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2 years ago

Cual es el diametro de la tierra

Physics

1 answer:

Fed [463]2 years ago

8 0

Plz write it in English

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

Lelu [443]

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

6 0

2 years ago

A biologist is researching antibodies, which are tiny natural proteins in the blood of animals, and she is using methods develop

Sedbober [7]

Answer:

Her friend disagrees because the biologist is not working with nanotechnology which would enable her to be a nanotechnologist.

The biologist is only working with nano antibodies.

8 0

3 years ago

Read 2 more answers

Calculate the force between two small charged spheres having charges of 3×10−7 C and 4×10−7 C placed 20 cm apart.

xxMikexx [17]

Answer:

27 . 10^-7 or 27/1000

Explanation:

We use the Coulomb Law

k = Coloumb Constant

q1 and q2 are the charges

d is the distance between the spheres

3 0

2 years ago

2. A jack exerts a vertical force of 4.5 X 103

skad [1K]

Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

newtons to raise a car 0.25 meter. How much

work is done by the jack?

$\\ \\$

Given :-

$\star \sf \small force = 4.5 \times {10}^{3} \: newton$

$\star \sf \small distance = 0.25 \: meter$

$\\ \\$

To find:-

$\sf \star \: work = \: ?$

$\\ \\$

Solution:-

we know :-

$\bf \dag \boxed{ \rm work = force \times distance}$

$\\ \\$

So:-

$\dashrightarrow \sf work = force \times distance$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{0 \cancel.5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \cancel \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{4\cancel.5}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{3 - 1} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{1} \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times \cancel{10}}{ \cancel2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times 5}{ 1} \\$

$\\ \\$

$\dashrightarrow \sf work =225 \times 10$

$\\ \\$

$\dashrightarrow \bf work =\red{2250\: joule}$

5 0

2 years ago

Some superconductors are capable of carrying a very large quantity of current. If the measured current is 1.00 ´ 105 A, how many

Zielflug [23.3K]

Answer:

The $6.25 \times 10^{23}$ electrons are moving through the superconductor per second.

Explanation:

Given :

Current $I = 1 \times 10^{5}$ A

Charge of electron $e = 1.6 \times 10^{-19}$ C

Time $t = 1$ sec

From the formula of current,

Current is the number of charges flowing per unit time.

$I = \frac{ne}{t}$

Where $n =$ number of charges means in our case number of electrons

$n = \frac{It}{e}$

$n = \frac{1 \times 10^{5} }{1.6 \times 10^{-19} }$

$n = 6.25 \times 10^{23}$

Therefore, $6.25 \times 10^{23}$ electrons are moving through the superconductor per second.

5 0

3 years ago