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3 years ago

15

A typical flying insect applies an average force equal to twice its weight during each downward stroke while hovering. Take the

mass of the insect to be 7.0g , and assume the wings move an average downward distance of 1.5cm during each stroke. Assuming 117 downward strokes per second, estimate the average power output of the insect.

1 answer:

Lelu [443]3 years ago

6 0

Answer:

Average power output of insect is 2.42W

Explanation:

Workdone by constant force during displacement is given by:

W= F× d cos theta

Where theta is angle between F and d.

Power output due to the force over the interval time is given by:

P= Workdone/change in time

Ginen:

Mass of insect,m= 7.0g= 7/1000 = 0.07kg

Downward force applied by insect,F= 2mg

Distance moved by the wing each stroke=1.5cm=1.5/100= 0.015m

W= F× d cos theta

Where theta=0° since force is in the same direction as the displacement.

W= 2mg×d

W= 2× 0.07 × 9.8 × 0.015

W= 0.02058J

Power output = W/ change in time

Since wings make 117strokes each second time interval is 1/117 = 8.5×10^-3seconds

Power= 0.02058/(8.5×10^-3)

Power= 2.42W

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4 years ago

A man is sitting on a chair with wheels. He grabs a 2.1 kg book from the desk and throws

geniusboy [140]

The speed of the man and the chair after the book is thrown is 0.2 m/s.

The given parameters:

<em>mass of the book, m₁ = 2.1 kg</em>
<em>speed of the book, u₁ = 7.2 m/s</em>
<em>mass of the man, M = 70 kg</em>
<em>mass of the book, m = 9.2 kg</em>

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The total mass of the man and the book is calculated as follows;

m₂ = 70 kg + 9.2 kg

m₂ = 79.2 kg

The speed of the man and the chair after the book is thrown is determined by applying the principle of conservation of linear momentum;

$m_1 u_1 = m_2u_2\\\\u_2 = \frac{m_1u_1}{m_2} \\\\u_2 = \frac{2.1 \times 7.2}{79.2} \\\\u_2 = 0.2 \ m/s$

Thus, the speed of the man and the chair after the book is thrown is 0.2 m/s.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

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3 years ago

What is the momentum of a 200 kg football player running west at a speed of 8m/s

lutik1710 [3]

Answer:

<h3>The answer is 1600 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 200 kg

velocity / speed = 8m/s

We have

momentum = 200 × 8

We have the final answer as

<h3>1600 kgm/s</h3>

Hope this helps you

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4 years ago

A grapefruit falls from a tree and hits the ground .75 seconds later. How far did the grapefruit drop? What was its speed?

Ivanshal [37]

1) The grapefruit is in free fall, so it moves by uniformly accelerated motion, with constant acceleration $g=9.81 m/s^2$ . Calling h its height at t=0, the height at time t is given by
$h(t)=h- \frac{1}{2}gt^2$
We are told thatn when $t=0.75 s$ the grapefruit hits the ground, so h(0.75 s)=0. If we substitute these data into the equation, we can find the initial height h of the grapefruit:
$0=h- \frac{1}{2}gt^2$
$h= \frac{1}{2}gt^2= \frac{1}{2}(9.81 m/s^2)(0.75 s)^2=2.76 m$

2) The speed of the grapefruit at time t is given by
$v(t)=v_0 +gt$
where $v_0=0$ is the initial speed of the grapefruit. Substituting t=0.75 s, we find the speed when the grapefruit hits the ground:
$v(0.75 s)=gt=(9.81 m/s^2)(0.75 s)=7.36 m/s$

3 0

4 years ago

A sanding disk with rotational inertia 3.8 x 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of magnit

Elis [28]

Answer:

(a) Angular momentum of disk is $1.343\, \frac{kg.m^{2}}{s}$

(b) Angular velocity of the disk is $353\frac{rad}{s}$

Explanation:

Given

Rotational inertia of the disk , $I=3.8\times 10^{-3}kg.m^{2}$

Torque delivered by the motor , $\tau =17N.m$

Torque is applied for duration , $\Delta t=79ms=0.079s$

(a)

Magnitude of angular momentum of the disk = Angular impulse produced by the torque

$\therefore L=\tau \Delta t=17\times 0.079\frac{kg.m^{2}}{s}$

=> $L=1.343\, \frac{kg.m^{2}}{s}$

Thus angular momentum of disk is $1.343\, \frac{kg.m^{2}}{s}$

(b)

Since Angular momentum , $L=I\omega$

where $\omega$ = Angular velocity of the disk

$=>\omega =\frac{L}{I}=\frac{1.343}{3.8\times 10^{-3}}\frac{rad}{s}$

$\therefore \omega =353\frac{rad}{s}$

Thus angular velocity of the disk is $353\frac{rad}{s}$

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3 years ago