Consider a bird that flies at an average speed of 10.7 m/s and releases energy from its body fat reserves at an average rate of 3.70 W (this rate represents the power consumption of the bird). Assume that the bird consumes 4.00g of fat to fly over a distance db without stopping for feeding. How far will the bird fly before feeding again?Fat is a good form of energy storage because it provides the most energy per unit mass: 1.00 grams of fat provides about 9.40 (food) Calories, compared to 4.20 (food) Calories per 1.00 grams of carbohydrate. Remember that Calories associated with food, which are always capitalized, are not exactly the same as calories used in physics or chemistry, even though they have the same name. More specifically, one food Calorie is equal to 1000 calories of mechanical work or 4186 joules. Therefore, in this problem use the conversion factor 1Cal=4186J.

Answers

Answer 1

Answer:

Answer:

The distance covered by the bird before feeding is $4.55 * 10^(5)$ m.

Explanation:

As the bird consumes 4 g of fat before flying, the amount of initial food energy ( $E_(F)$ ) stored by it is given by

$E_(F) = 4 g * 9.4 (food) cal = 37.6 (food) cal$

So the mechanical energy stored by the bird ( $E_(M)$ ) is given by

$E_(M) = E_(F) * 4186 J = 1.57 * 10^(5) J$

Given, the power consumed by the bird $P = 3.7 W$

So, the time ( $t$ ) required to consume this power by the bird is

$t = (E_(M))/(P) = (1.57 * 10^(5) J)/(3.7 W) = 4.24 * 10 ^(4) s$

As the bird flies at an average speed ( $v$ ) of $10.7 ms^(-1)$ , so the distance ( $d$ ) covered by the bird before feeding again is given by

$d = v * t = 10.7 ms^(-1) * 4.25 * 10 ^(4) s = 4.55 * * 10^(5) m$

Answer 2

Answer:

The distance of the bird'sflight before him/her feeds again is mathematically given as

d = 4.55* 10^{5} m

What is the distance of the bird's flight before him/her feeds again?

Question Parameter(s):

a bird that flies at an averagespeed of 10.7 m/s

its body fat reserves at an average rate of 3.70 W

the most energy per unit mass: 1.00 grams of fat provides about 9.40 (food) Calories,

Generally, the initial food energy is mathematically given as

Ex= 4 g*9.4

Ex= 37.6cal

Therefore, the mechanical energy

Em = Ex * 4186

Em = 1.57*10^{5} J

In conclusion, time of flight

$t = (E_(M))/(P) \n\n t=(1.57 *10^(5) J)/(3.7 W) \n$

t= 4.24*10 ^{4} s

Th distance hence is

d = v* t

d= 10.7 *4.25*10 ^{4}

d = 4.55* 10^{5} m

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A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and the sound of the whistle is then 290 Hz. What is the speed of the train before and after slowing down?

Answers

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

$f = f_0 ((v_0)/(v_0-v))$

Here,

$f_0$ = Frequency of Source

$v_s$ = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

$f = f_0 ((v_0)/(v_0-v))$

$300 = f_0 ((343)/(343-v))$

$(300*343) - 300v = 343f_0$

Now the second expression will be,

$f' = f_0 ((v_0)/(v_0-v/2))$

$290 = (343)((v_0)/(343-v/2))$

$290*343-145v = 343f_0$

Dividing the two expression we have,

$((300*343) - 300v)/(290*343-145v) = 1$

Solving for v, we have,

$v = 22.12m/s$

Therefore the speed of the train before and after slowing down is 22.12m/s

Final answer:

The speed of the train can be determined using the Doppler effect formula.

Explanation:

The question involves the Doppler effect, which is the change in frequency or wavelength of a wave as observed by an observer moving relative to the source of the wave. In this case, the train whistle's frequency changes from 300 Hz to 290 Hz as the train approaches the station.

To find the speed of the train before and after slowing down, we can use the formula for the Doppler effect:

f' = f((v + v_o)/(v - v_s))

Where:

f' is the observed frequency
f is the source frequency
v is the speed of sound
v_o is the speed of the observer (here it is the train)
v_s is the speed of the source (here it is the speed of sound)

By substituting the given values for observed frequency (290 Hz), source frequency (300 Hz), and the speed of sound (343 m/s), we can solve for the speed of the train before and after slowing down.

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Consider a proton travelling due west at a velocity of 5×10^5m/s. Assuming that the rth magnetic field has a strength of 5x10^-5Tand is directed due south calculate li) the magnitude of the force on the proton (q= 1.6x10^-9C)

Answers

Answer:

Explanation:

What is an inexpensive, portable, and common way to assess body fat in the fitness industry?DEXA
Bioelectrical impedance
Skinfold testing
Hydrostatic weighing

Answers

Answer: Skinfold testing

Explanation:

Skinfold testing, is also referred to as calliper testing and it's used to know the body fat percentage. Skinfold testing is an inexpensive, portable, and common way to assess body fat in the fitness industry.

It is typically done with the use of caliper tapes, marker pens which makes it cheap. Skinfold testing isn't usually accurate which is as a result of human errors.

An F-35 stealth jet takes off from the aircraft carrier Ronald Reagan. Starting from rest, the jet accelerated with a constant acceleration of 55.3 m/s2 along a straight line on the deck. What is the displacement of the jet when it reaches a speed of 181 m/s?

Answers

Answer:

When the jet reaches a speed of 181 m/s, its displacement is 296 m.

Explanation:

Hi there!

The equation of position and velocity of an object traveling with constant acceleration along a straight line are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the object at time t.

If we place the origin of the frame of reference at the point where the jet starts moving, then, x0 = 0. Since the jet starts from rest, v0 is also zero. Then the equations get reduced to the following:

x = 1/2 · a · t²

v = a · t

We know the acceleration and the final velocity of the jet. So, using the equation of velocity, we can find the time it takes the jet to reach that velocity. Then, we can calculate the position of the jet at that time. Since the initial position is zero, the final position of the jet will be equal to the displacement (because displacement = final position - initial position).

v = a · t

v/a = t

181 m/s / 55.3 m/s² = t

t = 3.27 s

The final position of the jet will be:

x = 1/2 · a · t²

x = 1/2 · 55.3 m/s² · (3.27 s)²

x = 296 m

When the jet reaches a speed of 181 m/s, its displacement is 296 m.

The displacement of the F-35 jet when it reaches a speed of 181 m/s is 16515 m.

To find displacement using constant acceleration,

we can use the following equation:

displacement = (final velocity)^2 - (initial velocity)^2 / 2 * acceleration.

In this case, the initial velocity is 0 m/s and the final velocity is 181 m/s.

The acceleration is given as 55.3 m/s^2.

Plugging in these values, we get:

displacement = (181)^2 - (0)^2 / 2 * 55.3 = 16515 m.

The displacement of the F-35 jet when it reaches a speed of 181 m/s is 16515 m.

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a weight is suspended from the ceiling by a spring (k = 20 ln/in) and is connected to the floor by a dashpot producing viscous damping. The damping force is 10 lb when the velocity of the dashpot plunger is 20 in/sec. The weight and plunger have W = 12 lb. What will be the frequency of the damped vibrations?

Answers

Answer:

The frequency of the damped vibrations is 3.82 Hz.

Explanation:

Given that,

Spring constant = 20 lb/in

Damping force = 10 lb

Velocity = 20 in/sec

Weight = 12 lb

We need to calculate the damping constant

Using formula of damping force

$b* v=F_(d)$

$b=(F_(d))/(v)$

Put the value into the formula

$b =(10)/(20)$

$b=0.5\ lb-sec/in$

$b=0.5*12 =6\ lb-sec/ft$

We need to calculate the frequency

Using formula of angular frequency

$\omga=\sqrt{\omega_(0)^2-((b)/(2m))^2}$

$\omega=\sqrt{(k)/(m)-((b)/(2m))^2}$

Put the value into the formula

$\omega=\sqrt{(20*12*32)/(12)-((6*32)/(2*12))^2}$

$\omega=24\ rad/s$

We need to calculate the frequency of the damped vibrations

Using formula of frequency

$f=(\omega)/(2\pi)$

Put the value into the formula

$f=(24)/(2\pi)$

$f=3.82\ Hz$

Hence, The frequency of the damped vibrations is 3.82 Hz.

A child is sitting on the outer edge of a merry-go-round that is 18 m in diameter. if the merry-go-round makes 5.9 rev/min, what is the velocity of the child in m/s?

Answers

Answer: The velocity of the kid is 5.6 m/s

Explanation: Ok, the velocity of the kid will be:

v = w*r

where r is the radius, and w is the frequency.

We know that the diameter is 18m, and the diameter is equal to two times the radius. So r = 18m/2 = 9m

Now, we know that the circumference of a circle is equal to c = 2pi*r, so each revolution has this length, if the kid does 5.9 revolutions in one minute, then the kid spins at v = 5.9*2pi*9m/min

But we want to write this in meters per second, this means that we need to divide it by 60.

v = (5.9*2pi*9/60)m/s = 5.56 m/s

Velocity of the child= 20008.1 m/s

Explanation;

diameter= 18 m

the linear velocity v is given by

v= r ω

r= radius=18/2= 9 m

ω= 5.9 rev/ min = 5.9 rev/min* [2π rad/ 1 rev] *[60 s/ 1 min]=2223.1 rad/s

so V= 9 (2223.1)

V= 20008.1 m/s

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