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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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40

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Suppose that, from measurements in a microscope, you determine that a certain bacterium covers an area of 1.50 μm2. Convert this
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1 m = 1 000 000 ym

converted other way we can say that:

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Distance versus Displacement Worksheet
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If the magnitude of the magnetic field is 6.50 mT at a distance of 12.8 cm from a long straight current carrying wire, what is t
Lunna [17]

Answer: magnitude of the magnetic field at a distance of 19.4 cm from the wire=4.29mT

Explanation:

According to  Biot-Savart law, A magnetic field generated by a current  carrying wire at a distance is represented as

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B = magnetic field intensity 1000 mT =1T, 6.50mT = 6.50 X 10^-3T

 μ₀ =permeability of free space  4π × 10−7 H/m

I = current intensity

r = radius, 100cm = 1m, 12.8 cm= 12.8 x 10^-2m

6.50 X 10^-3 =  μ₀ x I/ 2 π X 12.8 X 10^-2

I =6.50  X 10 ^-3 X 2π  X  X 12.8 X 10^-2/  4π × 10−7 H/m

I= 4160 A

when the magnetic field is at 19.4 cm from the wire

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=0.004288

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