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Genrish500 [490]
3 years ago
15

An engineer is designing a runway for an airport. Of the planes that will use the airport, the

Physics
1 answer:
Arada [10]3 years ago
6 0

Answer:

816m

Explanation:

70^2m/s=2(3m/ss)(x)

x=816

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In 3.0 seconds, a wave generator produces 15 pulses that spread over a distance of 45 centimeters. Calculate the wave's frequenc
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John is traveling north at 20 meters/second and his friend Betty is traveling south at 20 meters/second. If north is the positiv
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How much potential energy foes a 5kg mass have when its 2 meters above the ground?(Hint :PE=m*g*h)​
frez [133]

Answer:

<h2>98 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

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From the question we have

PE = 5 × 9.8 × 2

We have the final answer as

<h3>98 J</h3>

Hope this helps you

5 0
3 years ago
Explain why the types of technology valued can vary.
Elan Coil [88]
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4 0
3 years ago
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The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

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Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
3 years ago
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