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Natali5045456 [20]
3 years ago
10

Part a 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 r

ate 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?
Physics
1 answer:
Alex787 [66]3 years ago
5 0
<span>436 km The conversion factor between kilocalorie/hour and watts is 1.163 (1 kcal/hr = 1.163 watt). So let's convert the energy consumption of the bird from watts to kcal/hr 3.7 w / 1.163 w hr/kcal = 3.18 kcal /hr 1 gram of fat has 9 kcal, so the total number of kcals consumed will be 4 * 9 = 36. So the bird can fly for 36/3.18 = 11.32 hours The distance traveled will be 11.32 h * 3600 s/h * 10.7 m/s / 1000 m/km = 436 km</span>
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A hot air balloon is traveling vertically upward at a constant speed of 4.5 m/s. When it is 28 m above the ground, a package is
ella [17]

Answer:

1.97 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0

Solving the above equation we get

t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89

So, the time the package was in the air is 1.97 seconds

3 0
3 years ago
What is the momentum of a .005kg bumble bee that is traveling at a velocity of 3.0m/s?
stiks02 [169]

p=mv

p=0.005kg×3.0m/s

p= 0.015kgm/s

4 0
3 years ago
A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a
Zolol [24]
     Considering the unknown resistence as R and using the Ohm's First Law, we have:

i= \frac{V}{R_{eq}}  \\ 0.5= \frac{12}{R+10}  \\ R+10=24 \\ R=14-Ohm
 
     The equivalent resistence is given by the resistor series with the lamp resistence.

R_{eq}=R+10 \\ R_{eq}=14+10 \\ \boxed {R_{eq}=24-Ohm}

If you notice any mistake in my english, please let me know, because i am not native.

6 0
2 years ago
You can think of the work-kinetic energy theorem as the second theory of motion, parallel to Newton's laws in describing how out
kiruha [24]

Answer:

a) 4 289.8 J

b) 4 289.8 J

c) 6 620.1 N

d) 411 186.3 m/s^2

e) 6 620.1 N

Explanation:

Hi:

a)

The kinetic energy of the bullet is given by the following formula:

K = (1/2) m * v^2

With

    m = 16.1 g = 1.61 x 10^-2 kg

     v = 730 m/s

K = 4 289.8 J

b)

the work-kinetic energy theorem states that the work done on a system is the same as the differnce in kinetic energy of the same. Since the initial state of the bullet was at zero velocity (it was at rest)  Ki = 0, therefore:

W = ΔK = Kf - Ki  = 4 289.8 J

c)

The work done by a force is given by the line intergarl of the force along the trayectory of the system (in this case the bullet).

If we consider a constant force (and average net force) directed along the trayectory of the bullet, the work and the force will be realted by:

W = F * L

Where F is the net force and L is the length of the barrel, that is:

F = (4 289.8 J) / (64.8 cm) = (4 289.8 Nm) / (0.648 m) = 6620.1 N

d)

The acceleration can be found dividing the force by the mass:

a = F/m = (6620.1 N) /(16.1 g) = 411 186.3 m/s^2

e)

The force will have a magnitude equal to c) and direction along the barrel towards the exit

5 0
3 years ago
Which of the following is an example of a force?
strojnjashka [21]

The answer would be d!

6 0
3 years ago
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