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navik [9.2K]
2 years ago
7

Assume a planet is a uniform sphere of radius R that (somehow) has a narrow radial tunnel through its center. Also assume we can

position an apple anywhere along the tunnel or outside the sphere. Let FR be the magnitude of the gravitational force on the apple when it is located at the planet's surface.
a. How far from the surface is there a point where the magnitude is 1/2FR if we move the apple away from the planet?
b. How far from the surface is there a point where the magnitude is 1/2FR if we move the apple into the tunnel?
Physics
1 answer:
Drupady [299]2 years ago
5 0

Answer:

Explanation:

a )

Force at the surface = FR

Value of g at height h = g( 1 - 2 h / R )

force at height h = F_R ( 1 - 2 h / R)

F_R / 2 = F_R ( 1 - 2 h / R)

=  F_R   - 2  F_R xh / R

F_R / 2 = 2  F_R x h / R

1 / 2 = 2h / R

4h = R

h = R / 4 . Ans

b )

Value of g at depth  d = g( 1 -  d / R )

force at depth d  = F_R ( 1 - d / R)

F_R / 2 = F_R ( 1 - d / R)

=  F_R   -   F_R xd / R

F_R / 2 =   F_R x d / R

1 / 2 = d / R

2d  = R

d = R / 2

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Speed is constant. 50 miles = 1 hour. 600/50 = 12. 1hr(12) = 12 hours.
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According to Newton’s first law of motion what will an object in motion do when no external force acts on it?
PIT_PIT [208]
The object will continue moving in a straight line at constant speed.
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3 years ago
A string with a length of 4.00 m is held under a constant tension. The string has a linear mass density of \mu=0.000600~\text{kg
yulyashka [42]

Answer:

T=245.76N

Explanation:

We know that the frequency of the nth harmonic is given by f_n=nf, where f is the fundamental harmonic. Since we have the values of two consecutive frequencies, we can do:

f_{n+1}-f_n=(n+1)f-nf=nf+f-nf=f

Which for our values means (we do not need the value of <em>n</em>, that is, which harmonics are the frequencies given):

f=f_{n+1}-f_n=480Hz-400Hz=80Hz

Now we turn to the formula for the vibration frequency of a string (for the fundamental harmonic):

f=\frac{1}{2L} \sqrt{\frac{T}{\mu}}

So the tension is:

T=\mu(2Lf)^2

Which for our values is:

T=(0.0006kg/m)(2(4m)(80Hz))^2=245.76N

6 0
2 years ago
a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

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Alekssandra [29.7K]
Greater by 2 on the Richter Scale :)
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3 years ago
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