?

Sonbull [250]

Answer:

Because the Moon casts a smaller shadow than Earth does, eclipses of the Sun tightly constrain where you can see them. If the Moon completely hides the Sun, even for a moment, the eclipse is considered total.

Explanation:

8 0

3 years ago

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He __________ uses a series of gears to adjust the output speed of the engine.

Verizon [17]

Hey ／人 ◕ ‿‿ ◕ 人＼
The answer is transmission

uses a series of gears to transmit power to facilitate changes in speed .

GLAD TO HELP

~~~ ╔͎═͓═͙╗
~~~ ╚̨̈́═̈́﴾ ̥̂˖̫˖̥ ̂ )

4 0

3 years ago

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

a) 6738.27 J

b) 61.908 J

c) $\frac{4492.18}{v_{car} ^{2} }$

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = 6738.27 J

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s