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djverab [1.8K]
3 years ago
10

A honey bee flaps its wings 200 times per second. How much time is required for one wingbeat? Give your answer in milliseconds.

Physics
2 answers:
Schach [20]3 years ago
7 0

Answer:

The time is required for one wing beat is 5 millisecond.

Explanation:

Given that,

Frequency = 200 /sec

We need to calculate the time period

Using formula of time period

Time period is the reciprocal of the frequency.

T=\dfrac{1}{f}

T=\dfrac{1}{200}

T=0.005\ sec

T= 5\ miliseconds

Hence, The time is required for one wing beat is 5 millisecond.

kakasveta [241]3 years ago
3 0
A bees wings move so rapidly that studying them, even seeing them, has proved difficult.
The honeybees have a rapid wing beat honeybee flaps its wings 230 times every second.

Therefore, a bee flaps its wings over 200 times in about 1ms
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Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 80.5 N, Jill pulls with 81.7 N in the
Gnesinka [82]

Answer:

F = 233.52 N,  θ' = 351.41º

Explanation:

In this exercise we must find the net force applied on the donkey.

For this we use Newton's second law, where we create a reference frame with the horizontal x axis

let's decompose the forces

Jack

        = 80.5 N

Jill

       cos 45 = F_{2x} / F₂2

       sin 45 = F_{2y} / F₂2

       F_{2x} = F₂ cos 45

       F_{2y} = F₂ sin 45

       F_{2x} = 81.7 cos 45 = 57.77 N

       F_{2y} = 81.7 sin 45 = 57.77 N

Jane

      cos (270 + 45) = F_{3x} / F₃3

      sin 315 = F_{3y} / F₃

      F_{3x} = 131 cos 315 = 92.63 N

      F_{3y} = 131 sin 315 = -92.63 N

the force can be found in each axis

X axis

         F_{x} = F_{1x} + F_{2x} + F_{3x}

         F_{x} = 80.5 +57.77 + 92.63

         F_{x} = 230.9 N

Axis y

         F_{y} = F_{1y} + F_{2y} + F_{3y}

         F_{y} = 0 + 57.77 -92.63

         F_{y} = -34.86 N

we can give the result in two ways

a) F = (230.9 i ^ - 34.86 j ^) N

b) in the form of module and angle

we use the Pythagorean theorem

         F = √(Fₓ² + F_{y}²

        F = √(230.9² + 34.86²)

        F = 233.52 N

let's use trigonometry for the angle

        tan θ = \frac{F_y}{F_x} }

        θ = tan⁻¹ (\frac{F_y}{F_x} })

        θ = tan⁻¹ (-34.86 / 230.9)

        θ = -8.59º

if we measure this angle from the positive side of the x-axis counterclockwise

          θ' = 360 -θ

          θ‘= 360- 8.59

          θ' = 351.41º

5 0
3 years ago
A diver springs upward from a board that is 2.70 m above the water. At the instant she contacts the water her speed is 10.9 m/s
TiliK225 [7]

Answer:

vo=5.87m/s

Explanation:

Hello! In this problem we have a uniformly varied rectilinear movement.

Taking into account the data:

α =69.2

vf = 10m / s

h=2.7m

g=9.8m/s2

We know we want to know the speed on the y axis.

We calculate vfy

vfy = 10m / s * (sen69.2) = 9.35m / s

We can use the following equation.

vf^{2} =vo^{2}+2*g*h\\

We clear the vo (initial speed)

vo=\sqrt{vf^{2}-2*g*h }

v0=\sqrt{(9.35m/s)^{2}-2*9.8m/s^{2} *2.7m}

vo=5.87m/s

7 0
3 years ago
Is the star moving toward earth, away from earth, or is there not enough information provided to determine its motion?
ivanzaharov [21]

If a star is moving towards Earth, shift towards the blue end of the spectrum, this is called blue shift. If the star is moving away from Earth the light from that star will be red and is called red shift .

The faster a star moves towards the earth, the more its light is shifted to higher frequencies. In contrast, if a star is moving away from the earth, its light is shifted to lower frequencies on the color spectrum

if a star is moving towards Earth, it appears to emit light that is shorter in wavelength compared to a source of light that isn't moving. Because shorter wavelengths correspond to a shift towards the blue end of the spectrum, this is called blue shift.

If the star is moving away from Earth, its light will lose energy to reach Earth, therefore the light from that star will be red and is called red shift

learn more about blue shift :

brainly.com/question/5368237?referrer=searchResults

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8 0
2 years ago
A 0.500-V potential difference is maintained across a 1.50-m length of tungsten wire that has a cross-sectional area of 0.800 mm
Genrish500 [490]

Answer:

5.95 A

Explanation:

From the question

R = ρL/A..................... Equation 1

Where R = resistance of the tungsten wire, ρ = Resistivity of the tungsten wire, L = length, A = cross sectional area.

Given: L = 1.5 m, A = 0.8 mm² = 0.8×10⁻⁶ m, ρ = 5.60×10⁻⁸ Ω.m

Substitute these values into equation 1

R = 1.5(5.60×10⁻⁸)/0.8×10⁻⁶

R = 0.084 Ω.

Finally, using Ohm law,

V = IR

Where V = Voltage, I = current

Make I the subject of the equation

I = V/R............... Equation 2

I = 0.5/0.084

I = 5.95 A

4 0
3 years ago
Does any part of our body vibrate when we speak ? Name the part
Nadusha1986 [10]

When YOU speak, YOUR vocal cords vibrate, sending sound waves through the air and causing MY ear drums to vibrate.

When I speak, MY vocal cords vibrate, sending sound waves through the air an causing YOUR ear drums to vibrate.

7 0
4 years ago
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