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3 years ago

What would happen to a 100N gravitational force if both masses were doubled and the radius were

Physics

2 answers:

grin007 [14]3 years ago

7 0

Answer:

They are the only joints that can do 360 degrees and rotate with their own axis. But, because of its free-moving, it is prone to any dislocation compared to other movable joints.

Explanation:

Alik [6]3 years ago

6 0

The force winds up unchanged, at 100 N.

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NeTakaya

Answer:

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Explanation:

5 0

3 years ago

Line segment Q R , Line segment R S and Line segment S Q are midsegments of Î"WXY. Triangle R Q S is inside triangle X Y W. Poin

Whitepunk [10]

The perimeter of ΔWXY is : ( D ) 14.5 cm

Calculating the perimeter of ΔWXY

QR = WY / 2

RS = XW / 2

QS = XY / 2

Given that : QR = 2.93 cm , RS = 2.04 cm, QS = 2.28 cm

Therefore

Perimeter of ΔWXY = ∑ WY + XW + XY

= 2SR + 2QS + 2QR

= 2(2.04) + 2(2.28) + 2(2.93)

= 14.5 cm

Hence we can conclude that the perimeter of ΔWXY = 14.5 cm

learn more about perimeter calculations : brainly.com/question/24744445

8 0

3 years ago

Read 2 more answers

If you found a line of volcanic peaks a few hundred miles long in which the volcanoes were progressively older toward one end of

Basile [38]

The geological process that could be responsible for this phenomenon would be the plate tectonics. The plate tectonic theories are believed to be the cause of the landforms forming on the earth's surface. What is described in the questions would be an oceanic ridge which is formed from converging oceanic crusts.

5 0

3 years ago

A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a fricti

Ugo [173]

Use Newton's second law to determine the acceleration being applied to the sled. There are three forces at work on the sled (its weight, the force normal to the ground, and friction) but two of them cancel, leaving friction as the only effective force. This vector is pointed in the opposite direction of the sled's movement, so if we take the direction of its movement to be the positive axis, we would find the acceleration due to the friction to be

$\vec F_G+\vec F_N+\vec F_F=m\vec a\iff-12\,\mathrm N=(20\,\mathrm{kg})a\implies a=-0.6\,\dfrac{\rm m}{\mathrm s^2}$

Now we use the formula

${v_f}^2-{v_i}^2=2a(x_f-x_i)$

to find the distance it travels. The sled comes to a rest, so $v_f=0$ , and let's take the starting position $x_i=0$ to be the origin. Then the distance traveled $x_f-x_i=x_f$ is

$-\left(4.5\,\dfrac{\rm m}{\rm s}\right)^2=2\left(-0.6\,\dfrac{\rm m}{\mathrm s^2}\right)x_f\implies x_f\approx17\,\mathrm m$

6 0

3 years ago

Which of the following is an example of the strong nuclear force acting in an atom?

Pavel [41]

I believe the answer is
A. Neutrons attracting to protons

3 0

3 years ago

Read 2 more answers