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3 years ago

If a fly gets his wings cut of, is it still a fly?

Physics

2 answers:

jenyasd209 [6]3 years ago

4 0

Answer:

Explanation:

because if the fly get its wings cut off its obviously not going to be able to fly

svetoff [14.1K]3 years ago

3 0

Answer:

Yes I think... But the joke is it is a walk then.

You might be interested in

The sun is about 150 million km away from earth. The rays of sunlight that reach earth’s atmosphere are best represented as ____

tangare [24]

Five

The diagram that represents rays from the sun is given below.

The answer from the diagram is parallel lines.

Six

The formula for this is n_i * Sin(i) = n_r * sin(r)

n1 = 1.333 water

n2 = 1.309 ice

i (angle) = 40o

r (angle) = ?

Solution

1.333 * sin(40) = 1.309*sin(r)

1.333 * sin(40)/1.309 = sin(r)

0.6546 = sin(r)

sin-1(0.6546) = 40.88 degrees.

I guess my closest answer would be A. I'm not rounding at all and that's the answer I get.

6 0

3 years ago

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =

Sloan [31]

Answer:

The current is $I = 8.9 *10^{-5} \ A$

Explanation:

From the question we are told that

The radius is $r = 3.17 \ mm = 3.17 *10^{-3} \ m$

The current density is $J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2$

The distance we are considering is $r = 0.5 R = 0.001585$

Generally current density is mathematically represented as

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

=> $I = J * (\pi r^2 )$

Now the change in current per unit length is mathematically evaluated as

$dI = 2 J * \pi r dr$

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

$I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr$

$I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr$

$I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.$

$I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]$

substituting values

$I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]$

$I = 8.9 *10^{-5} \ A$

5 0

4 years ago

3N 6N what is the net force of this diagram?

11111nata11111 [884]

Answer:

where is the diagram..........

5 0

3 years ago

A string along which waves can travel is 4.36 m long and has a mass of 222 g. The tension in the string is 60.0 N. What must be

lora16 [44]

Answer:

frequency is 195.467 Hz

Explanation:

given data

length L = 4.36 m

mass m = 222 g = 0.222 kg

tension T = 60 N

amplitude A = 6.43 mm = 6.43 × $10^{-3}$ m

power P = 54 W

to find out

frequency f

solution

first we find here density of string that is

density ( μ )= m/L ................1

μ = 0.222 / 4.36

density μ is 0.050 kg/m

and speed of travelling wave

speed v = √(T/μ) ...............2

speed v = √(60/0.050)

speed v = 34.64 m/s

and we find wavelength by power that is

power = μ×A²×ω²×v / 2 ....................3

here ω is wavelength put value

54 = ( 0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 ) / 2

0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 = 108

ω² = 108 / 7.160 × $10^{-5}$

ω = 1228.16 rad/s

so frequency will be

frequency = ω / 2π

frequency = 1228.16 / 2π

frequency is 195.467 Hz

7 0

4 years ago

How would the atmosphere of earth differ if the earth was much larger

Juli2301 [7.4K]

If the earth just doubles its size in an instant, the oxygen present in the atmosphere could be much larger. Due to these, we could theorize that insects could grow much larger because the size of their body is heavily dependent on the maximum amount of oxygen present in the atmosphere. Air pressure also increases thus those aerial enthusiasts would enjoy their professions.

4 0

4 years ago

Read 2 more answers