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4 years ago

9

What is the difference between distance and position????? ANSWER FAST PLEASE IF YOU KNOW

1 answer:

Vesnalui [34]4 years ago

4 0

Distance is the length of the path an object moves

Position is an object’s distance and direction from a reference points

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You place a box weighing 289.9 N on an inclined plane that makes a 35.9 degree angle with the horizontal. Compute the component

Ksivusya [100]

Answer:

Fgparallel = 170N

Explanation:

Acting down the incline would be the paralell force. The Force of gravity on an incline for the parallel portion is mgsinθ.

Fg parallel = mgsinθ

mg is 289.9, as that is the weight. θ is 35.9

Fgpar = 289.9sin35.9

Fgparallel = 170N

5 0

3 years ago

A force of 10 N acts on a mass of 5 kg producing an acceleration of 1.5 m/s2. What is the magnitude of

yulyashka [42]

The answer is

2.5 N

B

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5 0

3 years ago

The force or push that causes electric charges to move is called

Nikolay [14]

Answer:

electrostatic force

Explanation:

Electrostatic force is the force that exists between electric charges on stationary charged bodies

7 0

3 years ago

Read 2 more answers

Venus has an average distance to the sun of 0.723 AU. In two or more complete sentences, explain how to calculate the orbital pe

Mice21 [21]

As per the question the distance of venus from sun is given as 0.723 AU

We have been asked to calculate the time period of the planet venus.

As per kepler's laws of planetary motion the square of time period of planet is directly proportional to the cube of semi major axis. mathematically

$T^{2} \alpha R^{3}$

⇒ $T^{2} = KR^{3}$ where is k is the proportionality constant

We may solve this problem by comparing with the time period of the earth . We know that time period of earth is 365.5 days

Hence $T_{1} =365.5 days$

The distance of sun from earth is taken as 1 AU i.e the mean distance of earth from sun

Hence $R_{1} =1 AU$

The distance of venus from sun is 0.723 AU i.e $R_{2} =0.723$

From keplers law we know that- $\frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{3} }$

⇒ $T_{2} ^{2} =T_{1} ^{2} *\frac{R_{2} ^{3} }{R_{1} ^{3} }$

Putting the values mentioned above we get-

$T_{2} ^{2} =50,350.132851075$

⇒ $T_{2} =\sqrt{50,350.132851075}$

⇒ $T_{2} = 224.388352752710 days.$

Hence the time period of venus is 224.388352752710 days

7 0

4 years ago

Read 2 more answers

A particle moves along the x axis. It is initially at the position 0.180 m, moving with velocity 0.060 m/s and acceleration -0.3

shusha [124]

Answer:

a

$x_2 = -2.3356$

b

$v = -1.384 \ m/s$

Explanation:

From the question we are told that

The initial position of the particle is $x_1 = 0.180 \ m$

The initial velocity of the particle is $u = 0.060 \ m/s$

The acceleration is $a = -0.380 \ m/s^2$

The time duration is $t = 3.80 \ s$

Generally from kinematic equation

$v = u + at$

=> $v = 0.060 + (-0.380 * 3.80)$

=> $v = -1.384 \ m/s$

Generally from kinematic equation

$v^2 = u^2 + 2as$

Here s is the distance covered by the particle, so

$(-1.384)^2 = (0.060)^2 + 2(-0.380)* s$

=> $s = -2.5156 \ m$

Generally the final position of the particle is

$x_2 = x_1 + s$

=> $x_2 = 0.180 + (-2.5156)$

=> $x_2 = -2.3356$

6 0

3 years ago