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kkurt [141]
3 years ago
8

When you jump off the earth, your momentum changes, but the Earth does not move. 1)If momentum is always conserved, why do we no

t feel the earth moving every time someone jumps? 2) If we were to get everyone to jump at once, could we change the momentum of the earth?
Physics
1 answer:
Mazyrski [523]3 years ago
8 0

This question is off-base and misleading from the beginning.

When you jump off the Earth, your momentum changes, <em>and the Earth moves away from you with an equal change of momentum in the opposite direction</em>.

1). Momentum is conserved when you jump.  But we don't feel the Earth moving. Since the Earth's mass is a bazillion times greater than YOUR mass, the speed with which the Earth moves away from you is only one bazillionth of your speed.  That way, the product of (mass) x (speed) is the SAME for you and for the Earth, and momentum is conserved.

2). <em>Of course !</em>  If everyone jumped at the same time, the Earth's momentum would change.  In answer-(1), I explained that the Earth's momentum changes whenever <em>ONE PERSON</em> jumps.  So 7 billion people all jumping at the same time would certainly make it change.

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Answer:

g = 8.61 m/s²

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using formula

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bearhunter [10]

Answer:

B = 38.2μT

Explanation:

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B=\frac{\mu_o I_r}{2\pi r}     (1)

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In this case, you first calculate the current Ir, by using the following relation:

I_r=JA_r

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Ar: cross sectional area for r in the hollow cylinder

Ar is given by  A_r=\pi(r^2-R_1^2)

The current density is given by the total area and the total current:

J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current  = 4 A

Then, the current in the wire for a distance r is:

I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}  (2)

You replace the last result of equation (2) into the equation (1):

B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})

Finally. you replace the values of all parameters:

B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T

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Brainliest?


Hope I helped! ^-^
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Answer:

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Explanation:

HOPE IT HELPS!!

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