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2 years ago

What is the term for the overall change in position?

2 answers:

8 0

Hi!

When we're looking at a basic overview of a change in position, this is known as displacement. Displacement is a vector quantity, compared to distance, which is a scalar quantity.

What this means is that displacement is focusing on magnitude and direction, instead of just the magnitude.

Hopefully, this helps! =)

mart [117]2 years ago

6 0

That's "displacement". It only depends on the beginning and ending locations, and doesn't care about the route between them.

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The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

Margaret [11]

Let us consider two bodies having masses m and m' respectively.

Let they are separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - $F = G\frac{mm'}{r^{2} }$ where G is the gravitational force constant.

From the above we see that F ∝ mm' and $F\alpha \frac{1}{r^{2} }$

Let the orbital radius of planet A is $r_{1}$ = r and mass of planet is $m_{1}$ .

Let the mass of central star is m .

Hence the gravitational force for planet A is $f_{1} =G \frac{m_{1}*m }{r^{2} }$

For planet B the orbital radius $r_{2} =2r_{1}$ and mass $m_{2} = 3 m_{1}$

Hence the gravitational force $f_{2} =G\frac{m m_{2} }{r^{2} }$

$f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }$

$= \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }$

Hence the ratio is $\frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2} }{Gmm_{1}/r_{1} ^2 }$

$=\frac{3}{4}$ [ ans]

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2 years ago

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fenix001 [56]

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You are watching a magic show. For one trick the magician rolls a ball down a hill. Suddenly the ball stops moving down the hill

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The forces (what causes the ball to accelerate) are gravity, friction, and the normal force. In this case, gravity is a downward force caused by the gigantic mass of the Earth and the mass of the ball. Keep in mind that a force is acceleration. Acceleration is a change in velocity. The ball speeds up. Than it stops speeding up at a certain point where the frictional force (along with air friction) equals the parallel component of gravity.

Newton's Second Law States- The greater mass of an object, the more force it will take to accelerate the object.

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2 years ago

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Answer:true

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2 years ago

Determine the velocity that a car should have while traveling around a frictionless curve of radius 100m and that is banked 20 d

alex41 [277]

Answer:

$v=18.89\frac{m}{s}$

Explanation:

From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:

$\sum F_x:Nsin(20^\circ)=ma_c(1)\\\sum F_y:Ncos(20^\circ)=mg(2)$

Solving N from (2) and replacing in (1):

$N=\frac{mg}{cos(20^\circ)}\\(\frac{mg}{cos(20^\circ)})sin(20^\circ)=ma_c\\gtan(20^\circ)=a_c$

The centripetal acceleration is given by:

$a_c=\frac{v^2}{r}$

Replacing and solving for v:

$gtan(20^\circ)=\frac{v^2}{r}\\v=\sqrt{grtan(20^\circ)}\\v=\sqrt{9.8\frac{m}{s^2}(100m)tan(20^\circ)}\\v=18.89\frac{m}{s}$

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2 years ago