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ahrayia [7]
4 years ago
10

. Where does a body have more weight at the pole on

Physics
2 answers:
kherson [118]4 years ago
7 0

Answer:

a body weigh more at the equator since earth is not completely round it is an oblate spheroid – a sphere with a bulge around the equator. that is why the places at the equator are more in contact with the core rather than the places at the poles.

hram777 [196]4 years ago
6 0

Answer:the body has more weight because all of your organs make you have a lot of weight

Explanation:

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qual charges, one at rest, the other having a velocity of 104 m/s, are released in a uniform magnetic field. Which charge has th
Yakvenalex [24]

Answer:

The correct option is D

Explanation:

From the question we are told that

    The speed of the first charge is  v  = 0 \ m/s

This is because it is at rest

     The speed of the second charge is  v  =  10^{4} \ m/s

Generally the force exerted by a magnetic field on a charge is mathematically represented as

      F_q  =  q * v  *  B sin \theta

Now looking at this above equation we can see that F_q can only be maximum at  \theta  =  90 ^o and this only obtained when the direction of the charge (i.e its velocity  ) is perpendicular to the direction of the magnetic field

so the correct option for this question is D

     

4 0
4 years ago
Which of the following most logically completes the argument?A photograph of the night sky was taken with the camera shutter ope
Nikolay [14]

Answer:

(D) the spot could have been caused by an object that emitted a flash that lasted for only a fraction of the time that the camera shutter was open

Explanation:

(A) the spot was not the brightest object in the photograph: The effect described and the brightness of the objects have no relation. Stars of very different brightness will be shown.

(B) the photograph contains many streaks that astronomers can identify as caused by noncelestial objects: Yes, but that doesn't explain the effect described. A plane could leave a streak.

(C) stars in the night sky do not appear to shift position relative to each other: True, at least for relative short times, but that has nothing to do with the effect described, which happens in a very short period of time.

(E) if the camera shutter had not been open for an extended period, it would have recorded substantially fewer celestial objects: True, but quantity of objects does not relate with the particular case described.

(D) the spot could have been caused by an object that emitted a flash that lasted for only a fraction of the time that the camera shutter was open: True, this can happen, for example, with Iridium satellites, they emit a flash (reflect solar light) that lasts a very short time as seen from one point on the surface (the place where the camera is), and something like this could have been captured by the camera shutter, appearing like a point compared to the streaks left by the stars.

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3 years ago
Why did the producers and director decide to have the girls run outside to the water during the climactic moments of Act Three?
jolli1 [7]
Is this a book and most likely because the were cute
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3 years ago
Which is not a high-level radioactive waste ?
Airida [17]
What is the multiple choice???
3 0
3 years ago
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At the nose of a missile in flight, the pressure and temperature are 5.6 atm and 850°R, respectively. Calculate the density and
Contact [7]

To solve this problem we will apply the definition of the ideal gas equation, where we will clear the density variable. In turn, the specific volume is the inverse of the density, so once the first term has been completed, we will simply proceed to divide it by 1. According to the definition of 1 atmosphere, this is equivalent in the English system to

1atm = 2116lb/ft^2

The ideal gas equation said us that,

PV = nRT

Here,

P = pressure

V = Volume

R = Gas ideal constant

T = Temperature

n = Amount of substance (at this case the mass)

Then

\frac{n}{V} = \frac{P}{RT}

The amount of substance per volume is the density, then

\rho = \frac{P}{RT}

Replacing with our values,

\rho = \frac{5.6*2116}{1716*850}

\rho = 0.00812slug/ft^3

Finally the specific volume would be

v = \frac{1}{\rho}

v = 123ft^3/slug

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3 years ago
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