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Serjik [45]
3 years ago
14

Which kind of electromagnetic waves has a wavelengths greater than 10 cm?

Physics
1 answer:
bonufazy [111]3 years ago
3 0
Radio waves
Hope this helps
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Enlarged brain ventricles are presumed to signify __________. A. an excess of dopamine activity B. frontal lobe abnormalities C.
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The correct answer is d. deterioration of brain tissue. A brain ventricle is a fluid filled structure located in the parenchyma. The fluid in the ventricles is called cerebro-spinal fluid. An increase in the volume of this fluid can lead to a condition known as hydrocephalus.
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A long uniform wooden board (a playground see-saw) has a pivot point at its center. An older child of mass M=32 kg is sitting a
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Answer:\frac{5L}{6}

Explanation:

Given

Wooden board is pivoted at center and

Older child of mass M=32\ kg is sitting at a distance of L from  center

if two child of mass \frac{M}{2} is sitting at a distance \frac{L}{6} and x(say) from pivot then net torque about pivot is zero

i.e.

\Rightarrow \tau_{net}=MgL-\frac{M}{2}g\frac{L}{6}-\frac{M}{2}gx

as \tau_{net}=0

Therefore

MgL=\frac{M}{2}g\frac{L}{6}+\frac{M}{2}gx

L-\frac{L}{6}=x

x=\frac{5L}{6}

Therefore another child is sitting at a distance of \frac{5L}{6}

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3 years ago
How does a planet’s size affect the length of a rotation of that planet
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There doesn't seem to be any direct connection.
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g In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constr
Lorico [155]

To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.

In other words the acceleration can be described as

a = \frac{GM}{r^2}

Where

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius of Earth

This equation can be differentiated with respect to the radius of change, that is

\frac{da}{dr} = -2\frac{GM}{r^3}

da = -2\frac{GM}{r^3}dr

At the same time since Newton's second law we know that:

F_w = ma

Where,

m = mass

a =Acceleration

From the previous value given for acceleration we have to

F_W = m (\frac{GM}{r^2} ) = 600N

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:

dF_W = mda

dF_W = m(-2\frac{GM}{r^3}dr)

dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})

dF_W = -2F_W(\frac{dr}{r})

But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:

dF_W = -2(600)(\frac{1.6*10^3}{6.37*10^6})

dF_W = -0.3N

Therefore there is a weight loss of 0.3N every kilometer.

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