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3 years ago

If the distance between two objects is decreased to 1/10 of the original

2 answers:

mihalych1998 [28]3 years ago

5 0

F = GMm/r²

The force follows inverse square law.

If the distance is reduced to (1/10) of the original value,

F = GMm/(1/10r)²

F = GMm/(1/100)r<span>²

F = </span> 100* GMm/r<span>²

Hence the force is 100 times the original value.</span>

GaryK [48]3 years ago

4 0

The force of attraction between two objects is proportional to the mass of the objects and inversely proportional to the square of teh distance between them. In this casem if the distance is decreased to 1/10 of the original, then teh force of attraction increases by 100 times (10^2).

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Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each

AURORKA [14]

Answer: A ( $\sqrt{61}$ ,309.8°)

B (2 $\sqrt{2}$ , 315°)

C ( $3\sqrt{5}$ , 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

$r=\sqrt{x^{2}+y^{2}}$ and $\theta=tan^{-1}(\frac{y}{x})$

For point A:

$r=\sqrt{(-5)^{2}+6^{2}}$

$r=\sqrt{61}$

$\theta=tan^{-1}(\frac{6}{-5})$

$\theta=tan^{-1}(-1.2)$

$\theta=-50.2$ °

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

$\theta=360-50.2$

$\theta=$ 309.8°

Polar coordinates for point A is ( $\sqrt{61}$ , 309.8°)

For point B:

$r=\sqrt{2^{2}+(-2)^{2}}$

$r=\sqrt{8}$

$r=2\sqrt{2}$

$\theta=tan^{-1}(\frac{-2}{2} )$

$\theta=tan^{-1}(1)$

$\theta=-45$ °

Point B is in IV quadrant, so:

$\theta=360-45$

$\theta=$ 315°

Polar coordinates for point B is ( $2\sqrt{2}$ , 315°)

For point C:

$r=\sqrt{(-6)^{2}+(-3)^{2}}$

$r=\sqrt{45}$

$r=3\sqrt{5}$

$\theta=tan^{-1}(\frac{-3}{-6} )$

$\theta=tan^{-1}(0.5)$

$\theta=$ 26.56°

Polar coordinates for point C is ( $3\sqrt{5}$ , 26.56°)

3 0

3 years ago

Forces that are equal in strength and opposite indirection are

Otrada [13]

Hey there! My name is Christy and I'm gladly to help you out!

The three main forces that stop moving objects are friction, gravity and wind resistance. Equal forces acting inopposite directions are called balanced forces. Balanced forcesacting on an object will not change the object's motion. When you add equal forces in opposite direction, the netforce is zero.

Hope this helped!

4 0

3 years ago

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Which fundamental force has a small range and is always an attractive force?

erica [24]

Answer:

gravitational force is a fundamental force and also , it does have a small range and is always an attractive force.

Explanation:

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3 years ago

An insect meanders across a sidewalk. The insect moves 15cm to the right, 10cm up the sidewalk and 8cm to the left. What is the

IRISSAK [1]

Answer:12.206 cm, $\theta =54.99^{\circ}$

Explanation:

Given

Insect walks 15 cm to the right

so its position vector is $r_1=15i$

Now it moves 10 cm up so its new position vector

$r_2=15i+10j$

Now it moves 8 cm left so its final position vector is

$r_3=15\hat{i}+10\hat{j}-8\hat{i}=7\hat{i}+10\hat{j}$

so its displacement is given by

$|r_3|=\sqrt{7^2+10^2}=\sqrt{149}=12.206 cm$

For direction, let \theta is the angle made by its position vector with x axis

$tan\theta =\frac{10}{7}=1.428$

$\theta =54.99^{\circ}$

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2 years ago

Hardness is indirectly related to: specific gravity surface roughness strength of chemical bonds weight

Zepler [3.9K]

The strength of chemical bonds would be the best answer to this question. Hardness is indirectly related to the strength of chemical bonds. Hardness is the resistance from the destruction of an object itself from a material. Hardness does not need to strength of chemical bonds in order to have this resistance.

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3 years ago

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