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AveGali [126]
2 years ago
5

If the moon's acceleration due to gravity caused by its gravitational field is one-sixth that of the earth, what is its accelera

tion at a point in space 3R (three moon radii) from its center? (Use the earth's value of "g" as your significant figure reference; i.e., use 2 SF.)
_______ m/s^2
Physics
1 answer:
EastWind [94]2 years ago
7 0

To make it easy, let's call the moon's surface gravity 'Q'.  We know that Q is 1/6 of 'g', but let's just hold onto that for a minute.   Let's first work out what it is at 3 moon radii from the moon's center, and once we have that, relate it back to the Earth.

We know that the strength of gravity is inversely proportional to the square of the distance between the centers of the two objects.  On the moon's surface, you're 1 Moon radius from the center.  At 3 Moon radii from the center, you're 3 times as far from the center, so the gravity out there is (1/3²) = 1/9 of the gravity on the surface.

So at 3 Moon radii from the surface, the Moon's gravity is ( Q/9 ) .  

So far, so good. Now, we know that Q is (1/6) x (Earth 'g').

Moon's gravity at 3 Moon radii = Q/9

Substitute g/6 for Q.

Moon's gravity at 3 Moon radii = (g/6) / 9 .

Moon's gravity at 3 Moon radii = g/54

(9.8 m/s) / 54 = <em>0.18 m/s²</em>

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A 0.75 kg book is pushed across the table with an acceleration of 0.3 m/s2. What force is being applied to the
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Answer:

\boxed {\boxed {\sf 0.225 \ Newtons}}

Explanation:

We are asked to find the force being applied to a book. According to Newton's Second Law of Motion, force is the product of mass and acceleration.

F= ma

The mass of the book is 0.75 kilograms and the acceleration is 0.3 meters per square second. Substitute these values into the formula.

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F= 0.75 \ kg * 0.3 \ m/s^2

Multiply.

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F= 0.225  \ N

<u>0.225 Newtons of force</u> are applied to the book.

5 0
3 years ago
A charge of -2.65 nC is placed at the origin of an xy-coordinate system, and a charge of 2.00 nC is placed on the y axis at y =
stiks02 [169]

Answer:

A. Fnx = 5.71*10⁻⁵ N  ,  Fny= -3.67*10⁻⁵ N

B. Fn= 6.78 *10⁻⁵ N

C. α= 32.4° counterclockwise with the positive x+ axis

Explanation:

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Equivalences

1nC= 10⁻⁹C

1cm = 10⁻²m

Known data

k= 9*10⁹N*m²/C²

q₁= -2.65 nC =-2.65*10⁻⁹C

q₂= +2.00 nC = 2*10⁻⁹C

q₃= +5.00 nC= =+5*10⁻⁹C

d_{13} = \sqrt{(3.2)^{2} +(3.8)^{2} }

d_{13} =\sqrt{24.68} * 10⁻²m    = 4.9678* 10⁻²m

(d₁₃)² = 24.68*10⁻⁴m²

d₂₃ = 3.2 cm = 3.2*10⁻²m  

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.

The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.

Magnitudes of F₁₃ and F₂₃

F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)

F₁₃ = 4.8 *10⁻⁵ N

F₂₃ = (k*q₂*q₃)/(d₂₃)² =  ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)

F₂₃ = 8.8 *10⁻⁵ N

x-y components of F₁₃ and F₂₃

F₁₃x= -4.8 *10⁻⁵ *cos β= - 4.8 *10⁻⁵(3.2/ (4.9678)= - 3.09*10⁻⁵ N

F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) =  - 3.67*10⁻⁵ N

F₂₃x  = F₂₃ =  +8.8 *10⁻⁵ N

F₂₃y = 0

x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)

Fnx= F₁₃x+F₂₃x =  - 3.09*10⁻⁵ N+8.8 *10⁻⁵ N= 5.71*10⁻⁵ N

Fny= F₁₃y+F₂₃y = - 3.67*10⁻⁵ N+0= - 3.67*10⁻⁵ N

Fn magnitude

F_{n} =\sqrt{(Fn_{x})^{2}+(Fn_{y})^{2}  }

F_{n} = \sqrt{(5.71)^{2}+(3.67)^{2}  } *10⁻⁵ N

Fn= 6.78 *10⁻⁵ N

Fn direction  (α)

\alpha =tan^{-1}( \frac{Fn_{y} }{Fn_{x} } )

\alpha =tan^{-1}( \frac{-3.67 }{5.71} )

α= -32.4°

α= 32.4° counterclockwise with the positive x+ axis

4 0
3 years ago
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